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MODULE 1

D.C. CIRCUITS

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LECTURE NO.1

PREREQUISITE

OBJECTIVES

• To get familiar with and to understand the basic elements encountered in electric networks.

• To understand the meaning of circuit ground and the voltages referenced to ground.

• To understand the basic principles of voltage dividers and current dividers.

• To learn Ohm’s law and series, parallel combination of resistors.

1.1 Electric charge

Electric charge is the physical property of matter that causes it to experience a force when close to other electrically charged matter. There are two types of electric charges, called positive and negative. Positively charged substances are repelled from other positively charged substances, but attracted to negatively charged substances; negatively charged substances are repelled from negative and attracted to positive.

The charged body has either deficiency or excess of electrons, an object will be negatively charged if it has an excess of electrons, and will otherwise be positively charged or uncharged. The SI unit of electric charge is the coulomb (C), although in electrical engineering it is also common to use the ampere-hour (Ah).

1 Coulomb = charge on 6.28 * 1018 electrons.

Fig 1.1 Electric field of positive and negative charge.

1.2 Electric Potential

Potential energy can be defined as the capacity for doing work which arises from position or configuration. In the electrical case, a charge will exert a force on any other charge and potential energy arises from any collection of charges.

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For example, if a positive charge Q is fixed at some point in space, any other positive charge which is brought close to it will experience a repulsive force and will therefore have potential energy.

The electric potential can be calculated at a point in either a static (time-invariant) electric field or in a dynamic (varying with time) electric field at a specific time, and has the units of joules per coulomb or volts.

Electric potential = , V = , SI: Volts = V = J/C

1.3 Potential difference

Consider the task of moving a positive test charge within a uniform electric field from location A to location B as shown in the diagram at the right. In moving the charge against the electric field from location A to location B, work will have to be done on the charge by an external force. The work done on the charge changes its potential energy to a higher value; and the amount of work that is done is equal to the change in the potential energy. As a result of this change in potential energy, there is also a difference in electric potential between locations A and B.

This difference in electric potential is represented by the symbol V and is formally referred to as the electric potential difference.

By definition, the electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy. In equation form, the electric potential difference is

Fig 1.2 Potential difference between two bodies

The standard metric unit on electric potential difference is the volt, abbreviated V. One Volt is equivalent to one Joule per Coulomb. If the electric potential difference between two locations is 1 volt, then one Coulomb of charge will gain 1 joule of potential energy when moved between those two locations. If the electric potential difference between two locations is 3 volts, then one coulomb of charge will gain 3 joules of potential energy when moved between those two locations.

1.4 Electromotive Force (EMF)

A conductor has abundant of free electrons. The free electrons in conductor are always moving in random direction. A small electric effort externally applied to the conductor makes all such

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free electrons to drift along the conductor in particular direction. This direction depends on how the external electric effort is applied to the conductor. The electric effort required to drift the free electrons in one particular direction in a conductor is called EMF. Such an electric effort may be generated by an electric cell commonly called battery. A metal consists of charged particles. Like charges repel each other, while unlike charge attracts each other. As external electric effort is applied the free electrons being negatively charged get attracted by positive terminal of the cell connected.

Hence electrons get aligned in one particular direction when emf is applied.

The work done on a unit of electric charge, or the energy thereby gained per unit electric charge,

is the electromotive force. Electromotive force is the characteristic of any energy source capable

of driving electric charge around a circuit. It is abbreviated E in the international metric system

but also, popularly, as emf.

Fig 1.3 Electromotive force

1.5 Electric current

An electric current is a flow of electric charge. Electric charge flows when there is voltage present across a conductor.

In electric circuits this charge is often carried by moving electrons in a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in plasma.

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The SI unit for measuring an electric current is the ampere, which is the flow of electric charges through a surface at the rate of one coulomb per second. Electric current can be measured using an ammeter.

Electric currents cause many effects, notably heating, but also induce magnetic fields, which are widely used for motors, inductors and generators.

The actual direction of current is from negative terminal to positive terminal through the circuit. But conventionally direction of current is assumed from positive terminal to negative terminal.

Current is defined as rate of flow of electrons i.e. charge per unit time.

Current, I =

Fig 1.4 Flow of electric current

1.6 Resistance

The opposition offered by a substance to the flow of electric current is called resistance.

Since current is the flow free electrons, resistance is opposition offered by the substance to the flow of free electrons.

This opposition occurs because atoms and molecules of the substance obstruct flow of free electrons. Depending upon the resistance offered by the substance there are two type of substances

Conductor: the substance which offer very little opposition to the flow of current

e.g : metal like copper, aluminum etc

Insulator: offer high opposition to the flow of electric current.

E.g. glass, rubber, mica etc. Resistance is dissipative elements which converts electric energy into heat, when current flows.

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Unit of resistance is Ohm (Ω).

Definition: A wire is said to have a resistance of 1 Ω if a potential difference of 1V across its ends causes 1A current to flow through it.

Resistor is a two terminal element, it conduct current in any direction hence bidirectional element.

Symbol :

Fig 1.5 Symbol of Resistor

Factors influencing resistance:

Resistance depends on following factors:

1. Length: If the length of conductor is increased then the distance that the free electrons must travel

is increased and hence the resistance of conductor is increased. i.e. R α l

2) Cross sectional area:

For a given number of free electrons to flow, if area of cross section of conductor is increased, the resistance offered to its flow is decreased.

It is because free electrons have larger area of path to flow & hence resistance decrease. i.e. R α 1/A

3) Nature of material:

Different material offer different resistance to flow of electric current. E.g. steel wire offer more resistance than copper wire of same length & cross sectional

area. :. Resistance depends upon nature of material

4) Temperature:

With change in temperature, the value of resistance also changes. From first three points, we have

i.e. R α t

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i.e. R = ρ

where, ρ (rho) is a constant and is known as resistivity or specific resistance of the material.

ρ depend upon the nature of material.

1.7 Effect of Temperature on Resistance

The effect of temperature on resistance varies according to the type of material as discussed below:

1. The resistance of pure metals (e.g. Cu, Al) increases with the increase in temperature. The resistance of metals increases with rise in temperature , they have positive temperature coefficient of resistance. The temperature resistance graph is a straight line as shown.

Fig 1.6 Temperature- resistance graph

2. The resistance of electrolytes, insulators and semiconductors decreases with the increase

in temperature. Hence, these materials have negative temperature coefficient of resistance.

3. The resistance of alloys increase with the rise in temperature but this increase is very small and irregular. For some high resistance alloys (e.g. eureka) the change is negligible over a wide range of temperatures.

The range of resistance of a material with rise in temperature can be expressed by means of temperature coefficient of resistance. Consider a conductor having resistance Ro at 0˚C and Rt at t˚C. Then for the normal range of temperature, the increase in resistance i.e. (Rt – Ro),

• Is directly proportional to initial resistance.

• Is directly proportional to the rise in temperature

• Depends on nature of the material. (Rt – Ro) α Rot

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Or, (Rt – Ro) = αo Rot where , Ro :- resistance at 0˚C αo :- temperature coefficient of resistance at 0˚C αo = (Rt – Ro) / Rot

Definition: Temperature coefficient of material may be defined as the increase in resistance per ohm original resistance per 0˚C in temperature. Rt = Ro (1+ αot)

1.8 Ohm’s Law

This law gives relationship between voltage (V) , current (I) and resistance (R) in dc circuits.

This relationship is called Ohms law and it is defined as the ratio of the potential difference (V) between the ends of a conductor to the current (I) flowing between them is constant, provided the physical conditions do not change i.e.

= constant = R

Example: suppose the voltage between two points A and B is V volt and the current flowing

between them is I ampere, then will be constant and equal to R, then resistance between A

and B.

Fig 1.7 Ohm’s law

Unit of resistance is Ohm (Ω).

Note: If the voltage is doubled, the current will also double, so that the ratio remains

constant.

Limitation of Ohm’s law:

1. Ohm’s law is true for dc circuits. In general, it is not valid for ac circuits.

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2. Ohm’s law is true for metal conductors at constant temperature. If temperature changes, Ohm’s law is not applicable.

3. There are many conductors to which ohm’s law is not applicable even if temperature is constant. It is because such material has the property of changing their resistance as the current through them is changed.

Ohm’s law can be expressed in 3 forms:-

I =

V = IR

R =

1.9 Electric Power and Electric Energy

Electric Power: The rate at which work is done in an electric circuit is called electric power .

i.e. electric power =

The total charge that flows in t sec is

Q = I * t

Also, V =

W = VQ

W = V.I.t

Electric power, P =

P =

P = VI Joule/sec or watt

P = VI

P = (IR) I

P = I2R

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P = V2 / R…………………….. Unit of P is watt (W)

Electrical Energy: The total work done in an electric circuit is called electrical energy.

Electrical energy = electrical power * time

Electrical energy = V.I.t

Electrical energy =I2Rt

Electrical energy = t

Note: - Unit of electrical energy will depends upon unit of P. watt- sec, Kw-hr etc.

Power Dissipation in a Resistor

We know that V = I.R

When a current flows through any resistor, power is absorbed by the resistor which is given by

P = V.I

The power dissipated in the resistor is converted to heat which is given by

W =

W =

W = I2R t

1.10 Series Resistor Circuit

Resistances R1 and R2 are said to be connected in series when the same current flows through each of the resistance.

Fig 1.8 resistors connected in series

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Voltage across R1 = V1 = R1 .I

Voltage across R2 = V2 = R2 .I

The total voltage applied should be balanced by the sum of voltage drops around the circuit.

V = V1 + V2

V = R1.I + R2 I

V = (R1+ R2) I

V = Rt. I, where Rt = R1 + R2

When the number of resistances is connected in series the total resistance is the sum of all individual resistance.

Note:

1. Same current flows through each resistance 2. Voltage drops are additive. 3. Resistances are additive. 4. Power is additive. 5. The applied voltage equals the sum of different voltage drops.

Voltage Division in Series circuit:-

The source voltage (V) as shown above divides among the resistors R1 and R2. The voltage drop across the resistance can be obtained as :-

V1 = I.R1

= * R1

Similarly, V2 = * R2

Fig 1.9 Voltage division in series circuit

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1.11 Parallel resistor circuit

Resistances R1 and R2 are said to be connected in parallel when the potential difference across each resistance is same.

Fig 1.10 resistors connected in parallel

I1 = ;

I2 = ;

I = I1 + I2 ;

I = + ;

I = V ( + );

I =

Where Rt = +

Hence, when a number of resistances are connected in parallel, the reciprocal of the total resistance is equal to the sum of reciprocal of individual resistance.

Note:-

1. Same voltage appears across each resistor. 2. Branch currents are additive. 3. Conductance is additive. 4. Power is additive.

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Current Division Rule:-

Fig 1.11 Current division in parallel circuit

We have the total resistance Rt as ,

Rt = R1.R2 / (R1+R2) …….. ( R1 and R2 are parallel to each other.) So , V = I Rt V = I. (R1.R2)/(R1+R2) ……………… (1.1) Current through R1, I1 = V/R1 I1 = ( I. (R1.R2)/(R1+R2) ) / R1 ................. from equation (1.1) I1 = I. R2/(R1+R2)

Similarly current through R2

I2 =

= I. R1/(R1+R2) ……………. from equation (1.1)

1.12 Solved Problems

Q1. Calculate the effective resistance of the circuit shown below and the current through 8Ω resistance, when potential difference of 60 V is applied between points A and B.

Solution : -

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Potential difference of 60 v is applied between points A and B The circuit become

6Ω and 3Ω resistances are in parallel. After combing them, circuit become

Now resistor 2Ω and 18Ω are in series 2 +18 = 20Ω And then 20Ω and 5 Ω are in parallel

20Ω 5Ω = 4Ω

Resistors 8Ω and 4Ω in series 8 + 4 = 12Ω

Effective resistance = 12 Ω

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I = = = 5A …………………..by ohm’s law

Current through 8Ω resistor

I8Ω = 5A (→)

Q2. Calculate battery current and the effective resistance of the network.

Solution:-

Taking the link CE outside, the cross connection can be avoided. Thus, the circuit becomes,

Nodes B and F are same. so by joining them, we get,

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8Ω and 6Ω resistors are in parallel. Also 4Ω and 12Ω resistor are in parallel. So, 6 || 8 = 3.43 Ω 4 || 12 = 3 Ω

Now, resistor 3 Ω and 3.43 Ω are in parallel 3 || 3.43 = 1.6 Ω

Effective resistance = 1.6 Ω Battery current = I = 24/1.6 = 15A

Q3. A resistance of 10 Ω is connected in series with two resistances each of 15 Ω arranged in parallel. What resistance must be shunted across parallel combination so that the total current taken shall be 1.5A with 20V applied ?

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Solution :-

Given : V= 20 V and I = 1.5 A Req = 20/1.5 = 13.33 Ω Simplifying the circuit :- 15 || 15 = 7.5 Ω Req = 10 + (7.5 || R)

1.33 = 10 + ( )

( ) = 3.33

7.5R = 25 + 3.33R 4.17 R = 25 R = 6 Ω

MULTIPLE CHOICE QUESTIONS

1. Which of these parameters remains same in series connected resistances.

a. Current b. Voltage drop c. Power d. All of these.

2. The resistance is inversely proportional to its

a. Length b. Area of cross section c. Resistivity d. both a & c.

3. One coulomb of electrical charge is contributed by how many electrons?

a. 0.625 X 1019. b. 1.6 X 1019. c. 1019. d. 1.6 X 1012.

4. A circuit contains two unequal resistances in parallel

a. Voltage drop across both are same. b. current flowing through both are same .

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c. heat losses in both are same. d. voltage drops are according to their resistive value.

5. Two resistances R1 and R2 give combined resistances 4.5Ω and 1Ω when they are connected in series and parallel respectively. What would be the values of these resistances ?

a. 3Ω and 6Ω b. 1.5Ω and 3Ω c. 3Ω and 9Ω d. 6Ω and 9Ω

6. Specific resistance is measured in

a. mho b. ohm c. ohm – cm d. ohm/cm

7. A wire of resistance R has it length and cross – section both doubled. Its resistance will become

a. 0.5R b. R. c.2R d. 4R

8. Three elements having conductance G1, G2 and G3 are connected in parallel. their combined conductance will be

a. (G1 + G2 + G3) – 1 b. G1+G2+G3

c. 1/G1 + 1/G2 + 1/G3 d. (1/G1 + 1/G2 + 1/G3) – 1

9. The elements which are not capable of delivering energy by its own are known as

a. passive elements b. unilateral elements c. bilateral elements d. active elements

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LECTURE 2

KIRCHHOFF’S LAWS

OBJECTIVES

• To understand the Kirchhoff’s voltage and current laws and their applications to circuits.

• To solve numericals based on Kirchhoff’s laws.

2.1 Kirchoff’s Laws

2.1.1 Kirchoff’s Current Law

Total current flowing towards junction is equal to total current flowing away from junction.

Algebraic sum of all the currents at a junction is zero.

Fig 2.1

At junction A,

I1, I3, I4 are incoming currents.

I2, I5 are outgoing currents.

If we consider incoming currents positive and outgoing currents negative,

By KCL,

I1 + I3 + I4 + (-I2) + (-I5) = 0

I1 + I3 + I4 – I2 – I5 = 0

I1 + I3 + I4 = I2 + I5

2.1.2 Kirchoff’s Voltage Law :

A

I2

I5

I1

I3

I4

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In any network, the algebraic sum of the voltage drops across the circuit elements of any closed

path is equal to the algebraic sum of emfs in a path.

Algebraic sum of the emfs and voltage drops in a closed path of circuit is equal to zero.

Fig 2.2

By KVL,

E1 + E2 = I1R1 + I2R2 + I3R3 + I4R 4

E1 + E2 – I1R1 – I2R2 – I3R3 – I4R4 = 0

2.2 Sign conventions

1) If we go from positive terminal of battery to negative terminal there will be fall in potential so

emf should be assigned negative sign.

Emf = -E2 Emf = +E1

Fig 2.3

Arrow is showing direction of trace.

If we go from negative terminal of battery to positive there is rise in potential so emf is given by

positive sign.

2) When current flows through a resistor there is a voltage drop across it. If we go through

resistance in the same direction as the current there is fall in potential so voltage drop is negative.

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Voltage drop = -I1R1 Voltage drop = I2R2

Fig 2.4

If we go through resistance in opposite direction of current flow, there is rise in potential and

hence voltage drop should be given positive sign.

2.3 Steps to apply Kirchoff’s laws

1) Draw the circuit diagram from given information. Mark the polarities of sources.

2) Mark all branch currents with some assumed directions using KCL at various nodes and

junction points. Keep the unknown currents minimum as far as possible to limit

mathematical calculations.

If sign of calculated current is negative then our assumed direction is wrong. Magnitude of

current is same. Only we should correct the direction.

A particular current leaving a particular source has some magnitude then same magnitude of

current should enter that source after travelling through various branches of network.

3) Mark all polarities of voltage drops and rise as per direction of assumed branch currents

flowing through various branch resistances of the network.

4) Apply KVL to different closed paths in the network and obtain corresponding equations.

Solve simultaneous equations for the unknown currents. From these currents other

parameters can be calculated.

5) If there is current source in the network then complete the current distribution considering

the current source. While applying KVL, loops should not be considered involving current

source. The loop equations must be written to those loops which do not include any current

source.

2.4 Solved Problems

Q1. Find Ix if Iy = 2A and Iz = 0A for following circuit

I2 I1

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Solution:

Applying KCL at node O,

5 + Iy + Iz = Ix + 3

Ix + 3 = 5 + 2 + 0

Ix = 7- 3 = 4ª

Q2. Find currents in all the branches of the network shown below by using Kirchoff’s laws.

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Solution:

Let IAF = x = 41A

Applying KVL to closed path ABCDEFA,

-0.02 - 0.01(x-30) - 0.01(x+40) - 0.03(x-80) - 0.01(x-20) - 0.02(x-80) = 0 x = 41A

IAF = x = 41A

IFE = x – 30 = 41 – 30 = 11A

IED = x + 40 = 41 + 40 = 81A

IDC = x – 80 = 41 +80 = -39A

ICB = x – 20 = 41 – 80 = -21A

IBA = x – 80 = 41 – 80 = -39A

Q3. By using Kirchoff’s laws calculate branch currents in circuit shown below.

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Solution

Applying KVL to ABDA,

35 – 3I1 – 2(I1+I2) = 0

5I1 + 2I2 = 35

Applying KVL to BCDB,

40 - 4I1 –(2I1 + I2) = 0

2I1 + 6I2 = 40

I1 + 3I2 = 20

I1 = 5A I2 = 5A

I3Ω = 5A

I2Ω = 5+5 = 10A

I4Ω = 5A

Q4. Find Vaa’ in following network.

Solution

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KVL to path abcda’

Vaa’ = -3 -2 – (-1) = - 4V

or

KVL to path afda’

Vaa’ = -5 – (-1) = - 4V

or

KVL to path abcea’

Vaa’ = -3 -1 = - 4V


1) According to KCL,

a. Total incoming current at any junction is 0.

b. Total outgoing current at any junction is 0.

c. Total incoming current and outgoing current at any junction is ∞.

d. Total incoming current is equal to total outgoing current at any junction.

2) At any junction A there are 4 currents, current I1=2A and current I2= 4A are incoming,

current I3=1A is outgoing, then current I4 outgoing from junction will be

a. 2A b. 4A

c. 1A d. 5A

3) According to KVL, algebraic sum of the emfs ∑ V in any closed path is

a. 0V b. 10V c. 1V d. ∞ V

4) According to KVL, while tracing a closed path if we go from positive terminal to

negative terminal that voltage must be taken as

a. 0V b. ∞V c. Positive d. Negative

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5) In following circuit Vaa’ is equal to

a. 4V b. 8V

c. -4V d. -8V

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LECTURE 3

MESH ANALYSIS

OBJECTIVES

• To learn the meaning of circuit analysis and distinguish between the terms mesh and loop.

• To provide more general and powerful circuit analysis tool based on Kirchhoff’s voltage law (KVL) only.

3.1 Maxwell’s Current Mesh Method

In this method, the Kirchhoff’s voltage law is applied to each mesh in terms of mesh currents

instead of branch currents. Each mesh is assigned separate mesh current. This mesh current is

assumed to be in clockwise direction around the perimeter of the mesh without splitting at a

junction into branch currents. Kirchhoff’s voltage law is applied to write equation in terms of

unknown mesh currents. Once the mesh currents are known, the branch currents can be easily

determined.

Maxwell’s mesh currents method consists of the following steps:

1. Each mesh is assigned a separate mesh current. For convenience, all mesh currents are

assumed to flow in clockwise direction. For example in fig meshes ABDA and BCDB have been

assigned mesh currents I1 and I2 respectively.

2. if two mesh currents are flowing through a circuit element, the actual current in the circuit

element is the algebraic sum of two. Thus in fig there are two mesh currents I1 and I2 flowing in

R2. If we go from B to D current is I1 - I2 and if we go in the other direction (i.e. from D to B)

currents is I2 – I1.

3. Kirchhoff’s voltage law is applied to write equation for each mesh in terms of unknown mesh

currents.

4. if the value of any mesh current comes out to be negative in the solution, it means that true

direction of that mesh current is anticlockwise i.e., opposite to the assumed clockwise direction.

Consider a circuit as shown in Fig 3.1

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Fig 3.1

Applying Kirchhoff’s voltage law to mesh ABDA,

- I1 R1- (I1-I2)R2+ E1=0

Or

- I1 R1- (I1-I2)R2 = E1 ….. ( 3.1 )

Applying Kirchhoff’s voltage law to mesh BCDB

- I2 R3- E2 -(I2-I1)R2 =0

Or

- I1 R2 +(R2+ R3)I2 = - E2 ….. (3.2 )

Solving equations 1 and 2 simultaneously, mesh currents I1 and I2 can be calculated. Once the

mesh currents are known, the branch current can be readily obtained.

Note: Branch currents are real current’s because they actually flow in the branches and can be

measured. However mesh, mesh currents is concept rather than a reality.

3.2 Solved Problems

Q1. Determine current through 1Ω resistor.

Fig 3.2

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Solution:

Assume current in mesh 1 as I1(clockwise) and current in mesh 2 as I2(clockwise).

Apply KVL to mesh 1

12.5 – 1I1 – 2I1 −3(I1 −I2) = 0

−6I1 + 3I2 = −12.5 …. (3.3)

Apply KVL to mesh 2

−3(I2 – I1) – 2.5I2 – 5I2 = 0

3I1 – 10.5I2 = 0 …. (3.4)

Solving equation 3.3 and 3.4 simultaneously

I1 = 2.43 A

I2 = 0.69 A

I1Ω = 2.43 A.

Q2. Determine all the Mesh currents in the given circuit

Solution:

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Assume current in mesh 1 as I1(clockwise), current in mesh 2 as I2(clockwise), current in mesh 3

as I3(clockwise).

Apply KVL to mesh 1

−2 – 6I1 + 19 – 1(I1 – I2) = 0

−7I1 +I2 = −17 ….. (3.5)

Apply KVL to mesh 2

25 – 2I2 – 1(I2 – I1) – 3(I2 – I3) = 0

I1 – 6I2 + 3I3 = −25 …. (3.6)

Apply KVL to mesh 3

−3(I3 − I2) – 19 – 1I3 = 0

3I2 – 4I3 = 19 …. (3.7)

Solving equation (1) (2) and (3)

I1 = 2.95 A

I2 = 3.65 A

I3 = −2.01 A

Q3. Find the current through 2Ω resistor.

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Solution: Assume current in mesh 1 as I1(clockwise), current in mesh 2 as I2(clockwise),

current in mesh 3 as I3(clockwise).

Mesh 1 contains a current source of 6A. KVL equation is not valid for mesh 1. So directly the

value of I1 is the value of the current source.

Direction of current source and mesh current I1 is the same.

I1 = 6A …. (3.8)

Apply KVL to mesh 2

36 – 12(I2 – I1) – 6(I2 – I3) = 0

12I1 − 18I2 + 6I3 = −36

But I1 = 6A

12(6) – 18I2 + 6I3 = −36

−18I2 + 6I3 = −108 …. (3.9)

Apply KVL to mesh 3

−6(I3 – I2) − 3I3 – 2I3 – 9 = 0

6I2 – 11I3 = 9 …. (3.10)

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Solving equation (2) and (3) we get

I2 = 7A

I3 = 3A

I2Ω = 3A

Q4. Determine current supplied by the battery.

Solution: Assume current in mesh 1 as I1(clockwise), current in mesh 2 as I2(clockwise),

current in mesh 3 as I3(clockwise).

Apply KVL to mesh 1

4 – 3I1 – 1(I1 – I2) – 4(I1 – I3) = 0

−8I1 + I2 + 4I3 = − 4 …. (3.11)

Apply KVL to mesh 2

−1(I2 – I1) – 2I2 – 5(I2 – I3) = 0

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I1 – 8I2 + 5I3 = 0 …. (3.12)

Apply KVL to mesh 3

−4(I3 – I1) – 5(I3 – I2) – 6I3 = 0

4I1 + 5I2 – 15I3 = 0 …. (3.13)

Solving equation (1) (2) and (3) we get

I1 = 0.66 A

I2 = 0.24 A

I3 = 0.25 A

Current supplied by the battery = I1 = 0.66 A

Supermesh Analysis

(i) Meshes that share a current source with other meshes, none of which contains a current

source in the outer loop forms a supermesh.

(ii) The total number of equations required for a supermesh is equal to the number of meshes

contained in the supermesh.

Q5. Find current through 5Ω resistor.

Solution:

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Note in the figure current source and 2Ω resistance are in series. Therefore the resistance

becomes redundant.

Mesh 1 and mesh 3 will form a supermesh as these two meshes share a common current source

of 2A

I1 – I3 = 2 …. (3.14)

Apply KVL to the outer path of the supermesh.

50 – 10(I1 – I2) – 5(I3 – I2) – 1I3 = 0

−10I1 + 15I2 – 6I3 = −50 …. (3.15)

Apply KVL to mesh 2

−10(I2 – I1) – 2I2 – 5(I2 – I3) = 0

10I1 – 17I2 + 5I3 = 0 …. (3.16)

Solving equation (3.14), (3.15) and (3.16) we get

I1 = 19.23A

I2 = 16.38A

I3 = 17.23A

I5Ω = I3 – I2 = 17.23 – 16.38 = 0.85A()

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1. Every mesh is a loop.

a. True b. False

2. In the circuit shown below, calculate current through 5 ohms.

a. 6A b. 8A c. 10A d. 12A

3. In the circuit shown , the current through 28 ohms

a. 1A b. 2A c. 4A d. 8A

4. The voltage across the 1kΩ for the network shown below is

36

a. 2V b. 3V c. 4V d. 8V

5. The value of R is

a. 3.5 Ω b. 2.5 Ω c. 1 Ω d. 4.5 Ω

37

LECTURE 4

NODAL ANALYSIS

OBJECTIVES

• To learn the meaning of circuit analysis and distinguish between the terms node and junctions.

• To provide more general and powerful circuit analysis tool based on Kirchhoff’s current law (KCL) only.

4.1 Nodal Analysis

• Based on Kirchhoff’s current law which states that the algebraic sum of currents meeting

at a point is zero.

• Every junction where two or more branches meet is regarded as a node.

• One of the nodes in the network is taken as reference node or datum node.

• If there are ‘n’ nodes in any network, the number of simultaneous equations to be solved

will be (n-1).

Steps to be followed:

1. Assuming that the network has n nodes, assign a reference node and the reference

directions.

2. Assign a current and a voltage name for each branch and node respectively.

3. Apply KCL at each node except for the reference node and apply Ohm’s law to the

branch currents.

4. Solve simultaneous equations for the unknown node voltages.

5. Using these voltages, find any branch currents required.

4.2 Solved Problems

Q1. Calculate current through 5Ω.

Solution:

38

Marking the different nodes, we obtain-

Fig 4.2

Assign voltages V1, V2, V3 at nodes 1,2 and 3 and reference node as node4(V4=0).

Assume that the currents are moving away from the nodes.

Applying KCL at node 1 we get,

4 +

24+3V1+2V1-2V2 = 0

5V1-2V2= -24 -----------------(4.1)


-10V1+31V2-6V3= - 300 -----------------(4.2)


V1 V3 V2

V4=0V

39

-4V2+9V3=160 ------------------------(4.3)

Solving (4.1), (4.2) & (4.3) we get,

V1= -8.77 V

V2= -9.92V

V3= 13.37V

Current through 5Ω =

= = - 4.658 A () = 4.658 A()

Q2. Find Va and Vb

Solution:


Applying KCL at node a

8Va – 2Vb = 50 ---------------(4.4)

Applying KCL at node b

Ref. node (0V)

40

-3Va + 9Vb =85 ----------------- (4.5)

Solving equations (4.4) and (4.5) we get

Va = 9.39 V

Vb = 12.575V

Q3. Find Va and Vb

Solution:


Applying KCL at node a

4Va – 2Vb = 5 ---------------(4.6)

Applying KCL at node b

-2Va + 3Vb =4 -----------(4.7)

Solving equations (4.6) and (4.7) we get

Va= 2.875 V Vb= 3.25V

Ref. node (0V)

41

Q4. Find voltage across 5Ω.

Solution:

Assign voltages V1, V2, V3 at nodes 1,2 and 3 as shown in the figure above.


Applying KCL at node 1

= -36

4V1 – 2V2-V3 = -24 ----------------(4.8)


-50V1+71V2-20V3=0 ----------(4.9)


V1 V2

V3

Ref. node (0V)

42

-5V1-4V2+10V3=180 ----------------(4.10)

Solving equations (4.8), (4.9) and (4.10) we get

V1= 6.35V

V2= 11.76V

V3= 25.88V

Voltage through 5Ω= V3-V2=25.88 - 11.76 = 14.12V

4.2 Advantages of Nodal Method

1. It can be used to find currents through all the branches or any one branch of given circuit.

2. It can be applied for circuits containing one or more constant current as well as constant

voltage sources.

Supernode Analysis

Nodes that are connected to each other by voltage sources, but not to the reference node by a

path of voltage sources, forms a supernode.

Problem on Supernode

Calculate current through 4Ω using Nodal analysis.

Solution

43

Here nodes 1 & 3 form a supernode.

V1 - V3 = 8 ……………………………………………….(4.11)


Applying KCL to node 2 we get,

(V2 – V1)/ 2 + V2/5 = 8

5 V2 – 5 V1 + 2 V2 = 80

-5 V1 + 7 V2 = 80 …………………………………………(4.12)

Applying KCL to supernode we get,

(V1 – V2)/ 2 + 8 + V3/4 = 5

2 V1 – 2 V2 + V3 = -12 ……………………………………(4.13)

Solving 1, 2 and 3 we get

V1 = 12V

V2= 20V

V3= 4V

Current through 4Ω = V3/4 = 4/4 = 1A


1. Find the node voltage VA.

a. 6.28V b. 6.12V c. 4.28V d. 3V

2. Nodal voltage method of analysis is applicable to network containing: a. only voltage sources b. only current sources

44

c. both voltage and current sources d. none of above

3. What is the current IR2

a. 3.19 A b. 319mA c. 1.73 A d. 173 mA

4. Find the current IR2

a. 0.2 mA b. 2mA c. 2A d. 200mA

5. Nodal Analysis is based on

a. KCL b. KVL c. KCL and KVL both d. None of these.

45

LECTURE 5

SUPERPOSITION THEOREM

OBJECTIVES

• Statement of superposition theorem and its application to a resistive dc network containing more than one source in order to find a current through a branch or to find a voltage across the branch.

• To understand the advantages and disadvantages of Superposition theorem.

5.1 Superposition Theorem

It is stated as, “In a linear network containing more than one active source (i.e. constant e.m.f

and constant current source.), the resultant current in any branch is algebraic sum of currents

that would be produced by each source acting alone, all other sources are being replaced by their

internal resistances.”

The constant e.m.f. sources are replaced by their internal resistance if given or simply with zero

resistance. i.e. short circuit if internal resistance is not given.

The constant current sources are replaced by infinite resistances i.e. open circuit.

A linear network is one whose parameters are constant i.e. they do not change with current and

voltage, in other words, it obeys the ohm’s law i.e. the relation between voltage and current is

linear.

Advantages:

1. The current through particular branch can be found easily.

2. Can be used for circuits with constant voltages as well as constant current sources.

Disadvantages:

1. If currents through all the branches are required then this method is lengthy.

2. The circuit must contain more than one voltages source.

3. It can be applied only to linear circuits. i.e. including R,L and C not applicable for non-

linear i.e. electronic circuits.

4. If the no. of independent sources increases then the method becomes more lengthy.

5.2 Steps to solve problems using Superposition theorem

Consider a network shown in the figure having voltage sources V1 and V2. Let us calculate the

current in branch AB of the network by using superposition theorem.

46

Fig 5.1

Case (1) According to superposition theorem, each source acts independently. Consider source

V1 acting alone. At this time, other sources must be replaced by internal resistances. But as

internal resistance of V2 is not given i.e. it is assumed to be zero, V2 must be assumed as short

circuit. Thus, the circuit becomes as shown in figure below. Using any of the techniques, obtain

the current through branch AB i.e. IAB due to source V1 alone.

Fig 5.2

Case (2) Now, consider source V2 alone, with V1 replaced by a short circuit, to obtain the

current through branch AB. The corresponding circuit is shown in figure. Obtain IAB due to V2

alone by using any of the techniques.

Fig 5.3

Therefore by Superposition Theorem, the resultant current through branch AB is the sum of the

current through branch AB produced by each source acting alone.

IAB = IAB due to V1 + IAB due to V2

47

5.3 Solved Problems

(Note: DO NOT USE SOURCE TRANSFORMATION IN ANY SUMS OF

SUPERPOSITION THEOREM)


Solution:

Case(1) : Consider 10 V source acting alone and replacing 4A source by its internal

resistance(O.C)

Apply KVL to mesh 1

10 – 1I1 – 10(I1 – I2) = 0

−11I1 + 10I2 = −10 …. (5.1)

Apply KVL to mesh 2

−10(I2 – I1) – 2I2 – 5I2 = 0

10I1 − 17I2 = 0 …. (5.2)

Solving equation (5.1) and (5.2) we get

I1 = 1.95A

I2 = 1.149A

48

I’10Ω = I1 – I2 = 1.95 – 1.149 = 0.8A( )

Case(2) : Consider 4A source acting alone and replace 10V source with its internal

resistance(S.C)

Applying KCL at node 1(Assuming all currents are moving away)

+ + = 0

16V1 – 5V2 = 0 …. (5.3)


− 4 + = 0

−5V1 + 7V2 = 40 …. (5.4)

Solving equation (5.3) and (5.4) we get

V1 = 2.298V

V2 = 7.356V

I’’10Ω = = 0.2298A( )

Direction of current is downward because for KCL we have assumed that all currents are moving

away from the nodes.

By Superposition Theorem

I10Ω = I’10Ω + I’’10Ω

= 0.8A( ) + 0.2298A( )

= 1.0298A( )

49


Solution:

Case(1) Consider 4V source acting alone. Replacing 1A and 3A current sources by O.C we get

Apply KVL to mesh

4 – 2I1 – 1I1 = 0

I1 = 1.33A( )

I’1Ω = 1.33A( )

Case(2) Consider 1A source acting alone. Replacing 4V source by S.C and 3A source by O.C

50

I1 = 1A (as mesh 1 contains a current source which is not shared by other meshes)

Apply KVL to mesh 2

−2(I2 – I1) – 1I1 = 0

2I1 – 3I2 = 0

2(1) – 3I2 = 0

I2 = 0.667A( )

I’’1Ω = 0.667A ( )

Case(3) Consider 3A source acting alone. Replacing 4V source by S.C and 1A source by

O.C

Applying KVL to mesh 1

−2I1 – 1(I1 – I2) = 0

−3I1 + I2 = 0 ….. (5.5)

I2 = −3A …. (5.6)

Substituing value of I2 in equation (5.5)

51

I1 = −1A …. (5.7)

I’’’1Ω = I1 – I2

= −1 – (−3)

= 2A( )

By superposition theorem

I1Ω = I’1Ω +I’’1Ω +I’’’1Ω

= 1.33 + 0.667 +2

= 3.997A( )

Q3. Find current in 3Ω using superposition Theorem.

Solution:

Case(1) Consider 12V source acting alone. Replacing 15A and 5A by O.C


– 5(I1 – I2) –10I1– 10(I1 – I2) = 0

–25I1 + 15I2 = 0 …. (5.8)

52


12 – 2I2 – 5(I2 – I1) – 10(I2 – I1) – 3I2 = 0

15I1 – 20I2 = –12 …. (5.9)

I1 = 0.6545A

I2 = 1.091A

I3 = 1.091A

I’3Ω = I3 = 1.091A( )

Case(2) Consider 15A source acting above. Replacing 5A and 12V source with O.C and S.C.


–2I3 – 5(I3 – I1) – 10(I3 – I2) – 3I3 = 0

5I1 + 10I2 – 20I3 = 0 …. (5.10)

For Supermesh 1 and 2

I2 – I1 = 15 …. (5.11)

–10I1 – 10(I2 – I3) – 5(I1 – I3) = 0

–15I1 – 10I2 + 15I3 = 0 …. (5.12)

Solving equation (5.10), (5.11) and (5.12) we get

I1 = –2.7272A

I2 = 12.2727A

I3 = 5.4545A

I’’3Ω = I3 = 5.4545A( )

53

Case(3) Consider 5A source acting above. Replacing 15A and 12V source with O.C and S.C.

Apply KVL to mesh 1

–10I1 – 10(I1– I3) – 5(I1 – I2) = 0

–25I1 + 5I2 + 10I3 = 0 …. (5.13)

For Supermesh 2 and 3

I2 – I3 = 5 …. (5.14)

Apply KVL to supermesh 2 and 3

–2I2 – 5(I2 – I1) – 10(I3 – I1) = 0

15I1 – 7I2 – 13I3 = 0 …. (5.15)

Solving equation (5.13), (5.14) and (5.15)

I1 = –0.0909A

I2 = 3.1818A

I3 = –1.8182A

I’’’3Ω = I3 = –1.8182( )

By Superposition theorem

I3Ω = I’3Ω + I’’3Ω + I’’3Ω

= 1.091( ) + 5.4545( ) –1.8182( )

= 4.7275A( )

54


1. Superposition theorem is not applicable to network containing

a. Nonlinear elements b. dependant voltages

c. Dependant current sources d. transformers

2. Superposition theorem is applicable to a linear network in determining

a. Current responses b. Voltage responses

c. Both a and b d. None of these

3. Choose the write option

a. Superposition is valid when there is atleast one source.

b. Superposition is valid for linear devices.

c. Superposition is valid for linear and bilateral devices.

d. None of these.

4. Ideal internal resistance for a voltage and current source is

a. 0, b. c. d. 0,0

5. A network is said to be a linear if and only if

a. Response is proportional to the excitation function.

b. Principle of Superposition applies.

c. Both a and b.

d. None of these.

55

LECTURE 6

SOURCE TRANSFORMATION

OBJECTIVES:

• To understand the fundamental differences between ideal and practical voltage and current sources and their mathematical models to represent these source models in electric circuits.

• To understand how to convert voltage source into current source and vice-versa and use this method to simplify the circuit.

6.1 Source Transformation

CASE I

A voltage source having resistance in series can be converted into current source having

resistance in parallel. Current supplied by that current is I = V/R and direction depends on

polarity of voltage source.

Fig 6.1

CASE II

Current source with resistance in parallel can be converted into voltage source V= IR with

resistance R in series. And polarity depends on direction of current as shown in figure.

I = V/R

I = V/R

56

Fig 6.2

Ideal voltage source :

Fig 6.3

Ideal voltage source is the one whose voltage remains constant whatever the change in load

current. It has zero internal resistance i.e Rint=0.

Ideal current source :

Fig 6.4

Ideal current source is the one which gives constant output current irrespective of changes in

voltage across its terminals. It has ∞ internal resistance.

I

I

V = IR

V = IR

Rint = 0

R int= ∞

57

6.2 Solved Problems

Q1. By source transformation find current in 4Ω .

Solution

Converting 5A current source with parallel resistance of 2Ω into equivalent voltage

source with series resistance, we get

V= I X R

= 5 X 2 = 10V.

Converting 4V voltage source with series resistance of 2Ω into equivalent current source with

parallel resistance, we get

I = V/R

= 4/2 = 2A.

58

I4Ω = 4 × 2 / (2+4);

I4Ω = 1.33A

Q2. Replace the circuit between A and B with voltage source in series with single resistor.

Solution

Converting 20V voltage source with series resistance of 5Ω into equivalent current source with

parallel resistance, we get

I = V/R

= 20/5 = 4 A.

59

Converting 7A current source with parallel resistance of 2.38Ω into equivalent voltage source

with series resistance, we get

V = I X R

= 7 X 2.38 = 16.67 V

60


1. If voltage source having voltage V volt is connected in series with resistance R then by

source transformation it can be converted into equivalent current source having current

a. I = VR c. I = V/R

b. I= V2 R d. I = V

2. If we convert following current source into equivalent voltage source then voltage of

voltage source will be

a. 20V c. 50V

b. 2V d. 10V

3. For the above circuit what will be resistance of equivalent voltage source

a. 10Ω c. 20Ω

b. 5Ω d. 2Ω

4. What will current of equivalent current source of following voltage source

a. 12A c. 10A b. 1A d. 144A

5. For the same problem no. 4 what will be resistance of equivalent current source

a. 12Ω c. 2Ω

b. 10Ω d. 24Ω

61

LECTURE 7

STAR AND DELTA TRANSFORMATION

OBJECTIVES

• To simplify a three terminal network which may be a star ( ) into equivalent delta (∆)

network and vice-versa.

• To study the proof of star to delta and delta to star conversion.

• Application of these transformations will be studied by solving resistive circuits.

By using series/parallel circuit rules, we can reduce or simplify the circuit. But there are some

networks in which the resistances are neither in series nor in parallel and are connected in star or

delta connection. In such situation, it is not possible to simplify the network by series/parallel

circuit rules. However, converting delta connection in to equivalent star connection and vice

versa, a network can be simplified and application of series/parallel circuit rules is made

possible.

7.1 DELTA TO STAR CONVERSION (∆ )

Consider the three resistors R12, R23 and R31 connected in Delta as shown below.

It is required to replace them by three resistors R1, R2 and R3 connected in star as shown below-

Fig. 7.1

I. From fig (a), resistance between terminals 1 and 2 = R12 ll (R23+R31)

= (R12 (R23+R31))/( R12+ R23+R31) -----------(7.1)

62

From fig (b), resistance between terminals 1 and 2 = R1 + R2 -------------- (7.2) II. Two networks are electrically equal. Hence, (1)=(2)

R1 + R2 = (R12 (R23+R31))/( R12+ R23+R31) -----------(7.3)

Similarly, R2 + R3 = = (R23 (R12+R31))/( R12+ R23+R31) -----------(7.4) and

R3 + R1 = -------------(7.5)

III. Performing (3)-(4)+(5), we get

R1 = (R12R31)/( R12+ R23+R31) R2 = (R12R23)/( R12+ R23+R31) R3 = (R31R23)/( R12+ R23+R31) These are the expressions for delta to star conversions.

7.2 STAR TO DELTA CONVERSION ( ∆)

IV. Performing (6) – (7), we get

R1/R2 = R31/R23

Therefore,

V. Performing (6) – (8)

R1/R3 = R12/R23 Therefore, R12=(R1.R23)/R3 VI. Substituting values of and

R1=

Therefore, after re-arranging and simplifying we get,

R23= R2 + R3 + (R2.R3 /R1) Similarly,

R31= R3 + R1 + (R3.R1 /R2) And,

R12= R1 + R2 + (R1.R2 /R3) These are the expressions for star to delta conversions.

63

6.3 Solved Problems

Q1. Find RXY

Solution:

Converting the delta connection formed by resistor 4Ω, 6Ω and 2Ω into equivalent star network.

∆ABC ΥABC

R1 = = 0.67Ω

R2 = = 2Ω

R3 = = 1Ω

64

Resistor 6Ω and 2Ω are in series. Also resistors 1Ω and 7Ω are in series.

Resistor 8Ω and 8Ω are in parallel

.RXY = 8.67Ω.

65

Q2. Find RAB

Solution:

Converting delta connection formed by three resistor of 2RΩ each into equivalent star network.

i.e. ∆CBD ΥCBD

R1 = = Ω

Similarly R2 = R3 = Ω

In branch ACE, RΩ and Ω are in series. Also branch ADE, RΩ and Ω resistor are in

series.

66

The resistor Ω and Ω are in parallel.

The resistor 2RΩ and Ω are in parallel.

67

RAB = Ω

Q3. Find the current I in the network.

Solution

The resistor 5Ω, 10Ω and 10Ω form a star network.. Converting into equivalent delta network we

get.

RA = = 20Ω

68

RB = = 20Ω

RC = = 40Ω

The circuit becomes

The resistors 20Ω and 5Ω are in parallel. Also 40Ω and 15Ω are in parallel.

Now converting the delta network formed by the resistor 6Ω, 8Ω and 20Ω into equivalent star

network.

R1 = = 1.411Ω

69

R2 = = 3.53Ω

R3 = = 4.7Ω

The circuit becomes

Simplifying the circuit further we get

= 15.408A

70

Q4. Find RAB

Solution

The resistor 3Ω, 5Ω and 2Ω form a star network. Converting into equivalent delta network we

get.

RA = = 15.5Ω

RB = = 6.2Ω

RC = = 10.33Ω

Resistor 5Ω and 6.2Ω are in parallel. Also 4Ω and 10.33Ω are in parallel.

Redrawing the same circuit ,

71

Now converting the delta network formed by the resistor 15.5Ω, 2.767Ω and 2.88Ω into

equivalent star network.

R1 = = 0.37Ω

R2 = = 2.02Ω

R3 = = 2.11Ω

Resistor 8.02Ω and 6.11Ω are in parallel.

72

RAB = 3.83Ω


1. If all the resistance in a delta connected network are same say R, then the equivalent

resistance of star connected network is-

a. 3R b. R/3 c. 3/R d. R2

2. Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k, k> 0, the elements of the corresponding star equivalent will be scaled by a factor of

a. 2 k b. k c. 1/k d. k

3. In above figure, when Ra = Rb = Rc = 3Ω then equivalent star will have R1 = R2 = R3

equal to

a. 1Ω b. 3Ω c. 9Ω d. 12Ω

73

4. In above figure ,when R1=R2=R3=3Ω then equivalent delta will have Ra = Rb = Rc equal to

a. 1Ω b. 3Ω c. 9Ω d. 12Ω

5. Consider the following circuit. What is the value of current I in the circuit shown ?

a. 0.5 A b. 1 A c. 2 A d. 3 A

74

LECTURE 8

THEVENIN’S THEOREM

OBJECTIVES:

• To understand the basic concept of Thevenin’s theorem and its application to solve dc circuits.

• Explain the advantage of Thevenin’s theorem over conventional circuit reduction techniques in situations where load changes.

8.1 Thevenin’s Theorem

The Thevenin’s theorem applied to D.C. circuits can be stated as follows:

“Any linear network having terminals A and B can be replaced by single source of emf

VTH(called the Thevenin’s voltage ) in series with single resistance RTH (called the Thevenin’s

resistance ).

(i) The emf VTH is the voltages obtained across terminals A and B with load, if any, removed

i.e. it is open circuited voltage between A and B.

(ii) The resistance RTH is the resistance of network measured between A and B with load, if

any, removed and constant voltage sources being replaced by their internal resistance ( or

simply by zero resistance i.e. short circuit if internal resistance is not given) and constant

current sources are replaced by infinity resistance. i.e. open circuit.

Thus according to Thevenin’s theorem, any two terminal network, however complex, can be

replaced by single source of e.m.f. VTH called Thevenin’s voltage source is in series with single

resistance RTH called Thevenin’s resistance.

Fig 8.1

8.2 Advantages

1. It reduces a complex circuit into a simple circuit of single voltage source in series with

single resistance.

75

2. Current through a particular branch can be found easily.

3. Can be used for the circuits having one or more constant voltages as well as constant

current sources.

4. This theorem is best suited for finding current through load resistance which taken up

several finite values viz RL1 , RL2

8.3 Disadvantages

If currents through all the branches are required then this method is lengthy.

8.4 Steps to apply Thevenin’s Theorem

Step 1:

Remove the branch resistance through which current is to be calculated & mark the open circuit

terminals as A & B.

Step 2:

Calculate RTH as viewed through the open circuited two terminals A & B, by replacing all the

sources by their internal resistances (i.e. voltage source by internal resistance if mentioned in the

problem or else by zero resistance i.e. short circuit & current source by infinite resistance i.e.

open circuit.)

Step 3:

Calculate the voltage across the open circuit terminals A & B by one of the network

simplification technique. This is VTH.

Step 4:

Draw the Thevenin’s equivalent circuit showing source VTH with a resistance RTH in series with

it.

Step 5:

Reconnect the branch resistance. Let it be RL . The required current through the branch is given

by

8.5 Solved Problems

Q1. Determine current through 1Ω resistance.

76

Solution:

Step 1:Remove the branch resistance RL through which current is to be calculated & mark the

open circuit terminals as A & B.

Step 2: Calculation of RTh

Replace Voltage Source 12.5V by its internal resistance i.e. short circuit

In the above figure, we observe that resistances 2.5Ω & 5Ω are in series.

RS= 2.5+5= 7.5Ω

Resistances 7.5Ω & 3Ω are in parallel

RP= ((7.5Ω)-1+(3Ω)-1) -1=2.143 Ω

In the figure 1.8.6, we observe that resistances 2.143Ω & 1Ω are in series.

RAB=RTH= 2.143+1= 3.143Ω

RTH= 3.143Ω

Step 3: Calculation of VTh

77

VTh= 3 I1 + 12.5

For mesh1 , there is no source of EMF present therefore I1 =0 A ( Applying KVL in Mesh1, 3

I1+2.5 I1+5 I1=0

10.5 I1=0

I1= 0 A

VTh= 3 (0) + 12.5=12.5V

VTH = 12.5V

Step 3: Calculation of Load current:

Thevenin’s equivalent model can be drawn as shown below

By Ohm’s Law,

Hence, IL=I20Ω= ( )

Q2. Find Current through Galvanometer

78

Solution:

Step 1: Remove the galvanometer with RL=5Ω through which current is to be calculated

& mark the open circuit terminals as A & B.


Replace Voltage Source 10V by its internal resistance i.e. short circuit

In the above figure, we observe that resistances 10Ω & 15Ω are in parallel as the connected

between the same terminals A& C, similarly Resistances 12Ω & 16Ω are in parallel as the

connected between the same terminals B & C

RP1= ((15Ω)-1+(10Ω)-1) -1=6 Ω

79

RP2= ((12Ω)-1+(16Ω)-1) -1=6.857 Ω

Resistances 6Ω & 6.857Ω are in series

RS= 6Ω + 6.857Ω = 12.867Ω

RTH= 12.867Ω


VTh= 16I1 -16I2+15I1=31I1 -16I2

Applying KVL in Mesh1,

53 I1-28I2 =0,


-28I1+28I2 =10,

I1 =0.4 A

I2=0.757A

VTh= 31I1 -16I2=31×0.4 -16×0.757=0.288V

VTh=0.288Volts

Step 4: Calculation of Load current


80

By Ohm’s Law,

Hence, IL=I5Ω= ( )

I5Ω =0.01613 A=16.13mA ( )

The current through the galvanometer is 16.13mA ( )

Q3. Find Thevenin’s equivalent circuit across AB [Vth = 32V and Rth = 4Ω]

Solution:


Replace Voltage Source 48V by its internal resistance i.e. short circuit & mark all the essential

nodes

81

In the figure, we observe terminals A & C are the same.

Resistances 6Ω & 12Ω are in parallel as the connected between the same terminals A & D

RP1= ((6Ω)-1+ (12Ω)-1) -1= 4 Ω

RAD & RBD are in series RS= 4+4= 8 Ω

Resistances 8Ω & 8Ω are in parallel

RAB=RP= ((8Ω)-1+ (8Ω)-1) -1= 4 Ω

RTH= 4Ω


VTh= 6I1 +4I2 ………………..(8.1)


-6I1-12(I1-I2) +48 =0,

18I1-12I2 =48 ………………..(8.2)


-12(I2-I1) -4I1-8I1=0,

-12I1+24I2 =0 ………………...(8.3)

Solving equations (8.2) & (8.3) simultaneously

I1=4A ,

I2=2A

VTh= 6I1 +4I2=6×4+4×2=24+8=32V

VTH= 32 Volts

Step 3: Draw Thevenin’s Equivalent model

82

Q4. Find power drawn by 20Ω resistor

Solution:

Step 1:Remove the branch resistance RL through which current is to be calculated & mark

the open circuit terminals as A & B.


Replace Voltage Source 12V,8V by its internal resistance i.e. short circuit & 4A by its internal

resistance i.e. open circuit

Applying series parallel reduction techniques, equivalent resistance across the load resistance A

& B is called the Thevenin’s resistance

In the above figure, we observe

Resistances 10Ω & 2Ω are in series. Also 25Ω & 5Ω resistors are in series

83

Resistances 12Ω & 15Ω are in parallel

RAB=RP= ((12Ω)-1+(15Ω)-1) -1= 6.67 Ω

RTH= 6.67Ω


The voltage which appears across the open circuit terminals A & b is called as Thevenin’s

Voltage(VTh). For Calculation of VTh i.e. VAB, the selected path from A to B is marked by dotted

line in the figure.

VTh= 12-2I1 -10I1=12-12I1

As this path contains the resistances 2Ω & 10Ω, current through these resistances are required.

By using Mesh analysis, these currents can be calculated.

Mesh 1 & mesh 2 forms a supermesh

By expressing the current in the common branch, we get the current equation as

I1-I2=4 --------------(8.4)

Applying KVL to the supermesh, we get the voltage equation as

-10I1 -5I2 -15(I2-I3) +12 -2I1 =0,

-12I1 -20I2 -15I3= -12 --------------(8.5)


-25I3 -8 -5I3-15(I3-I2) =0,

15I2 -45I3 = 8 --------------(8.6)

84

Solving equations (8.4), (8.5) & (8.6) simultaneously,

I1 =2.57A

I2 = -1.432 A

I2=-0.655A

VTh=12-12I1=12-12×2.57= -18.84V

VTH = -18.84 Volts

Step 4: Calculation of Load current


By Ohm’s Law,

Hence, IL=I5Ω= ( )

I20Ω = -0.706A ( ) = 0.706A( )

Power drawn by 20Ω resistor is

P20Ω= =(0.706)2 20= 9.97W


1. Thevenins theorem replaces a complicated circuit facing a load by an

a. Ideal voltage source & parallel resistor

b. ideal current source & parallel resistor

c. Ideal current source & series resistor

d. Ideal voltage source & series resistor

2. To get Thevenin’s voltage , you have to

a. short the load resistance c. open the load resistance

b. short the voltage source d. open the voltage source

85

3. For the circuit given in the figure, the Thevenin voltage and resistance as seen at AB are

represented by

a. 5 V ,10 Ω b. 10 V ,10 Ω c. 5 V ,5 Ω d. 54 V ,15 Ω

4. To get Thevenin’s equivalent resistance we have to a. Short circuit the current source. b. Open circuit the current source. c. Short circuit the voltage source. d. both (b) & (c)

5. What is the Thevenin’s Voltage across load terminals A & B?

a. 68 V b. -68V c. 8V d. -8V

86

LECTURE 9

NORTON’S THEOREM

OBJECTIVES

• To understand the basic concept of Norton’s theorem and its application to solve dc circuits.

• Explain the advantage of Norton’s theorem over conventional circuit reduction techniques in situations where load changes.

9.1 Norton’s Theorem

The Norton’s Theorem applied to D.C. circuits can be stated as follows:

Any network containing terminal A and B can be replaced by single source of current IN in

parallel with single resistance RN.

(i) The current IN is the current that would flow through AB when A and B are short circuited.

(ii) The resistance RN is the resistance of a network measured between A and B with load, if

any, removed and constant voltage sources being replaced by their internal resistance ( or

simply by zero resistance i.e. short circuit if internal resistance is not given) and constant

current sources are replaced by infinity resistance. i.e. open circuit.

(iii) Thus according to Norton’s theorem, any two terminal network, however complex, can be

replaced by single current source IN called Norton’s current inn parallel with single

resistance RN called Norton’s resistance.

Fig 9.1

9.2 Advantages

1. It reduces a complex circuit into a simple circuit of single current source in parallel with

single resistance. 2. Current through a particular branch can be found easily.

IN

RN

87

3. Can be used for the circuits having one or more constant voltages as well as constant current sources.

4. This theorem is best suited for finding current through load resistance which taken up several finite values viz RL1 , RL2

9.3 Disadvantages

1. If currents through all the branches are required then this method is lengthy.

9.4 Steps to be followed for solving a problem using Norton’s Theorem

1) Short the branch resistance through which current is to be calculated.

2) Obtain the current through this short circuited branch using any one of the network

simplification techniques. This current is Norton’s current IN

3) Calculate RN as viewed through the two terminals of the branch from which current is

to be calculated by removing that branch resistance and replacing all sources by their

internal resistances.(for a given circuit ,RN=RTH)

4) Draw the Norton’s equivalent circuit showing current source IN, with resistance RN in

parallel with it.

5) Reconnect the branch resistance. Let it be RL. The required current through the branch

is given as

IL=IN

9.5 Solved Problems

Q1. Find current through 10Ω resistor using Norton’s theorem.

Solution:

12 V

88

Current through 10Ω resistor is required. The resistance can be called load resistance RL.

Its terminals are called load terminals.

1) Calculation of IN

Removing load resistance from the network and short circuiting the load terminals, we get

Applying KVL to mesh1

-5 I1+ 20 – 2( I1-I2 )= 0 -7 I1 +2 I2 = -20 ----------------------------------------(9.1)


-2 ( I2 - I1 ) – 8 I2 – 12 = 0 2 I1 – 10 I2 = 12 ----------------------------------------(9.2)

Solving equation (9.1) and (9.2) simultaneously

I1= 2.67 A I2 = IN = -0.67 A

2) Calculation of RN

Removing the load resistance from the network and replacing the voltage source by short circuit we get-

RN= RAB = (5||2)+8

89

= 1.428+8 = 9.428Ω

3) Calculation of Load Current

Norton’s equivalent can be drawn as-

IL= I10Ω= IN

= -0.67 x

= -0.325 A ( )


Solution:

Current through 4Ω resistor is required. The resistance can be called load resistance RL. Its

terminals are called load terminals.



90

I1= 6 A -------------------------(9.3) Applying KVL to mesh 2

-2 (I2-I1)- 10(I2-I3) – 1( I2-I4 )= 0 -13 I2 +10 I3 + I4 = -12 -------------------------(9.4)


-10 ( I3 – I2 ) - 10 - 2 I3 – 3(I3-I4) = 0 10 I2 – 15 I3 + 3 I4 = 10 ------------------------(9.5)


-1 (I4-I2)- 3(I4-I3) +24= 0 I2 +3 I3 - 4 I4 = -24 ----------------------------(9.6)

Solving equation (9.4), (9.5) and (9.6) simultaneously

I2 = 6.71 A I3 = 6.29 A I4 = IN = 12.39 A


Removing the load resistance from the network and replacing the voltage sources and current sources by their respective internal resistances we get-

91

Replacing the star network formed by 10Ω, 1Ω and 3Ω into delta network, we get-

RA= (3 + 30 + 10 )/3=14.33 Ω

RB= (3 + 30 +10)/1 = 43 Ω

RC= (30 + 10 +1)/10=4.3 Ω

Simplifying the circuit

92

RN = 4.3 ll (1.75+1.91) = 1.977 Ω



IL= IN

= 12.39 x

= 4.098 A


Solution:

Current through 5Ω resistor is required. The resistance can be called load resistance RL.

Its terminals are called load terminals.



93



---------------------------(9.7)


-------------------------(9.8)


-------------------------(9.9)

Solving equations (9.7), (9.8) and (9.9) we get

V1= 19.54V

V2= 12.16V

V3= 0.089V

IN= = 4.77 A( )

94


Removing the load resistance from the network and replacing the voltage sources and current sources by their respective internal resistances we get-

RN = 6 + (10 || (6+3+4)) + 2 = (6 + 5.65 + 2) Ω = 13.65 Ω



IL = IN

= 4.77 x

= 3.49 A

95


1. Applications of Norton’s theorem in circuit yields a. equivalent current source and impedance in series b. equivalent current source and impedance in parallel c. equivalent impedance d. equivalent current source 2. To convert a Norton equivalent circuit to a Thevenin’s equivalent circuit,

a. Let the Norton resistance equal the Thevenin resistance b. The product of Norton current and Norton resistance equals the Thevenin voltage c. Both a and b d. None of the above

3. A circuit has a voltage source of 15 volts and three 15 Ω resistors connected in parallel across

the source. What Norton resistance (RN) would a load "see" when connected to this circuit. a. 0 b. 15 c. 45 d. 5 4. A voltage source of 100 volts is connected in series with three resistors;

R1 = 10 Ω, R2 = 10 Ω, R3 = 20 Ω. A load is placed in parallel with R3. Determine the Norton current (IN).

a. 5 amp b. 50 amp c. 10 amp d. 2.5 amp 5. The Norton resistance is placed in series with the Norton current source a. True b. False

96

LECTURE 10

MAXIMUM POWER TRANSFER THEOREM

OBJECTIVES

• To derive the condition for maximum power transfer in dc circuit. • To calculate maximum power transferred through the load resistor using the above

condition. • To know the efficiency at the condition of maximum power transfer.

10.1 Maximum Power Transfer Theorem

The Maximum Power Transfer Theorem applied to D.C. circuits can be stated as follows:

“The maximum transfer of power to a load resistor occurs when the load resistor has a value

equal to the series equivalent or internal resistance of the source.”

Fig 10.1

OR

“A resistive load would abstract maximum power from a network when load resistance is equal

to resistance of network as viewed from output terminals, with all energy sources removed

leaving behind their internal resistances.”

In Fig 10.2(a), a load resistance RL is connected across the terminal A and B of a network which

consists of a generator of emf E and the internal resistance Rs and a series resistance R, which

represents the resistance of the connecting wires.

97

Fig 10.2

Let Ri=Rs+R = internal resistance of the network viewed from A and B, as shown in the

Figure 1.10.2(b).

According to this theorem RL will abstract maximum power from the network when RL = Ri.

Proof:

(i) Load current IL= E/(RL+Ri)

(ii) Power drawn by the load PL = IL2. RL =( E/(RL+Ri))

2 . RL

(iii) For a given source or circuit, E and Ri are constant, the power drawn by the

load PL depends upon RL only.

Thus for PL to be maximum, = 0

Differentiating equation (i) above and equating to zero, we get

= 0

E2 ≠ 0, also the denominator (

The numerator ,

Or

or

and hence the proof.

(iv) The maximum power transferred to the load will be given by substituting

(RL=Ri )in the equation (i) above

Thus,

98

Note: For solving the problems on maximum power transfer , we replace a two terminal network

by Thevenin’s equivalent circuit with a single voltage source (VTH) and a resistance in series

(RTH) .

The maximum power transferred to the load is given by

(v) Under the condition of maximum power transfer,

(vi) Under the condition of maximum power transfer,

Load voltage

Due to low efficiency and greater voltage drop under maximum power transfer , electric power systems never apply maximum power transfer theorem.

It is because in electronic power system the goal is higher efficiency rather than maximum power transfer.

However, in the case of electronic & communication networks, the goal is either to receive or transmit maximum power though at reduced efficiency as the power involved is only in mW or µW & therefore maximum power transfer is used in the case of electronic & telecommunication fields.


Q1. Find value of RL for maximum power and find Pmax

Solution:

99

According to maximum power transfer theorem, maximum power will be transferred from the

circuit to the load when RL is made equal to RTH, the Thevenin’s resistance at terminals A & B.

Load terminals are marked as A & B.

Step 1:Remove the branch resistance RL & mark the open circuit terminals as A & B.


Replace Voltage Source 20V by its internal resistance i.e. short circuit & 2A Current source by

its internal resistance i.e. open circuit.

In the above figure, we observe that resistances 6Ω & 5Ω are in series.

RS= 6+5= 11Ω

Also resistances 12Ω & 8Ω are in parallel as they are connected between the same terminals A

& C.

RP = ((12Ω)-1+(8Ω)-1) -1 = 4.8 Ω

RAB = RTH = 11+4.8= 15.8Ω

RTH = 15.8 Ω

Step 3: Calculation of VTH


Voltage (VTh). For Calculation of VTh i.e. VAB, the selected path from A to B is marked by dotted

line in the figure

100

VTh = 5I +6I2 +8I1 (but branch AB is open it is no more a mesh I=0 , the voltage across 5Ω

is zero)

VTH = 8I1 +6I2

As this path contains the resistances 8Ω & 6Ω, current through these resistances are required. By

using Mesh analysis, these currents can be calculated.

Applying KVL to the mesh 1, we get the voltage equation as

-12I1 -8I1 +20=0

20I1=20

I1=1 A

For mesh 2, I2= -2A

VTH = 8I1 +6I2 =8× 1 +6× -2=8-12 = - 4Volts

VTH = - 4 Volts

Now

0.253 Watts

Q2. Find RL for maximum power transfer and calculate maximum power transferred to RL

Solution:

101


circuit to the load when RL is made equal to RTH, the Thevenin’s resistance at terminals A & B.


Step 1:Remove the branch resistance RL

.


Replace Voltage Source 50V by its internal resistance i.e. short circuit. In the figure . equivalent

resistance across the load terminals A & B is called Thevenin’s resistance RTH. For

simplification circuit can be redrawn as shown in fig

Converting the delta formed by 4Ω, 5Ω & 2Ω resistors (∆ CAD) into equivalent star network,

i.e., ∆ CAD Y CAD, we obtain modified network as shown in the figure

By Series-Parallel reduction techniques , we get the following network

102

RAB=RTH= 3.27Ω Hence when RL =3.27Ω maximum power is transferred to the load from the

circuit

RL=RTH= 15.8 Ω



Voltage(VTh). For Calculation of VTh i.e. VAB, the selected path from A to B is marked by dotted

line in the figure

VTh= 1I2 -4I1

As this path contains the resitances 1Ω & 4Ω,

current through these resistances are required.

The actual directions of the currents are

marked . The 50V source produces the total

current IA , which divides at node C. Let

Current through branch CBD is I2. By Series-

Parallel reduction techniques , we get the

circuits as shown in figure

From figure the total circuit current , I =

From figure the total circuit current gets divided at node C, BY CDR,

103

I1 =

I2 =

VTH = 1I2 -4I1= 1 ×3.845 - 4 ×3.845 = -11.54 V

VTH = - 11.54 Volts

Hence,

10.18Watts

Q3. Find value of RL for maximum power and find Pmax

Solution:

Step 1:Remove the branch resistance RL & mark the open circuit terminals as A & B.

104


Replace Voltage Source 180V by its internal resistance i.e. short circuit.

In the figure . equivalent resistance across the load terminals A & B is called Thevenin’s

resistance RTH. For simplification circuit can be redrawn as shown in fig


RS= 8+4= 12Ω

Also resistances 17Ω & 13Ω are in series

RS= 17+3= 30Ω

105

Converting the delta connections formed by three 12Ω resistors (∆ CAD) and three 30Ω resistors

(∆ EFB) into equivalent star network i.e. (∆ CAD) (Y CAD )& (∆ EFB) (Y EFB) , we get

the network as shown in figure

R1=R2=R3= = 4Ω

Rx=Ry=Rz= = 10Ω

The simplified network is as shown in figure

In Branch SDES’ , 4Ω, 34Ω, 10Ω are in series. Also in branch SCFS’ 4Ω, 10Ω are in series

In Figure 48 Ω & 14 Ω resistors are in parallel

106

RAB=RTH= 24.84 Ω

RTH= 24.84 Ω


The voltage which appears across the open circuit terminals A & B is called as Thevenin’s

Voltage (VTh).


RS= 8+4= 12Ω


RS= 17+3= 20Ω

107

Converting the delta connections formed by three 12Ω resistors (∆ CAD) and three 30Ω resistors

(∆ EFB) into equivalent star network i.e. (∆ CAD) (Y CAD )& (∆ EFB) (Y EFB) , we get

the netqwork as shown in figure

R1=R2=R3= = 4Ω

Rx=Ry=Rz= = 10Ω

The simplified network is as shown in figure

In Branch SDES’ , 4Ω, 34Ω, 10Ω are in series.

Let the current delivered by the battery is I A

108

For calculation of VTh i.e. VAB, the selected path from A to B is marked by dotted line in the

figure. As branch SA & Branch S’B are open circuited , ISA=0 & IS’B=0. By Ohm’s law the total

circuit current can be calculated as

I =

Hence, VTH =VAB= 10

= 10

VTH= 139.34Volts

Now

195.41Watts

Q4. Find maximum power delivered across P-Q

Solution:


circuit to the load when RL is made equal to RTH, the Thevenin’s resistance at terminals P & Q.


109

Step 1: Calculation of RTh across the open load terminals P & Q

Replace Voltage Source 2V by its internal resistance i.e.1Ω, & 4V by its internal resistance i.e.

2Ω short circuit & 2A, 8A Current source by its internal resistance i.e. open circuit.

In the above figure, we observe that resistances 4Ω , 2Ω & 2Ω are in series.

RS= 4+2+2= 8Ω


The Simplified circuit , 8Ω & 3Ω are in parallel

RP= ((8Ω)-1+(3Ω)-1) -1=2.182 Ω

RAB=RTH= 2.182Ω

RTH= 2.182 Ω

Step 2: Calculation of VTH

The voltage which appears across the open circuit terminals P & Q is called as Thevenin’s

Voltage (VTh). For Calculation of VTh i.e. VPQ, the selected path from P to Q is marked by dotted

line in the figure

VTh=-2I3 -1I3 + 2 = -3I3 + 2

110

As this path contains the resistance 2Ω current through this resistances is required. By using

Mesh analysis, this current can be calculated.

For Mesh 1, I1=8A

Mesh 1 & 2 form a supermesh , the current equation for the supermesh is

I2 -I1 =2

I2 = I1 + 2

But I1=8A I2 =2+8 = 10 A

Applying KVL to the mesh 3, we get the voltage equation as

-4(I3 –I1) - 2(I3 –I1) -4 -2 (I3 –I2)-2I3+2 -1I3 =0

6I1 +2 I2 -11I3= 2

I3=6 A

VTh= -3I3 + 2 = -3 6 +2 =-18+2 = -16 Volts

VTH= - 16 Volts

Now

29.331 Watts


1. Maximum power is transferred if load resistance is equal to _ _ _ _ _of the source.

a. half the internal resistance. b. internal resistance

c. twice the internal resistance d. none of the above

2. Maximum power will be transferred from a source of 10Ω resistance to a load of

a. 5Ω b. 20Ω c. 10Ω d. 40Ω

3. The open-circuited the voltage at the terminals of load RL in a network is 30V. Under the

condition of maximum power transfer, the load voltage will be

a. 30V b. 10V c. 5V d. 15V

111

4. Under the condition of maximum power transfer , a voltage source is delivering a power of

30W to the load .the power produced by the source is

a. 45W b. 60W c. 30W d. 90W

5. The maximum power is transfer theorem is used in

a. Electronics circuits b. power system

c. home lighting circuits d. none of the above.

6. What is the efficiency of the system under maximum power transfer?

a. 100% b. 10% c. 50% d. none of these

EXERCISES

Q1. Using Source Transformation, find I

Q2. Using Norton’s theorem, find I

Q3. Derive the expression for star-delta and vice-versa conversion of three resistances.

Q4. Using Superposition theorem, find I

112

Q5. Derive the condition for maximum power transfer.

Q6. State and explain Superposition, Thevenin’s, Norton’s theorem.

Q7. Show that the efficiency is 50% at maximum power transfer condition.

Q8. Using Nodal analysis, find the voltage at X.

Q9. Determine the value of resistance R as shown below using KCL and KVL

Q10. Determine the current through 2.5Ω resistor using Norton’s and Thevenin’s theorem in the

network shown below.

113

Q11. Calculate the voltage across branch AB shown in figure by using mesh analysis

Q12. Explain with suitable example, how to obtain

a) An equivalent current source from a given voltage source.

b) An equivalent voltage source from a given current source.

Q13. Determine the current through 10Ω resistor in the network by i) Star-delta transformation

and ii) Thevenin’s theorem.

Q14. Find maximum power delivered across P-Q

114

Q15. Calculate using Thevenin’s theorem the current flowing through 5Ω resistor connected

across terminals A and B as shown.

ANSWER KEY

Lec 1 Lec 2 Lec 3 Lec 4 Lec 5 Lec 6 Lec 7 Lec 8 Lec 9 Lec

10

Q1. a d a c A c b d b c

Q2. b d a c C c b c c c

Q3. b a b d C b a a a b

Q4. a d b b A b c d a c

Q5. b a a a C a d b b c

Q6. a -- -- -- -- -- -- -- -- --

Q7. b -- -- -- -- -- -- -- -- --

Q8. b -- -- -- -- -- -- -- -- --

Q9. a -- -- -- -- -- -- -- -- --

Q10. -- -- -- -- -- -- -- -- -- --

115

MODULE 2

A.C. CIRCUITS

116

LECTURE 11

INTRODUCTION TO AC

OBJECTIVES

• To learn what is an alternating quantity

• Generation of alternating voltages and currents

• To understand parameters associated with alternating quantities

• Numerical solving

11.1 What is dc current /voltage

We have known that current drawn from a battery is uni-directional. The polarities of the battery

are marked positive and negative. When a load is connected across the battery, current flows

through the lamp in a particular direction. The magnitude of the current as well as its direction

remains constant with respect to time as long as the battery voltage is constant. Such a current is

known as steady state current or direct current.

The fig 11.1 below demonstrates this. As long as the switch is connected in the given position as

in figure (a), the current would flow in the same direction through the load, figure (b).

Fig 11.1

11.2 What is an alternating quantity?

To generate ac from a given dc source, we need an automatic switching arrangement. This can be made available by using electronic circuitry, which can be switched at regular intervals. However, alternating current on a large scale can be obtained from ac generators installed in power houses. Fig 11.2(b) shows the output waveform when switch S is moved alternately between positions 1 and 2(figure a). So the current through the load keeps alternating for every half cycle.

117

Fig 11.2

DEFINITION: An alternating quantity changes continuously in magnitude and alternates in direction at regular intervals of time. This is shown in figure below.

Fig 11.3

11.3 Important terms associated with an alternating quantity 1. Amplitude It is the maximum value attained by an alternating quantity. Also called as maximum or peak value ( Em or -Em)

Fig 11.4

2. Time Period (T) It is the Time Taken in seconds to complete one cycle of an alternating quantity 3. Instantaneous Value It is the value of the quantity at any instant

118

4. Frequency (f) It is the number of cycles that occur in one second. The unit for frequency is Hz or cycles/sec. The relationship between frequency and time period can be derived as follows.

Time taken to complete f cycles = 1 second Time taken to complete 1 cycle = 1/f second

T = 1/f

11.4 Why AC over DC

1. AC voltages can be efficiently stepped up/down using transformer 2. AC motors are cheaper and simpler in construction than DC motors 3. Switchgear for AC system is simpler than DC system

11.5 Generation of ac voltages/currents As stated earlier, large scale ac generation takes place by using ac generators or alternators. Let’s try to understand the underlying principle of ac generation.

Fig 11.5 Now let’s consider the rotation of the conductor by about 360o (in steps of 90 o)

The generator shown in Fig 11.5 consists of:

• North and south pole magnets which

create magnetic flux

• A single loop conductor called armature.

• Slip rings mounted on a shaft to which

each end wire is connected.

• Brushes which enable current to be

carried to the external circuit.

119

Fig 11.6

Fig 11.7

Fig. 11.8

With respect to Fig 11.6

• The armature conductors move parallel

to the lines of flux.

• Though the conductor is linked with

maximum flux, rate of change of flux is

zero.

• Hence emf induced is zero in this

position


• The armature conductors have rotated by

90o and cut maximum lines of flux.

• Rate of change of flux and hence, emf

induced is maximum in this position.

• This is clear from the maximum value of

the sine wave.


• The positive voltage decreases as the flux

lines are no longer at right angles.


induced is minimum in this position.

• This is clear from the zero value of the

sine wave at 180o .

120

Fig 11.9

Fig 11.10 Thus it is seen that in one revolution of the armature, we get one cycle of an alternating sinusoidal waveform. The equation of the sine wave is given by

wtEe m sin=

Where e = instantaneous value of voltage Em = maximum value of the sine wave (or) its amplitude ω = the angular velocity of rotation of the coil in rad/s and is given by

fT

w Π=Π

= 22

Other forms of writing the same equation are:

T

tEftEwtEe mmm

Π=Π==

2sin2sinsin

With respect to Fig. 11.9

• The armature conductors have rotated by

270o and cut maximum lines of flux.


induced is maximum in this position, but

with the opposite direction.

• This is clear from the negative peak

maximum value of the sine wave.


• The negative voltage decreases as the

flux lines are no longer cut at right

angles.

• When the armature loop has made full

rotation (360o), none of the flux lines are

cut and zero emf is induced.

121

Where f is frequency of the sine wave and T is its time period

11.6 Average and rms values

Average Value The arithmetic average of all the values of an alternating quantity over one cycle is called its average value.

n

VVVV n

avg

+++=

.............21 ---------(mid –ordinate method – useful for non- sinusoidal

waveform)

)()sin(1

0

wtdwtVT

V

T

mavg ∫= ------------------------------ (analytical method)

The average value is taken over the time period T for un-symmetrical waveforms whereas for symmetrical waveforms, the average value is calculated for only half the cycle, because for full cycle, its value would be zero. A waveform is said to be symmetrical if it has equal /similar positive and negative cycles. The Fig 11.11 below illustrate this:

Fig 11.11

Fig 11.11(a) shows a symmetrical waveform, whose average value over the interval 2π will be zero. Hence its average value is calculated only over the interval π. Fig (b) and (c) are un-symmetrical waveforms with time periods of 2π and π respectively.

Rms value or effective value : The effective or RMS value of an alternating quantity is that steady current (dc) which when flowing through a given resistance for a given time produces the same amount of heat produced by the alternating current flowing through the same resistance for the same time.

Hence for symmetrical waveforms the average value is taken for only one half cycle.

But, for un- symmetrical waveforms as shown below, the average value should always

be taken for the full cycle/ time period.

122

Fig 11.12

RMS stands for root mean square i.e. the root of the mean of the squared values of the function: be it current /voltage.

n

VVVV n

rms

22

2

2

1 ............. +++= ----------------------------------------- ( Mid –ordinate method)

(or)

base

curvesquaredtheunderAreaVrms = ----------------------------------- (Analytical

method)

∫=T

dtvT

0

21

11.7 Solved Problems Q1. An alternating current i is given by i = 141.4 sin 314t Find i) The maximum value ii) Frequency iii) Time Period iv) The instantaneous value

when t=3ms Solution:

Comparing the given equation with the standard equation

i) Maximum value Im=141.4 V ii) ω = 314 rad/sec f = ω/2π = 50 Hz iii) T=1/f = 0.02 sec

123

iv) i=141.4 sin(314x0.003) = 114.35A Q2. Find the average value of the sinusoidal current as shown in the figure below. Solution:

Q3. Find the average value of the sinusoidal current as shown in the figure below. Solution:

Q4. Find the rms value of the sinusoidal current as shown in the figure below. Solution:

124

11.8 Form factor and Peak factor Now, we define two factors, the form factor and the peak factor. Form factor Its defined as the ratio of the rms value to the average value of an alternating quantity.

valueAverage

valueRMSFF =

Peak factor/ Crest factor Its defined as the ratio of maximum value to rms value of an alternating quantity.

valueRMS

valueMaximumPF =

For a sinusoidal waveform,

414.1707.0

11.1637.0

707.0

707.02

637.02

===

===

==

=Π

=

m

m

rms

m

m

m

avg

rms

mm

rms

mm

av

I

I

I

IPF

I

I

I

IFF

II

I

II

I

Significance of form factor and peak factor:

1. The form factor gives a measure of the peakiness of the waveform. The peakier

the wave, greater is its form factor, and greater the form factor, greater is the

hysteresis loss such a current produces in a magnetic material.

2. The peak factor plays an important role in dielectric insulation testing – the

dielectric stress produced depends on the peak value of the waveform.

125


1. If Vrms=10V for a sine wave, then Vmax= a. 10V b. 7.07V c. 20V d. 14.14V 2. Form factor is defined as

a. avg value /rms value b. rms value/ avg value c. max value/ avg value d. max value/ rms value

3. If w=100 rad/s, then time period of the waveform is

a. 6.283ms b. 62.83ms c. 628.3 ms d. 15.91 ms

4. The peak value of a sine wave is 200V. Its average value is

a. 12.74V b. 200V c. 100V d. 127.4V

5. The voltage of domestic supply is 220V. This figure represents

a. mean value b. average value c. rms value d. peak value

6. The rms and the mean value is the same in the case of a

a. triangular wave b. sine wave c. square wave d. half wave rectified sine wave

7. The rms value of a sinusoidal ac current is equal to its value at an angle of ______ degrees.

a. 90 b. 60 c. 30 d. 45

8. The effective or RMS value of an alternating quantity is that steady current (dc) which when

flowing through a given resistance for a given time produces

126

a. more heat than produced by AC flowing through the same circuit b. the same amount of heat produced by AC flowing through the same circuit c. less amount of heat than produced by AC flowing through the same circuit d. none of these.

9. For un- symmetrical alternating waveforms, the average value should always be taken over

a. full cycle

b. quarter cycle

c. half cycle

d. unsymmetrical part of the waveform.

10. If a sine wave has a frequency of 50Hz with 30A rms current, which of the equations below represents this wave

a. 42.42 sin 314t b. 60 sin 25t c. 30 sin 50t d. 84.84 sin 25t

127

LECTURE 12

RMS AND AVERAGE VALUES, PHASOR REPRESENTATION

OF SINUSOIDAL QUANTITIES

OBJECTIVES

• To solve numericals on rms and average values

• To understand the concept of phasor

• Polar and rectangular representation of a phasor

• Numericals based on polar- rectangular conversion

12.1 Solved numericals

Q1. Compute the average value, rms value , form factor and peak factor for the figure shown

below.

128

Q2. Compute average value, rms value, form factor and peak factor for the figure shown below.

129

Q3. Compute the rms and average values of the waveform given below.

130

Solution:

12.2 Phasor and phasor diagrams

An alternating quantity can be represented using (i) Waveform (ii) Equations (iii) Phasor

When we want to add/ subtract/ multiply or divide two or more than two alternating voltages/

currents, it becomes cumbersome to handle equations which are in the standard form:

i.e. wtEe m sin=

Hence and alternative to this is to represent each alternating quantity as a phasor.

Definition

This is how a sinusoid can be represented as a phasor

1. A phasor is a vector which rotates anticlockwise at a speed equal

to the angular velocity of the sinusoid.

2. Its length represents the magnitude/ peak value of the sinusoid and

its direction is the angle it makes with respect to the positive x-axis.

3. In short- a phasor is a rotating vector. By the time the sinusoid

finishes one cycle (T), the phasor finishes one revolution (360o).

131

Fig 12.1

Draw a line OP of length equal to Im. This line OP rotates in the anticlockwise direction with a uniform angular velocity ω rad/sec and follows the circular trajectory shown in figure. At any instant, the projection of OP on the y-axis is given by OM=OPsinθ = Im sin ωt. Hence the line OP is the phasor representation of the sinusoidal current

LAG AND LEAD

The concept of lag and lead of a phasor can be explained with respect to the fig below.

Fig. 12.2

In general, given

θ

θ

θ

byvleadsvequationtheinsignuseweif

byvlagsvequationtheinsignuseweif

wtVvandwtVv mm

21

21

21

,

,

,)sin()sin(

−

+

±==

Example:

1. In the clock shown, taking 3 as the reference, it is

observed that 12 is leading 3 and 5 is lagging 3.

2. Always, when we talk about lead and lag, the reference

is extremely important.

3. If we take 5 as reference, then both 12 and 3 would be

leading 5.

4.On the other hand, if we take 12 as reference, both 3 and

5 would be lagging 12.

132

V1=10 sin wt V2=5 sin (wt-30)

Lag (V2 lags V1 by 30o)

V1=10 sin wt V2=5 sin (wt+30)

Lead (V2 leads V1 by 30o)

V1=10 sin wt V2=5 sin wt

V1 and V2 are in phase

12.3 Phasor Algebra

A phasor has two components, magnitude M and angle θ.

There are various forms of representing a phasor A , namely,

1. Polar form : θ∠= MA

2. Rectangular form: jbaA += (where a-real part of phasor, b - imaginary part of the phasor)

Fig 12.3

Here we introduce the operator j, representing the imaginary axis. Seen below is the vector

rotation of the operator ‘j’

θ

V2 V1

θ

V2

V1

V2

V1

30

V2

V1 30

V2 V1 V2

V1

133

Fig 12.4

To convert polar ( M, θ given) ---rectangular:

a=M cos θ and b=M sin θ.

To convert rectangular (a, b given) ---polar:

22baM += and θ =

Example:

To add/subtract two phasors together, we must convert them into rectangular form:

C1=A1±jB1.

C2=A2±jB2, then

C1 ± C2=( A1 ± A2 ) + j( ±B1 ± B2 )

To multiply two phasors, we must first convert them to polar form. The product in polar form is

simply the product of their magnitudes and the phase is the sum of their phases.

212121

222

111

** φφ

φ

φ

+∠=

∠=

∠=

MMCC

thenMC

MC

Division is similar to multiplication, except now we divide the magnitudes, and subtract the

phases

For addition and subtraction of phasors, use rectangular co-ordinators.

For multiplication and division of phasors, use polar co-ordinators.

134

21

2

121

222

111

/ φφ

φ

φ

−∠=

∠=

∠=

M

MCC

thenMC

MC

An important relationship worth understanding is the inversion property of phasors:

1800 ∠−=∠= cc MMC


Q1. Convert 306∠ into rectangular form:

Solution:

M=6, θ=30

a= M cos θ= 5.2

b=M sin θ= 3

Rectangular form= a+jb = 5.2+ j3

Q2. Represent the following equations in polar form:

Solution:

a) Vwtv )30sin(20 °+=

°∠=°∠= 3014.14302

20Vrms

b) Vwtv )45sin(35 °+=

°∠=°∠= 4575.24452

35Vrms

c) Vwtv )75sin(10 °−=

°−∠=°−∠= 7507.7752

10Vrms

d) Vwtv )sin(20=

°∠=°∠= 014.1402

20Vrms

135

Q3. Transform the following into phasors

136


1. Which of these statements is true about a phasor?

a. a vector which rotates

b. they are used to represent sinusoidal quantities.

c. both a and b.

d. none of these.

2. is

a. j2 b. –j2 c. 2/(-j) d. 2

3. =°∠°−∠°∠ 1906*604*253

a. 65.25+j30.42

b. -65.25+j30.42

c. -65.25-j30.42

d. 65.25-j30.42

4. Two sinusoidal currents are given by i1=10 sin (wt+ π /3) and i2= 15 sin (wt- π /4). What is the

angle between them.

a. 105o

b. 15 o

c. 30 o

d. 25 o

5. If bewillBofAvaluethethenBandA /,1554510 ∠=∠=

a. 5∠60

b. 2∠30

c. 50∠60

d. 0.2∠- 30

137

6. When a phasor is multiplied by –j, it gets rotated through _____ in the counterclockwise

direction.

a. 90o

b. 180 o

c. 270 o

d. -90 o.

7. The voltage v = √2 x 50 sin (377t-35o) V in rectangular form is:

a. 40.95-j28.67’

b. --40.95+j28.67

c. -40.95-j28.67

d. 40.95+j28.67

8. Find a single sinusoid that’s equal to 6.23 sin wt +9.34 cos wt

a. 15.57 sin wt

b. 15.57 cos wt

c. 11.22 sin (wt+56.32)

d. 9.34 sin 2wt

9. Find a single sinusoid that’s equal to 5 sin 377t +8 cos 387t.

a. 5 sin 377t

b. 8 cos 387t

c. 13 cos 754t

d. none

10. Two sinusoidal currents are given by i1=10 sin (wt+ π /3) and i2= 15 cos (wt- π /4). What is

the angle between them.

a. 105o

b. 15 o

c.30 o

d. 25 o

138

LECTURE 13

AC THROUGH PURE R, L, AND C

OBJECTIVES

• To understand the waveforms of current through various loads ( pure R, pure L, pure C)

when an alternating voltage is applied across it

• To solve numerical problems based on this concept

Today let’s try to understand what would be the current flowing through a load when a voltage

)sin(wtVv m= is applied across it. Of course, as you know, the current would depend on the

nature of the load. So, we will examine the current for three cases of the load:

a. Purely resistive

b. Purely inductive

c. Purely capacitive.

13.1 AC circuit with a pure resistance

Fig 13.1

The alternating voltage v is given by )sin(wtVv m= ----------------(13.1)

The current flowing in the circuit is i. The voltage across the resistor is given as VR which is the same as v. Using ohms law, we can write the following relations:

139

From equation (13.1) and (13.2) we conclude that in a pure resistive circuit, the voltage and current are in phase. Hence the voltage and current waveforms and phasors can be drawn as below

Fig 13.2 Instantaneous power The instantaneous power in the above circuit can be derived as follows

wtVmVm

p

wtVm

p

wtVmp

wtwtVmp

vip

2cos2

Im

2

Im

)2cos1(2

Im

sinIm

)sin)(Imsin(

2

−=

−=

=

=

=

The instantaneous power consists of two terms. The first term is called as the constant power term and the second term is called as the fluctuating power term. Average power From the instantaneous power we can find the average power over one cycle as follows

VIP

VmVmP

dwtwtVmVm

P

=

==

−Π

= ∫Π

2

Im

22

Im

2cos2

Im

2

Im

2

12

0

As seen above the average power is the product of the rms voltage and the rms current. The voltage, current and power waveforms of a purely resistive circuit is as shown in the figure below.

i

V

i V

140

Fig 13.3

Phasor Algebra for a pure resistive circuit

°∠=+=+

==

+=°∠=

000

00

IjIR

jV

R

VI

jVVV

13.2 AC circuit with a pure inductance

Fig 13.4

Consider an AC circuit with a pure inductance L as shown in the figure. The alternating voltage v is given by

)sin(wtVv m= ----------------(13.3)

We can find the current through the inductor as follows

141

From equation (13.3) and (13.4) we observe that in a pure inductive circuit, the current lags behind the voltage by 90o. Hence the voltage and current waveforms and phasors can be drawn as below.

Fig 13.5

Inductive reactance Here we define a term XL, the inductive reactance defined as:

L

L

X

Vm

fLwLX

=

Π==

Im

2

It is equivalent to resistance in a resistive circuit. The unit is ohms. Instantaneous power The instantaneous power in the above circuit can be derived as follows:

)2sin(2

Im

cossinIm

))2/sin()(Imsin(

wtVm

p

wtwtVmp

wtwtVmp

vip

−=

−=

Π−=

=

142

Average power From the instantaneous power we can find the average power over one cycle as follows

0

2sinIm2

12

0

=

−Π

= ∫Π

P

dwtwtVmP

Fig. 13.6 As seen from the power waveform, the instantaneous power is alternately positive and negative. When the power is positive, the power flows from the source to the inductor and when the power in negative, the power flows from the inductor to the source. The positive power is equal to the negative power and hence the average power in the circuit is equal to zero. The power just flows between the source and the inductor, but the inductor does not consume any power. Phasor algebra for a pure inductive circuit

)(

9090

0

090

00

L

L

jXIV

XI

V

I

V

jIII

jVVV

=

°∠=°−∠

∠=

−=°−∠=

+=∠=

13.3 AC circuit with a pure capacitance

Fig 13.7

143

Consider an AC circuit with a pure capacitance C as shown in the figure. The alternating voltage is given by

)sin(wtVv m= -----(1)

We can find the current through the capacitor as follows

wCVmwhere

wti

wtwCVmi

wtCVmi

dt

dqi

wtCVmq

Cvq

=

−−−−−−−−−−Π+=

Π+=

=

=

=

=

Im

)2()2/sin(Im

)2/sin(

cos

sin

From equation (1) and (2) we observe that in a pure capacitive circuit, the current leads the voltage by 90o. Hence the voltage and current waveforms and phasors can be drawn as below.

Fig 13.8

Capacitive reactance The capacitive reactance XC is given as

C

C

X

Vm

fCwCX

=

Π==

Im

2

11

It is equivalent to resistance in a resistive circuit. The unit is ohms. Instantaneous power The instantaneous power in the above circuit can be derived as follows

144

)2sin(2

Im

cossinIm

))2/sin()(Imsin(

wtVm

p

wtwtVmp

wtwtVmp

vip

=

=

Π+=

=

As seen from the above equation, the instantaneous power is fluctuating in nature. Average power From the instantaneous power we can find the average power over one cycle as follows

0

2sinIm2

12

0

=

Π= ∫

Π

P

dwtwtVmP

The average power in a pure capacitive circuit is zero. Or in other words, the power consumed by a pure capacitance is zero. The voltage, current and power waveforms of a purely capacitive circuit is as shown in the figure.

Fig 13.9

As seen from the power waveform, the instantaneous power is alternately positive and negative. When the power is positive, the power flows from the source to the capacitor and when the power in negative, the power flows from the capacitor to the source. The positive power is equal to the negative power and hence the average power in the circuit is equal to zero. The power just flows between the source and the capacitor, but the capacitor does not consume any power. Phasor algebra in a pure capacitive circuit

145

)(

9090

0

090

00

C

C

jXIV

XI

V

I

V

jIII

jVVV

−=

°−∠=°∠°∠

=

+=°∠=

+=°∠=

13.4 Solved Problems Q1. An ac circuit consists of a pure resistance of 10Ω and is connected to an ac supply of 230 V,

50 Hz. Calculate the (i) current (ii) power consumed and (iii) equations for voltage and current.

Solution:

Q2. A pure inductive coil allows a current of 10A to flow from a 230V, 50 Hz supply. Find (i)

inductance of the coil (ii) power absorbed and (iii) equations for voltage and current. Solution:

146

Q3. A 318µF capacitor is connected across a 230V, 50 Hz system. Find (i) the capacitive reactance (ii) rms value of current and (iii) equations for voltage and current.

Solution:


1. The current in a certain ac circuit is independent of the frequency at a given voltage. Which combination of elements is most likely to comprise the circuit?

a.Resistors only b. Inductors only c. Capacitors only d. none

2. The inductive reactance XL is

a.1/2πfL b. 2πfL c. L d. none

3. The capacitive reactance XC is a. 1/2πf b. 2πfC c. C d. none 4. Power dissipated in a pure inductive circuit is a. zero b. VI c. VXL. d. I XL.

147

5. In a purely inductive circuit, a. I lags V by 90 b. I leads V by 90 c. I and V are in phase d. none 6. In a purely capacitive circuit a. I lags V by 90 b. I leads V by 90 c. I and V are in phase d. none 7. The frequency of the instantaneous power in a pure L circuit is ___________ than the frequency of the supply voltage. a. half b. double c. same d. one fourth 8. If a voltage V=(10+j0)V is applied to a circuit and the resultant current is I= -j5 A, the nature of the circuit is: a. pure R b. pure L c. pure C d. none 9. If a voltage V=(10+j0)V is applied to a circuit and the resultant current is I= 5 A, the nature of the circuit is: a. pure R b. pure L c. pure C d. none 10. . If a voltage V=(20+j0)V is applied to a circuit and the resultant current is I= j5 A, then the power consumed by the circuit is: a. 100W b. 25W c. 0W d. none

148

LECTURE 14

AC THROUGH RL SERIES CIRCUIT

OBJECTIVES

• To understand the waveform of current through RL load when an alternating voltage is

applied across them

• To understand impedance and power triangle

• To solve numerical problems

14.1 RL Circuit

Let us try to understand what would be the current flowing through a R-L load when a voltage

)sin(wtVv m= is applied across it

Fig 14.1

The current flowing in the circuit is i. The voltage across the resistor is VR and that across the inductor is VL.

VR = IR is in phase with I VL = IXL leads current by 90 degrees

With the above information, the phasor diagram can be drawn as shown by taking current as the

reference.

149

Fig 14.2

The resultant voltage V can be drawn as shown in the figure. From the phasor diagram we observe that the voltage leads the current by an angle Φ or in other words the current lags behind the voltage by an angle Φ. The waveform and equations for an RL series circuit can be drawn as below.

Fig 14.3

From the phasor diagram, the expressions for the resultant voltage V and the angle φ can be derived as follows.

circuittheofimpedancethecalledisXRZwhere

ZIV

XRIV

IXIRV

VVV

L

L

L

LR

22

22

22

22

)()(

+=

=

+=

+=

+=

The impedance in an AC circuit is similar to a resistance in a DC circuit. The unit for impedance is ohms Ω.

150

Phase angle:

R

wL

R

X

RI

XI

V

V

L

L

R

L

1

1

1

1

tan

tan

tan

tan

−

−

−

−

=

=

=

=

φ

φ

φ

φ

Instantaneous power The instantaneous power in an RL series circuit can be derived as follows:

)2cos(2

cos2

)sin(sin

φφ

φ

−−=

−=

=

wtIVIV

p

wtwtIVp

vip

mmmm

mm

The instantaneous power consists of two terms. The first term is called as the constant power term and the second term is called as the fluctuating power term. Average value From the instantaneous power we can find the average power over one cycle as follows:

φ

φ

φ

φφ

cos

cos2

Im

2

cos2

Im

)]2cos(2

Imcos

2

Im[

2

12

0

IVP

VmP

VmP

dwtwtVmVm

P

=

=

=

−−Π

= ∫Π

The voltage, current and power waveforms of a RL series circuit is as shown in the figure below.

151

Fig 14.4

As seen from the power waveform, the instantaneous power is alternately positive and negative. When the power is positive, the power flows from the source to the load and when the power in negative, the power flows from the load to the source. The positive power is not equal to the negative power and hence the average power in the circuit is not equal to zero. From the phasor diagram,

RIP

Z

RIIZP

VIP

Z

R

ZI

RI

V

VR

2

**)(

cos

cos

=

=

Φ=

===Φ

Power Factor The power factor in an AC circuit is defined as the cosine of the angle between voltage and current ie cos φ P =VI cosφ The power in an AC circuit is equal to the product of voltage, current and power factor

Hence the power in an RL series circuit is consumed only in the resistance. The inductance does not consume any power.

152

14.2 Impedance Triangle We can derive a triangle called the impedance triangle from the phasor diagram of an RL series circuit as shown in figure below.

Fig 14.5

The impedance triangle is right angled triangle with R and XL as two sides and impedance as the hypotenuse. The angle between the base and hypotenuse is φ. The impedance triangle enables us to calculate the following things.

1. Impedance 22

LXRZ +=

2. Power factor Z

R=φcos

3. Phase angle

4. Whether the current leads or lags behind the voltage Power In an AC circuit, the various powers can be classified as 1. Real or Active power :P 2. Reactive power : Q 3. Apparent power: S

1. Real or active power in an AC circuit is the power that does useful work in the circuit. It is the power that is consumed by the resistance (Watts or KW).

2. Reactive power flows in an AC circuit but does not do any useful work. It is the circulating power in the L and C components. (VAR or KVAR)

3. Apparent power is the total power in an AC circuit.(VA or KVA)

153

Fig 14.6

From the phasor diagram of an RL series circuit shown above, the current can be divided into two components. One component along the voltage Icosφ, that is called as the active component of current and another component perpendicular to the voltage Isinφ that is called as the reactive component of current.

14.3 Power Triangle From the impedance triangle, another triangle called the power triangle can be derived as shown

Fig 14.7

From the above fig, we can have three different ways of defining the power factor of a circuit:

P = VI cosφ = V2/R=I2*R

Q = VI sinφ = I2*X ( where X could be XL or XC)

S= VI = I2*Z = 22 QP +

Cosine of angle between V and I Resistance/Impedance R/Z

Real Power/Apparent Power P/S

154

Phasor algebra in a RL series circuit

jQPVIS

Z

V

Z

VI

ZjXRZ

jVVV

L

+==

−∠==

∠=+=

+=∠=

*

00

φ

φ

14.4 Solved Problems Q1. A coil having a resistance of 7Ω and an inductance of 31.8mH is connected to 230V, 50Hz

supply. Calculate (i) the circuit current (ii) phase angle (iii) power factor (iv) power consumed.

Solution:

Q2. A 200 V, 50 Hz, inductive circuit takes a current of 10A, lagging 30 degree. Find (i) the

resistance (ii) reactance (iii) inductance of the coil. Solution:

155

.

156

MULTIPLE CHOICE QUESTIONS 1. As frequency increases, XL

a. Decreases b. Increases c. Remains the same d. none

2. In a series RL circuit, if XL>>>R then, the angle φ will be

a. Nearly 0 b. Nearly 90 c. Nearly 180 d. None

3. In a circuit, voltage and current are given by )30sin(10)30sin(10 °−=°+= wtiandwtv

The power consumed in the circuit is a. 100W b. 50W c. 25W d. 12.5W 4. Reactive power Q in RL series circuit is:

a. positive b. negative c. zero d. none

157

5. When f=50 Hz, XL=25Ω, find XL for a frequency of 40Hz. a. 10 Ω b. 25 Ω c. 20 Ω d. 15 Ω. 6. If P=10KW and pf=0.5 find S. a. 10KVA b. 20KVA c. 20VA d. 10VA 7. The pf of a series RL circuit is always a. leading b. lagging c. could be lead or lag d. none 8.The value of pf in an RL series circuit is a. 0< pf<1, lag b. 0< pf<1, lead c.1 d. >1 9. If R=3Ω and XL=4Ω , find the current phasor for an applied voltage of (100+j0)V. a. (12-j16)A b. (-12-j16)A c. (12+j16)A d. (-12+j16)A 10. Given applied voltage V=10V for a series RL circuit and VR=6V, what is VL. a. 5V b. 10V c. 8.66V d. 8V.

158

LECTURE 15

AC THROUGH RC SERIES CIRCUIT

OBJECTIVES

• To understand the current waveforms in RC series circuit with an alternating voltage

appled across it

• To solve numerical problems

15.1 RC Series Circuit

Fig 15.1

Consider an AC circuit with a resistance R and a capacitance C connected in series as shown in the figure above. The alternating voltage v is given by

)sin(wtVv m= ------------------(15.1)

The current flowing in the circuit is i. The voltage across the resistor is VR and that across the capacitor is VC. VR=I R is in phase with I VC=I XC lags behind the current by 90 degrees. With the above information, the phasor diagram, taking current as reference, can be drawn as shown.

Fig 15.2

159

The voltage VR is in phase with I and the voltage VC lags behind the current by 90o. The resultant voltage V can be drawn as shown in the figure. From the phasor diagram we observe that the voltage lags behind the current by an angle Φ or in other words the current leads the voltage by an angle Φ. The waveform and equations for an RC series circuit can be drawn as below.

Fig 15.3

From the phasor diagram, the expressions for the resultant voltage V and the angle φ can be derived as follows.

circuittheofimpedancethecalledisXRZwhere

ZIV

XRIV

IXIRV

VVV

C

C

C

CR

22

22

22

22

)()(

+=

=

+=

+=

+=

Phase angle :

wCR

R

X

RI

XI

V

Vc

C

C

R

1tan

tan

tan

tan

1

1

1

1

−

−

−

−

=

=

=

=

φ

φ

φ

φ

15.2 Impedance Triangle We can derive a triangle called the impedance triangle from the phasor diagram of an RC series circuit as shown in figure below.

160

Fig 15.4 Average power:

RIP

Z

RIIZP

VIP

2

**)(

cos

=

=

Φ=

15.3 Power triangle The power triangle is shown in fig below:

Fig 15.5

Since the current leads the voltage, the RC series circuit is said to have a leading pf.

Z

Rpf

S

Ppf

==

=

φcos

Hence the power in an RC series circuit is consumed only in the resistance. The capacitor does not consume any power.

161

Phasor algebra for RC series circuit

jQPVIS

Z

V

Z

VI

ZjXRZ

jVVV

C

−==

∠==

−∠=−=

+=∠=

*

00

φ

φ

15.4 Solved Problems Q1.

Solution:

Q2. A voltage of 120 V at 50 Hz is applied to a resistance, R in series with a capacitance C. The current drawn is 2 A, and the power loss in the resistance is 100 W. Calculate the resistance and the capacitance. Solution:

V=120V I = 2 A P = 100 W f= 50 Hz

162

Q3.

MULTIPLE CHOICE QUESTIONS 1. In a RC series circuit, pf is always a. lagging b. leading c.0 d.1 2. The value of pf in an RC series circuit is a. 0< pf<1,lag b. 0< pf <1, lead c. 1 d. >1 lead

163

3. The reactive power Q in a RC series circuit is a. positive b. negative c. zero d. none 4. In an RC series circuit, as the angle φ approaches 90o, the circuit becomes more a. resistive b. capacitive c. inductive d. none 5. If R=3Ω and XC=4Ω , find the current phasor for an applied voltage of (100+j0)V. a. (12-j16)A b. (-12-j16)A c. (12+j16)A d. (-12+j16)A 6. As frequency increases, XC_______

a. Decreases b. Increases c. Remains the same d. None

7. When f=50 Hz, XC=25Ω, find XC for a frequency of 40Hz. a. 10 Ω b. 25 Ω c. 20 Ω d. 80 Ω. 8. Given P=5KW, pf=0.5, applied voltage V=100V for RC series circuit, find C for f=50Hz. a. 36.7 µF b. 367.5 µF c. 3.67 mF d. 36756 µF 9. Given applied voltage V=10V for a series RC circuit and VR=5V, what is VC. a. 5V b. 10V c. 8.66V d. 7V. 10. Given Z=(2-j3) Ω, what impedance must be added so that the circuit has a unity power factor.

a. 2 Ω b. 1 Ω c. j3 Ω d. –j3 Ω

164

LECTURE 16

R – L – C SERIES CIRCUITS

OBJECTIVES

• How to compute the total reactance and impedance / admittance, of the series circuits, fed from single phase ac supply?

• How to compute the different currents and also voltage drops in the components, both in magnitude and phase, of the circuit?

• How to draw the complete phasor diagram, showing the currents and voltage drops?

• How to compute the total power and also power consumed in the different components, along with power factor?

16.1 RLC Series Circuit

Consider now the series RLC circuit shown in Figure 16.1

Fig 16.1: Series RLC Circuit

165

Impedance

We have already seen that the inductive reactance XL

=ωLand capacitance reactance

XC

=1/ωC play the role of an effective resistance in the purely inductive and capacitive circuits,

respectively. In the series RLC circuit, the effective resistance is the impedance, defined as

The relationship between Z, XL and XC can be represented by the diagram shown in

Figure 16.2 Diagrammatic representation of the relationship between Z, X

L and X

C .

166

Two cases are: (a) Inductive and (b) Capacitive (a) Inductive

In this case, the circuit is inductive, as total reactance is positive, under the

condition . The current lags the voltage by φ (taken as positive), with the current

phasor taken as reference. The power factor (lagging) is less than 1 (one), as . The complete phasor diagram, with the voltage drops is shown below.

Fig. 16.3 Phasor diagram for the series RLC circuit for (a) XL

>XC

(b) Capacitive

The circuit is now capacitive, as total reactance is negative, under the condition

. The current leads the voltage by φ, which is negative as per convention described earlier. The current phasor is taken as reference here. The complete phasor diagram, with the voltage drops across the components and input voltage, and also current, is

shown in Fig. 16.4. The power factor (leading) is less than 1 (one), as , φ being negative.

Fig.16.4: Phasor diagram for the series RLC circuit for (b) XL

<XC

.

From Figure 16.3, we see that VL0

>VC 0

in the inductive case and V0

leads I0

by a phase φ . On the

other hand, in the capacitive case shown in Figure 4, VC 0

>VL0

and I0

leadsV0

by a phaseφ .

When V

L0 = V

C 0, or φ= 0 , the circuit is at resonance. The corresponding resonant frequency is

167

ω0

= 1/ LC , and the power delivered to the resistor is a maximum.

Power in an AC circuit

In the series RLC circuit, the instantaneous power delivered by the AC generator is given by

where we have used the trigonometric identity

The time average of the power is

In terms of the rms quantities, the average power can be rewritten as

The quantity cosφ is called the power factor. From Figure 2, one can readily show that

Thus, we may rewrite

16.2 Problem-Solving Tips

In this chapter, we have seen how phasors provide a powerful tool for analyzing the AC circuits.

Below are some important tips:

168

1. Keep in mind the phase relationships for simple circuits

(1) For a resistor, the voltage and the phase are always in phase. (2) For an inductor, the current lags the voltage by90°. (3) For a capacitor, the current leads to voltage by 90°.

2. When circuit elements are connected in series, the instantaneous current is the same for

all elements, and the instantaneous voltages across the elements are out of phase.

3. For series connection, draw a phasor diagram for the voltages. The amplitudes of the voltage drop across all the circuit elements involved should be represented with phasors. In Figure 3 and 4, the phasor diagram for a series RLC circuit is shown for both the inductive case X

L >X

C and the capacitive case X

L <X

C .


Q1. A series RLC circuit with L =160 mH ,C = 100 µF , and R =40.0Ωis connected to a

sinusoidal voltage V(t) =(40.0) . V sin ωt, with ω=200 rad/s

(a) What is the impedance of the circuit?

(b) Let the current at any instant in the circuit be I(t) = I0

sin (ωt - φ). Find I0.

(c) What is the phase φ? Solution:

(a) The impedance of a series RLC circuit is given by

are the inductive reactance and the capacitive reactance, respectively. Since the general expression of the voltage source is Vt() =V

0 sin( ωt) , where V0 is the maximum output voltage

and ω is the angular frequency, we have V0

=40 V and ω=200 rad/s . Thus, the impedance Z

becomes

169

(b) With V

0 =40.0 V , the amplitude of the current is given by

(c) The phase between the current and the voltage is determined by


1. A series circuit consists of R = 20Ω, L = 20 mH, and ac supply 60V with f = 100 Hz. The voltage drop across L is

a. 39.1V b. 31.9V c. 45.5V d. 50.5 V

2. A series circuit consists of R = 20Ω, L = 20 mH, and ac supply 60V with f = 100 Hz. The current in R is

a. 2.54A b. 1.27A c. 5.08A d. 10.16 A

3. The reactance offered by a capacitor to alternating current of frequency 50 Hz is 20 Ohms. If

frequency is increased to 100 Hz, reactance becomes_____ohms.

170

a. 40 ohm b. 20 ohm c. 10 ohm d. 80 ohm

4. In a series RLC circuit that is operating above the resonant frequency, the current

a. lags the applied voltage b. leads the applied voltage c. is in phase the applied voltage d. is zero

5. XL and XC have opposing effects in an RLC circuit.

a. True b. False

171

LECTURE 17

R – L – C SERIES CIRCUITS : NUMERICALS

OBJECTIVES

• To clearly understand the difference between calculations of d.c. quantities and a.c. quantities.

• To practice few a.c. series circuit problems to find the impedance, current, voltage, power, power factor in the circuits.

• To be more familiar with the phasor representation of the a.c. quantities in the circuit.


Q1. Suppose an AC generator with Vt = (150 V) sin (100t) is connected to a series RLC circuit

with R =40.0 Ω, L =80.0 mH , and C =50.0 µF, as shown in Figure below.

(a) Calculate VR0

, VL0

and VC 0

, the maximum of the voltage drops across each circuit element.

(b) Calculate the maximum potential difference across the inductor and the capacitor between points b and d shown in Figure .

Solution:

(a) The inductive reactance, capacitive reactance and the impedance of the circuit are given by

172

respectively. Therefore, the corresponding maximum current amplitude is

The maximum voltage across the resistance would be just the product of maximum current and the resistance:

V = IR = ( 0.765 A) ( 40.0 Ω) = 30.6 V Similarly, the maximum voltage across the inductor is

VL0 = I0XL = (0.765 A ) ( 8.00 Ω) = 6.12V

and the maximum voltage across the capacitor is

VCO = I0XC = (0.765 A) (200 Ω) = 153 V

Note that the maximum input voltage V0 is related to VR0

, VL0

and VC0

by

(b) From b to d, the maximum voltage would be the difference between V

L0 and V

C 0:

|V

bd | = | V

L0 +V

C 0|

= | V

L0 −V

C 0 | = |6.12 V −153 V| = 147V

Q2. A resistance, R is connected in series with an iron-cored choke coil (r in series with L).

The circuit (Fig. 15.2a) draws a current of 5 A at 240 V, 50 Hz. The voltages across the resistance and the coil are 120 V and 200 V respectively. Calculate, (a) the resistance, reactance and impedance of the coil, (b) the power absorbed by the coil, and (c) the power factor (pf) of the input current

173

Solution

Q3. An inductive coil, having resistance of 8 Ω and inductance of 80 mH, is connected in series with a capacitance of 100Fµacross 150 V, 50 Hz supply (Fig. 15.3a). Calculate, (a) the current, (b) the power factor, and (c) the voltages drops in the coil and capacitance

174

respectively

175


1. A certain series RLC circuit with a 200 Hz, 15 V ac source has the following values: R = 12, C = 80 F, and L = 10 mH. The total impedance, expressed in polar form, is

a. 12.28 12.31° b. 12.57 12.31° c. 9.95 12.31° d. 12.62 12.31°

2. In a series RLC circuit, the larger reactance determines the net reactance of the circuit.

a. True b. False

3. A series circuit consists of R = 20Ω, L = 20 mH, and ac supply 60V with f = 100 Hz. The phase angle of current in respect of supply voltage will be

a. . 40.4o b. 32.1o c. 28.8o d. 20.2o

4. A 20 µF capacitor and 200W, 220V lamp is connected in series across a 220V alternating supply. In which frequency of the supply the lamp will glow dimmest?

a. 1000Hz b. 1Hz c. 10Hz d. 100 Hz

5. If a resistor and an inductor are connected in series across a voltage source. Which two parameters in that circuit increase if frequency of voltage source increases ?

a. VL&Z b. Z&I c. VL&I d. VL & VR

176

6. The maximum and minimum values of power factor in an ac circuit are

a. 1&0 b. ∞&0 c. ∞&1 d. 1 & 0.1

7. If a resistor and a capacitor are connected to form series R – C circuit across a voltage source. If frequency of voltage source increases

a .The current increases b. The current decreases c. The current remain unaltered d. The current decreases abruptly

177

LECTURE 18

PARALLEL AC CIRCUITS

OBJECTIVES

• To understand and compute the impedance/admittance, of the parallel and series-parallel

circuits, fed from single phase ac supply.

• To learn computation of currents and voltage drops different components of the parallel

a.c.circuits , both in magnitude and phase.

• To represent the parallel circuits with the help of complete phasor diagram, showing the

currents and voltage drops.

• To compute the total power and also power consumed in the different components, along

with the power factor.

18.1 Parallel RLC circuit

Consider the circuit with all three elements, R, L & C connected in parallel. The a.c. voltage

source is V(t) = Vm sin ωt = Vsin ωt.

Figure 18.1: Circuit diagram.

Unlike the series RLC circuit, the instantaneous voltages across all three circuit elements R, L,

and C are the same, and each voltage is in phase with the current through the resistor. However,

the currents through each element will be different.

The current from the supply can be computed by various methods, of which two are described

here.

First method :

178

The current in three branches are first computed and the total current drawn from the supply is

the phasor sum of all three branch currents, by using Kirchoff’s first law related to the currents at

the node. The voltage phasor (V) is taken as reference.

All currents, i.e. three branch currents and total current, in steady state, are sinusoidal in nature,

as the input supply voltage is sinusoidal of the form,

Three branch currents are obtained by the procedure given below in brief.

where,

similarly

So

Total supply current(instantaneous value) is

The two equations given here are obtained by expanding the trigonometric form appearing in the

last term on RHS, into components of cosωt and sinωt, and then equating the components of

cosωt and sinωt from the last term and last but one (previous).

179

_ _ _ _

I = IR + IL+ IC

From these equations, the magnitude and phase angle of the total (supply) current are,

Admittance (Y) of the circuit which is reciprocal of impedance, is a complex quantity. The angle

of admittance, is same as the phase angle, φ of the current I, with the supply voltage taken as

reference phasor.

The steps required to find the r.m.s. values of three branch currents and the total current, using

complex form of impedance, are given here.

Three branch currents are

Of the three branches, the first one consists of resistance only, the current, is in phase with the

voltage (V). In the second branch, the current IL, lags the voltage by 90o, as there is inductance

only, while in the third one having capacitance only, the current IC, leads the voltage by 90o . The

total current is

180

The relationship between Z , R , XL

and XC obtained from above equation is shown in

Admittance triangle (Figure18.2).

1/Z = Y (admittance)

1/R = G (conductance)

1/X = B (susceptance)

Figure 18. 2: Admittance triangle.

(A) Inductive circuit

Figure 18.3: Phasor diagram for Inductive circuit

181

In this case, the circuit being inductive, the current lags the voltage by φ (positive) as IL > IC i.e.

.This condition is in contrast to that derived in the case of

series circuit earlier. The power factor is less than 1 (one). The complete phasor diagram, with

the three branch currents along with total current, and also the voltage, is shown in Figure 18.3.

The voltage phasor is taken as reference in all cases. It may be observed there that

_ _ _ _

IR + IL + IC = I

The Kirchoff’s First law related to the currents at the node is applied, as stated above. The

expression for the average power is

The power is only consumed in the resistance, R, but not in inductance/capacitance (L/C), in all

three cases.

(B) Capacitive Circuit

Figure 18.4 : Phasor diagram for capacitive circuit

The circuit is capacitive, as

The current leads the voltage by φ (φ being negative), with the power factor less than 1 (one).

The complete phasor diagram, with the three branch currents along with total current, and also

the voltage, is shown in Figure 18. 4.

182

(C) Resistive Circuit

The third case is resistive as . It may be

noted from the phsor diagram (Figure 18.5) that two currents IL and IC are equal in magnitude as

shown, but opposite in sign (phase difference being 180O), and the sum of these currents IL+IC is

zero (0). The total current is in phase with the voltage (φ =0O) with , the power factor

being unity. This condition can be termed as resonance in the parallel circuit. The magnitude of

the impedance in the circuit is maximum (i.e. the magnitude of the admittance is minimum)

under this condition, with the magnitude of the total (supply) current being minimum.

Figure 18. 5: Phasor diagram for Resistive circuit

Second method

Before going into the details of this method, the term, admittance must be explained. In the case

of two resistances connected in series, the equivalent resistance is the sum of two resistances, the

resistance being scalar (positive). If two impedances are connected in series, the equivalent

impedance is the phasor sum of two impedances, all impedances being complex. Please note that

the two terms, real and imaginary, of two impedances and also of the equivalent impedance, may

be positive or negative. If two resistances are connected in parallel, the inverse of the equivalent

resistance is the sum of the inverse of the two resistances. If two impedances are connected in

parallel, the inverse of the equivalent impedance is the sum of the inverse of the two impedances.

The inverse or reciprocal of the impedance is termed ‘Admittance’, which is complex.

Mathematically, this is expressed as

= = + = +

As admittance (Y) is complex, its real and imaginary parts are called conductance (G) and

susceptance (B) respectively. So,

= G + jB

183

Unit of all the three quantities is ‘Siemens’.

If impedance, with X being positive, then the admittance is

where

Note the way in which the result of the division of two complex quantities is obtained. Both the

numerator and the denominator are multiplied by the complex conjugate of the denominator, so

as to make the denominator a real quantity.

The magnitude and phase angle of Z and Y are

To obtain the current in the circuit (Figure 18.1), the steps are given here.

The admittances of the three branches are

The total admittance, obtained by the phasor sum of the three branch admittances, is

where,

184

The total impedance of the circuit is

The total current in the circuit is obtained as

where the magnitude of current is I=V.Y=V/Z

As in a series RLC circuit, power is dissipated only through resistor. The average power is

P = VIR = (IR)2R = (V)2/R = VI (active power W)

Q = V(IL-IC) = (IL)2XL - (IC)2XC = VI (reactive power VAR)

S =VI = I2Z = V2Y (apparent power VA)

18.2 Solved Examples

Q1. Find the total admittance and power factor for the given circuit in figure 18.6

Solution:

For branch (1), = R1 + j XL1

Admittance of branch (1) is,

185

= =

=

= -

= G1 + jBL1 where

G1 = = and BL1 = Similarly the

admittance of branch (2) is ,

= =

= G2 + jBC2

Where G2 = and BC2 = =

Now total admittance of two parallel branches is,

= +

= (G1 - jBL1) (G2 + j BC2 )

= (G1 +G2) – j (BL1 - BC2)

Total conductance G = G1 +G2

Total susceptance B = (BL1 - BC1 )

Power factor of the circuit = =

Q2. The circuit, having two impedances of Z1=(8+j15) Ω and of Z2=(6-j8) Ω in parallel, is

connected to a single phase ac supply (Fig.16.7), and the current drawn is 10 A. Find each

branch current, both in magnitude and phase, and also the supply voltage.

186

Solution:

One method to solve this circuit is to find branch admittances and add them to get total

admittance. The current in each branch is calculated from the input voltage and the branch

impedance.

187

The phasor diagram with current as reference is shown below

Note that the names of the phasors in the diagram are mentioned with each quantity in the above

solution.

188

Alternative method:

(In this method total impedance is calculated from which total using current division formula)


1. An a.c. circuit with impedance Z = (R+jX) has conductance = _________and

susceptance =________.

a. R/(R2+X2), X/(R2+X2)

b. X/(R2+X2), R/(R2+X2)

c. R/(R2+X2), R/(R2+X2)

d. X/(R2+X2), X/(R2+X2)

2. Unit of admittance is _________.

a. Seimens b. ohms c. ohms/m d. seimens/m

3. If an RLC parallel circuit has 0.8 leading power factor that means____

a. IC > IL b. IC < IL c. IC = IL

189

4. A circuit having total admittance of (12.5- j9.1) siemens draws the current that ________

the supply voltage in phase.

a. Leads b. lags c. in phase

5. A circuit consisting of 15 Ω resistance, 12 Ω inductive reactance and 8 Ω capacitive

reactance, all connected in parallel to each other, has total impedance of

a. 1.552 Ω b. 12.7 Ω c. 15.52 Ω d. 127 Ω

6. Power factor of the circuit in Q. No. 5. is__________.

a. 0.848 in phase b. 0.848 leading c. 0848 lagging

7. An impedance of (8+j6) Ω is connected in parallel with a resistance across 100V a.c.

supply. Both the branches carry currents of equal magnitudes. The power consumption of

the circuit is_______

a. 1800 W b. 1600 W c. 2000 W d. 1200 W

190

LECTURE 19

AC PARALLEL CIRCUITS : NUMERICAL PROBLEMS

OBJECTIVES

• To practice solving different types of series-parallel a.c. circuits to find impedance/admittance, currents and voltage drops across the components, both in magnitude and phase, of the circuit.

• To learn to draw complete phasor diagram of a.c. circuits

• To understand the computation of real, reactive and apparent power and power factor of a.c. circuit.

19.1 Solved problems

Q1. An inductive load (R in series with L) is connected in parallel with a capacitance C of 12.5

µF .The input voltage to the circuit is 100 V at 31.8 Hz. The phase angle between the two

branch currents, (I1=IL) and (I2=IC) is 120o , and the current in the first branch is 0.5A .

Find the total current, and also the values of R & L.

191

Q2. Find the input voltage at 50 Hz to be applied to the circuit shown, such that the current in

the capacitor is 8 A.

192

Phasor diagram with I2 as reference is shown

193

Q3. A resistor of 50Ω in parallel with an inductor of 30 mH, is connected in series with a capacitor, C . A voltage of 220 V, 50 Hz is applied to the circuit. Find,

(a) the value of C to give unity power factor, (b) the total current, and (c) the current in the inductor

194

Note that the currents and voltages can be drawn with different scales in a single phasor diagram.

Q4. The power consumed in the inductive load is 2.5 kW at 0.71 lagging power factor. The

input voltage is 230 V, 50 Hz. Find the value of the capacitor C, such that the resultant

power factor of the input current is 0.866 lagging.

195

The phasor diagram with input voltage as reference is shown

196

Q5. In the circuit, the wattmeter reads 960 W and the ammeter reads 6 A. Calculate the values VS, VC, IC, IL, and XC.

197


1. In a circuit formed by two parallel branches, if active and reactive power consumption of

first branch is 500W and -200 VAR respectively and that of the whole circuit is 900W

and -100VAR respectively then the power factor of the second branch is __________.

a. 0.97 lagging b. 0.97 in phase c. 0.97 leading

2. An inductive load with 0.707 power factor draws 1kW power. The power factor of the

ckt. can be improved by connecting a _____________ across the load.

a. capacitor b. inductor c. resistor

3. In Q.no.2, the useful power consumption of the modified ckt. will __________.

a. increase b. decrease

c. remain unchanged d. either of (a) or (b) is possible

4. In Q. no.2,The current drawn in the modified ckt. in will __________.

a. increase b. decrease

c.remain unchanged d. either of (a) or (b) is possible

5. If the modified power factor of the ckt. in Q.2 is unity the relation between XL and XC of

the two branches should be _________.

a. XC = XL b. XC = 2XL c. XC = 4XL d. XL = XC

198

LECTURE 20

RESONANCE IN SERIES CIRCUIT OBJECTIVES

• To understand the concept of electrical resonance, bandwidth and quality factor.

• To discuss the effect of variable frequency on different parameters of series R-L-C circuit

and to derive resonance condition for the same.

• To compute the bandwidth and half power frequency, including power and power factor

under resonance condition.

• To understand the applications of resonance.

20.1 Introduction

Resonance occurs in electrical circuits as well, where it is used to select or ‘tune’ to specific

frequencies. Example the tuning circuit in radio receivers where resonance phenomenon is used

to select a particular radio station.

20.2 Series Resonance

Figure 20.1 Series RLC circuit

The circuit, with resistance R, inductance L, and a capacitor, C in series (Figure 20.1) is

connected to a single phase variable frequency (f) supply.

The total impedance of the circuit is Z = R+j (XL-XC)

The magnitude Z and angle Φ is

199

The inductive reactance is directly proportional to frequency (XL = ωL), capacitive reactance is

inversely proportional to frequency (Xc = 1/ωC) whereas resistance R is constant at all

frequencies. The variation of R, XL , XC and Z is shown in the diagram below. As the frequency

changes the impedance also varies and Z is minimum ( Z = R)at resonance frequency .

The the amplitude of the current I0

=V/Z reaches a maximum when Z is at a minimum. This

occurs when XL

=XC

, or ωL =1/ωC , leading to

The resonant frequency ( in Hz) is

The phenomenon at which I0

reaches a maximum is called a resonance, and the frequency ω0

is

called the resonant frequency. At resonance, XL

=XC

and the impedance becomes Z =R , the

amplitude of the current is

\

Figure 20.2: Amplitude of current as a function of frequency in RLC series circuit

200

At resonance RLC circuit behaves like a pure resistive circuit and the power factor is unity (cos

Φ=1).The magnitude of the voltage drop in the inductance L and capacitance C are equal, as the

reactance are equal (VL=VC) but this voltage is much greater than supply voltage and this voltage

magnification ratio is called the Quality factor of the coil.

The magnification of the voltage drop as a ratio of the input (supply) voltage is

Figure 20.3: Variation of current under variable frequency supply

The current is maximum at resonance as impedance is minimum (Z=R) at f = f0 .

The maximum value of the current is (V/R). If the magnitude of the current is reduced to (1/√2 )

of its maximum value, the power consumed in R will be half of that with the maximum current,

as power is I2 R. So, these points are termed as half power points f1 and f2.

Cut off frequency:

The frequency at which the power in the circuit is half the maximum power is called cut off

frequency.

Bandwidth:

The bandwidth of a resonant circuit is the range of frequencies for which there is more than half

the maximum power in the circuit. Therefore bandwidth is the difference between higher and

lower cut off frequencies.

201

Bandwidth indicates the selectivity of resonant circuit.

Bandwidth (BW) (rad/sec)

= ………rad/s

= ………..Hz

We can derive the equation for ω1 and ω2 as

Lower cut off frequency ω =

Higher cut off frequency ω =

Quality Factor:-

Quality factor ( Q factor) is the voltage magnification factor at resonance.

Q factor =

=

=

We define Q factor of the circuit as

The Quality factor of a coil can also be calculated using resonant frequency fo and bandwidth

( )

Q factor = f0 /

1 1o

o

w L LQ

R w R C R C

= = =

202

20.3 Solved Problems:

Q1. A sinusoidal voltage V(t)=200sin ωt is applied to a series RLC circuit with R = 20 Ω ,

L=100 mH, and C =10 nF. Find the following quantities:

(a) the resonant frequency,

(b) the amplitude of the current at resonance,

(c) the quality factor Q of the circuit, and

(d) the amplitude of the voltage across the inductor at the resonant frequency.

Solution:

(a) The resonant frequency for the circuit is given by

(b) At resonance, the current is

(c) The quality factor Q of the circuit is given by

(d) At resonance, the amplitude of the voltage across the inductor is

Q2. A constant voltage of frequency, 1 MHz is applied to a lossy inductor (r in series with L), in

series with a variable capacitor, C . The current drawn is maximum, when C = 400 pF; while

current is reduced to (1/√2) of the above value, when C = 450 pF. Find the values of r and L.

Calculate also the quality factor of the coil, and the bandwidth.

203

Solution

Given f = 1 MHz = 10 6Hz ω=2πf C=400 pF=400*10-12


1. A series RLC circuit at resonance will behave like

a. pure resistive circuit b. inductive circuit c. capacitive circuit

204

2. A series RLC circuit below resonance frequency (when f < f 0 ) is_________in nature.

a. inductive b. capacitive c. resistive

3. In RLC series circuit current at resonance is

a. maximum b. minimum c. zero

4. In RLC series circuit impedence at resonance is


5. In RLC series circuit at resonance it magnifies _________ .

a. voltage b. current

c. both voltage and current d. none of the above

6. Resonant frequency is independent of

a. L value b. C value

c. R value d. all the above three

205

LECTURE 21

RESONANCE IN PARALLEL CIRCUITS

OBJECTIVES

• To discuss the effect of variable frequency on different parameters of parallel R-L-C circuit and to derive resonance condition for the same.To compute the impedance and quality factor under resonance condition.

• To compare series and parallel resonance.

21.1 Parallel Resonance

Figure 21.1(a): Parallel RLC circuit

Resonance is the condition when the current drawn by the circuit is in phase with the applied

voltage.

In the given circuit Figure 21.1(a), an inductive branch (inductance L and resistance R) is in

parallel with capacitive branch with negligible resistance. Inductive branch draws current that

lags the voltage where as capacitive branch current leads the voltage.

Figure 21.1(b): Phasor diagram at resonance in parallel RLC circuit

206

The frequency at which the reactive components of two branch currents are equal and balance

each other, the current drawn from the source is only the real component which is in phase with

voltage. This is nothing but resonance. At resonance power factor is unity. As seen from the

phasor diagram Figure 21.1(b)

_

IL = ILcos …………… current in branch 1

_ IC = 0 + jIC …………………………………… current in branch 2

At resonance the reactive component of two branch current becomes equal.

ILsin = IC

= =

ZL2 = XLXC …………………………………….(21.1)

(ω0L) ( ) =

=

0 = ……………resonance frequency in rad/s

f0 = ………….. resonance frequency in hertz.

However if R of the inductive branch is negligible, then can be neglected.

f0 = (same as series resonance)

Current magnification: At resonance the current in the inductive and capacitive branches may

be many times greater than the current drawn from the source. The current magnification in the

parallel resonance circuit is,

207

Current magnification =

( ILcos is the resultant from IL & IC which is supplied by the source whereas IC = IL sin current

that is reactive component which circulates in the R-L-C loop.)

Current magnification = tan

= = = Q factor

Impedance in parallel resonance ckt: The resultant current drawn from source is

I = ILcos

= ILcos

= =

Now at resonance ZL2 = XLXC = from equation (21.1)

I =

is the impedance of the parallel resonant circuit which is purely resistive. It is termed as

dynamic impedance.

Dynamic impedance =

Figure 21.2: Variation of impedance and current with frequency

208

Figure 21.3: Parallel RLC circuit

A parallel circuit containing a resistance, R, an inductance, L and a capacitance, C will produce a parallel resonance circuit when the resultant current through the parallel combination is in phase with the supply voltage. At resonance there will be a large circulating current between the inductor and the capacitor due to the energy of the oscillations.

A parallel resonant circuit stores the circuit energy in the magnetic field of the inductor and the electric field of the capacitor. This energy is constantly being transferred back and forth between the inductor and the capacitor which results in zero current and energy being drawn from the supply. This is because the corresponding instantaneous values of IL and IC will always be equal and opposite and therefore the current drawn from the supply is the vector addition of these two currents and the current flowing in IR.

Series circuit Parallel circuit

Impedance at resonance Minimum (Zr = R) Maximum(Zr = L/CR)

Current at resonance Maximum(Ir=V/R) Minimum(Ir=V/Zr)

Resonant frequency f0 =

Hz

f0 =

Q-factor (VL or Vc )/ V IL or Ic )/ I

magnifies voltage Current

When f <f 0 Circuit is capacitive Circuit is inductive

Table 21.1: Comparion between series and parallel resonance

209

21.2 Summary

In an AC circuit with a sinusoidal voltage source V(t) =V0

sin ωt, the current is given by

I(t) =I0

sin(ωt-Φ) ,

where I0

is the amplitude and φ is the phase angle.

• For simple circuit with only one element (a resistor, a capacitor or an inductor) connected to the voltage source, the results are as follows:

Table 21.2

• For circuits which have more than one circuit element connected in series, the results are

Table 21.3

• The phase angle between the voltage and the current in an AC circuit is

210

• In the parallel RLC circuit, the impedance is given by

• The resonant frequency ω0

is

At resonance, the current in the series RLC circuit reaches the maximum, but the current in the

parallel RLC circuit is at a minimum.


1. A parallel ac circuit in resonance will

a. have a high voltage developed across each capacitive and inductive section

b. have a high impedance

c. current in each branch is equal to line current

2. At _________frequency a parallel RL circuit behaves as purely resistive.

a. low b. very low c. high d. very high

3. In an a.c. circuit power is dissipated only in

a. resistance b. inductance

c. capacitive only d. none of the above

4. A parallel RLC circuit below resonance frequency (when f < f 0 ) is

a. capacitive nature b. inductive nature c. resistive

5. In RLC parallel circuit current at resonance is


211

EXERCISES

1. An alternating current of frequency 60 Hz has a maximum value of 12A. (i) Write the

equation for the instantaneous value. (ii) Find the value of the current after 1/360 sec (iii)

find the time taken to reach 9.6A for the first time.

[Ans: (i) i=12 sin 377t (ii) 10.39A (iii) 2.46ms]

2. An alternating voltage is given by v=10sin (942t) V. Determine the time taken from t=0

for the voltage to reach +6V for the first and the second time.

[Ans: 0.683ms, 2.35ms]

3. An ac voltage is given by v=100 sin (1256t) V. Determine a)the frequency b)the time

period c) the time taken from t=0 for the voltage to reach a value of 60V for the first and

the second time and d) the energy dissipated when this current flows through a 25Ω

resistor for 30 min.

[ Ans: a) 200Hz b) 0.005s c) 0.512ms, 5.512ms d) 36KW]

4. 4. An alternating voltage is represented by v=141.4 sin (377t) V. Find its maximum

value, frequency, time period, instantaneous value at t=3ms.

[ Ans: 141.4V, 60 Hz, 16.67ms, 127.8V]

5. An alternating current of frequency 40 Hz takes 3.375ms to reach 15A for the first time.

Find its maximum value. [ Ans :20A ]

6. Find the rms and average values of the waveform given below.

[ Ans: I rms=8.16A, I avg==5A ]


[ Ans: 10.39V, 9V ]

T sec 8 6 4 0

20

A

i

2

3 T sec 19 6 0

18

v

212


[ Ans: 0.687Vm, 0.621Vm ]

9. Find average value of the waveform given below.

[ Ans: 0.5Vm ]

10. Find rms value of the waveform given below

[ Ans: 0.476Vm ]

11. Find rms value of the waveform given below. [ Ans: 0.697Vm ]

π π/2

0.866Vm

Vm

π/3

v

Ѳ

T/ T sec T T/2 0

V

m

v

-

V

2π

Vm

π π/4

Ѳ

Vm

-

2

Vm

π

0.5Vm

213

12. Find the rms and average values of the waveform given below

[ Ans: 0.584 Vm, 0.54 Vm]

13. Calculate the rms value, the form and the peak factors of a periodic voltage having the

following values for equal time intervals changing suddenly from one value to the other:

0,5,10,20,40,50,40,20,10,5,0,-5,………What would be the rms value of sine wave

having the same peak value?

[ Ans: V rms=25.98V, FF=1.3, PF= 1.924; 35.35V ]

14. Find vs if v1=10.2 sin (754t+30 o), v2=14.9 sin (754 t-10 o), v3=16.1 sin (754t-25 o)

[ Ans: vs = 8.78 sin (754t+5.77 o) V ]

15. Find i1 if

i2= 14.6 sin (377t-15 o) mA, i3= 21.3 sin (377t +30 o) mA and i4=13.7 cos (377t +15 o) mA.

[ Ans: 27.64 sin (377t-91.67 o) mA

Ѳ 2π π

0.707Vm

Vm

V2

-

+ -

+

vs

V1

- +

V3 - +

i3

i4

i2

i1

214

16. Two currents i1=15 sin (wt+π/3) A and i2=25 sin (wt +π/4) A are fed to a common

conductor of resistance 10 Ω. Find the total current. What will be the total energy loss in

the conductor for 24h.

[ Ans: 39.8 sin (wt+50.61 o) A. E=188.96KWh ]

17. Given v1=10.2 sin 440t , v2=10√2 sin (440 t-45 o), v3=20 cos 440t. Find an expression for

the resultant voltage, its rms value and its frequency.

[ Ans: v=22.36 sin (440t+26.56o)V , 15.81V, f=70.028 Hz ] 18. Four wires p, q, r, s are connected to a common point. The currents in them respectively

are 6 sin (wt+ π /3), 5 cos (wt+ π /3), 3 cos (wt + π /3) and i. Find i.

[ Ans: 10 sin (wt-66.77)] 19. A 50 Hz alternating voltage of 150V is applied independently to a) resistance of 10 Ω b)

inductance of 0.2H c) capacitor of 50 µF. Find the expression for the instantaneous

current in each case

[ Ans: a) i = 21.21 sin (314.16t) A b) i= 3.376 sin (314t- 90 o) A c) i= 3.33 sin (314t+ 90 o) A ]

20. A 10 mH inductor has a current of i=5 cos 2000t A. Obtain the voltage VL across it.

[ Ans: 100/√2 V ] 21. A voltage of 150V 50 Hz is applied to a coil of negligible resistance and inductance of

0.2 H.Write time equations for voltage and current.

[ Ans: v = 212.3 sin 314t, i=3.38 sin (314t-90 o) A ] 22. In the fig below, v=100 sin (5000t + π/4). Calculate the branch currents and the total

current.

23.

[ Ans: iR = 4 sin (5000t + π/4).

i C i L i R

i T

V ~ 25 Ω 2m H 30 µF

215

iL = -10 cos (5000t + π/4).

iC =15 cos (5000t + π/4). iT = iR + iL + iC . ]

24. Find the current flowing through a purely inductive circuit containing a voltage source

v=325 sin (100 π t) and an inductance of 2H. Ans; I=0.366A]

25. Find the current flowing through a purely capacitive circuit containing a voltage source

of 230V, 50 Hz and a capacitor of 1000 µF. [ Ans: 72.25A]

26. A coil having a resistance of 7 Ω and inductance of 3.18mH is connected to a 230V 50

Hz supply. Calculate current, phase angle, pf and power consumed.

[ Ans: °−∠ 5584.18 , 55 , 0.573 lag. 2.482 KW ] 27. A voltage of V= (150 + j180 ) V is applied across an impedance Z and the resultant

current I= ( 5-j4) A. Determine Z, R , X power consumed. Is the circuit inductive or

capacitive.

[ Ans: 36.61 Ω, 0.73 Ω, 36.6 Ω, 30.096W, inductive ] 28. The voltage applied to a series circuit consisting of two pure elements is given by v=180

sin wt. and the resultant current i= 2.5 sin ( wt-45o) . Find the average power and the

values of the elements in the circuit. [ Ans: 159.3W, R=50.85 Ω, XL= 50.85 Ω ]

29. A current of 5A flows through a non inductive resistance, in series with a choking coil

supplied at 250V, 50 Hz. If the voltage across the resistance is 125 V and across the coil

is 200 V, calculate a) impedance, reactance and resistance of the coil b) Power absorbed

by the coil c) pf

[ Ans: a) 40 Ω, 5.5 Ω, 39.62 Ω b) 137.5W c) 0.137 lag ] 30. When a resistor R and a coil in series are connected to a 240V supply, current of 3A

flows lagging the supply voltage by 37o. Voltage across coil is 171 V. Find calculate a)

impedance, reactance and resistance of the coil and R.

[ Ans : 57 Ω, 48.14 Ω, , 30.52 Ω ,R=33.37 Ω ] 31. A coil takes 2A at pf of 0.8 lag with 10V applied across it. A second coil B takes 2A at

0.7 pf lag with 5V applied across it. What voltage will be required to produce a total

current of 2A with both coils A and B connected in series. Find the overall pf in this case.

[ Ans: °∠ 79.3996.114 V, 0.768 lag ] 32. In an RL series circuit, i(t) = 5 sin (314t + 2π/3), v(t)= 15 sin (314t + 5π/6). Find Z, R, L

Pavg, pf of the circuit. [ Ans: 3 Ω, 2.6 Ω, 4.77mH, 32.468W, 0.866 lag ]

216

33. Two coils A and B are connected in series across a 240V, 50 Hz supply. RA=5 Ω and

LB=0.015H. If the input from the supply is 3KW and 2KVAR, sind LA and RB. Calculate

voltage across each coil.

[ Ans: LA = 0.0132 H, RB = 8.297 Ω, VA= 97.65V, VB= 143.3V ]

34. An emf of 50V, 50 Hz is applied to an impedance Z1= (12.5 + j21) Ω. An impedance Z2

is added in series with Z1, then the current becomes half of the original and leads it by

14.2o. Find Z2. [ Ans: (22.04+ j 13.59) Ω ]

35. Find r and L in the circuit below. [ Ans: 6 Ω, 0.0219 H ]

36. The potential difference measured across a coil is 4.5 V when it carries a direct current of

9A. The same coil when carries an alternating current of 9A at 25 Hz, the potential

difference is 24V. Find the current, power, pf when supplied by 50V, 50 Hz supply.

[ Ans: 9.505A, 45.15W0.095 lag] 37. A 120V, 100W lamp is to be connected to a 220V, 50 Hz ac supply. What value of pure

inductance should be connected in series in order to run the lamp at rated voltage?

[ Ans:0.705H]

38. Two coils having resistance of 25Ω and 15Ω and inductances of 0.04H and 0.2H

respectively are energized by a 230V, 50 Hz supply. Calculate the power dissipation in

each coil and the pf of the whole circuit. [ Ans; 181.57W, 108.95W, 0.4687 lag]

39. When a coil is connected to a 114V 60 Hz supply, current flowing is 3A. The current

rises to 4A when 116V, 25hz supply is connected to the same coil. Find the resistance

and inductance of the coil. [ Ans: 26.71Ω , 71.65mH]

40. Find R2 if the overall pf is 0.92 lag. [ Ans: 1.35 Ω ]

10 Ω

20.4V 20V

L r

R2

j2 Ω 1Ω

217

41. A resistor and a capacitor are connected in series to a 120V, 60 Hz supply. The

impedance of the circuit is 86Ω power consumed is also 86W. Find R and C.

[Ans: 44.17 Ω, 35.9 µF]

42. A 200 Ω resistor is in series with a capacitor across 240V, 60Hz supply. The power consumed is 100W. Find C and I. [ Ans: 9.67 µF, 0.707A]

43. A capacitor of 35 µF is connected in series with a variable resistor across 50 Hz supply. Find the value of the resistor for a condition when the voltage across the capacitor is half the supply voltage.

[Ans: 157.5Ω ] 44. A resistance and a capacitance connected in series across a 250V supply draw 5A at 50

Hz. When frequency is increased to 60 Hz, it draws 5.8A. Find R, C and the power in the second case.

[ Ans: 19.94 Ω, 69.42 µF, 670.78W] 45. A 240V, 50KVA generator operates at a leading pf of 0.8. Determine the equivalent

series impedance of the load. [ Ans: 0.922-j0.691] 46. The voltage applied to a series circuit consisting of two pure elements is given by v=200

sin wt and the resultant current i= 25 sin ( wt+15o) . Find the average power and the

values of the elements in the circuit. [ Ans: 2.414K3W, R=7.727 Ω, XC= 2.07 Ω ]

47. A voltage V = (150 + j180)V is applied across an impedance and the current is found to be I = (5 – j4)A. Determine (i) scalar impedance, (ii) reactance and (iii) power consumed.

48. A current of 5A flows through a non - inductive resistance in series with a choking coil supplied at 250V, 50c/sec. If the voltage across the resistance is 125V and across the coil 200V, calculate (i) Impedance, reactance and resistance of the coil. (ii) Power absorbed by the coil (iii) Power factor of the coil Also draw phasor diagram.

49. A capacitor of 35 µF is connected series with a variable resistor. The circuit is connected

to across 50Hz mains. Find the value of resistor for a condition when the voltage across

the capacitor is half the supply voltage. [Ans. R = 157.5 ohms]

50. An R-L-C series circuit has a current that lags behind the applied voltage by 45. The voltage across the inductance has a maximum value equal to twice the maximum value of voltage across the capacitor. Voltage across the inductance is 300 sin(1000t) and R = 20ohms. Find the values of inductance and capacitance. [L = 0.04 H, C = 50µF]

51. Find the impedance, Zab

in the following circuits

218

52. Calculate the current and power factor (lagging / leading) for the following circuits, fed

from an ac supply of 200 V. Also draw the phasor diagram in all cases.

53. A voltage of 200 V is applied to a pure resistor R, a pure capacitor C and a lossy

inductor coil, all of them connected in parallel. The total current is 2.4 A, while the

component currents are 1.5, 2.0 and 1.2 A respectively. Find the total power factor and

power factor of the coil. Also find the total power consumption of the circuit. Draw the

phasor diagram.

54. A 200 V, 50Hz supply is connected to a lamp having a rating of 100 V, 200 W in series

with a pure inductance L, such that the total power consumed remains the same( i.e.

200W). Find the value of L. A capacitance C is now connected across the supply. Find

219

value of C, to bring the supply power factor to unity. Draw the phasor diagram in the

second case.

55. . In a series-parallel circuit the two parallel branches A and B, are in series with the

branch C. The impedances in Ω are, ZA

= 5+j6, ZB

= 6-j8, and ZC

= 10+j8. The voltage

across the branch C is (150+j0) V. Find the branch currents, IA

and IB, and the phase

angle between them. Find also the input voltage. Draw the phasor diagram.

56. A total current of 1A is drawn by the circuit fed from an ac voltage, V of 50 Hz. Find the

input voltage. Draw the phasor diagram.

57. A coil having a resistance of 20 Ω and inductance of 20 mH, in series with a capacitor is

fed from a constant voltage variable frequency supply. The maximum current is 10 A at

100 Hz. Find the two cut-off frequencies, when the current is 0.71 A.

58. With the ac voltage source in the circuit shown operating a frequency of f, it was found

that I =1.0 ∠0° A. When the source frequency was doubled (2f), the current became I =

0.707 ∠ – 45° A. Find: a) The frequency f, and

b) The inductance L, and also the reactances, XL

and XC

at 2f

59. For the circuit shown , a) Find the resonant frequency f0, if R = 250 Ω, and also calculate

Q0

(quality factor), BW (band width) in Hz, and lower and upper cut-off frequencies (f1

220

and f2) of the circuit. b) Suppose it was desired to increase the selectivity, so that

bandwidth was 65 Hz. What value of R would accomplish this?

60. The circuit components of a parallel circuit shown in Fig. 17.10 are R = 60 kΩ, L = 5mH, and C = 50 pF. Find a) the resonant frequency, f

0, b) the quality factor, Q

0, and c) the

bandwidth.

61. A parallel circuit consist of a 2.5uF capacitor and a coil whose resistance and inductance

are 15Ω and 260 mH. Determine (i) resonant frequency (ii)Q factor (iii)dynamic

impedance.

[Ans: f0 = 197.19Hz, Q factor= 21.48 , dynamic impedance= 6933.33Ω]

62. A coil takes a current of 1A at 0.3 p.f when connected to a 100v,50Hz supply. Determine

the value of capacitance which when connected in parallel with the coil, will reduce the

line current to a minimum. Calculate the impedance of parallel circuit at 50Hz.

[Ans: C=30.37µF , Zr=333.33Ω]

221

UNIVERSITY QUESTIONS

1. Two coils A and B are connected in series across 240V, 50Hz supply. The resistance of A is

5 Ω and inductance of B is 0.015H.If the input from the supply is 3kW and 2kVAR.Find

inductance of A and resistance of B. Calculate voltage across each coil.

2. A 46mH inductive coil has a resistance of 10 Ω.(i)How much current will it draw if

connected across a 100V,60Hz supply.(ii)What is the power factor of the coil.(iii)Determine

the value of the capacitance that must be connected across the coil to make the power factor

of overall circuit unity.

3. An inductor having a resistance of 25 Ω and Q-factor of 10 at resonant frequency of 10 kHz

is fed from 100 ∠0 supply. Calculate (i) Value of series capacitance required to produce

resonance with the coil.(ii) The inductance of the coil (iii) Q-factor (iv) Voltage across

capacitor (v) Voltage across coil.

4. Two circuits have the same value of numerical impedance. The p.f of one is 0.8 and the

other is 0.6.What is the p.f of combination if they are connected in parallel.

5. A current of 5A flows through a pure resistance in series with a coil when supplied at

250V,50Hz.If the voltage across the resistance is 125V and across the coil is 200V calculate

(i)impedance, resistance and reactance of the coil(ii)Power absorbed by the coil (iii)Draw

the phasor diagram.

6. A resistor and a capacitor are in series with a variable inductor. When the circuit is

connected to a 200V,50Hz supply, the maximum current obtainable by varying the

inductance is 0.314.The voltage across the capacitor is then 300 V. Find the circuit

constants.

7. A circuit has L=0.2H and inductive resistance of 20 Ω is connected in parallel with 200uF

capacitor with variable frequency,230V supply. Find the resonant frequency at which the

total current taken from the supply is in phase supply voltage. Also find the value of this

current. Draw the phasor diagram.

8. A coil of 0.6 p.f is in series with a 100uF capacitor and is connected to a 50Hz supply.The

potential difference across the coil is equal to the potential difference across the capacitor.

Find the inductance and resistance of the coil.

9. A 33 Ω esistance,10mH inductor and 0.1uF capacitor are connected in series and excited by

a supply voltage Vs=13.2+j0 V.Find

(i)The resonant frequency in Hz

(ii)Q-factor

222

(iii)Polar forms of current and voltage across R,L and C at resonance.

(iv)Draw a phasor diagram showing all quantities in(iii)

10. Three impedances are connected in parallel across a voltage source Vs=60sin(wt).

Z1=50 Ω, Z2=j40 Ω , Z3=-j80 Ω. Find currents in each impedance ,show that total

current=voltage * total admittance. Also draw the phasor diagram showing VS,total current

IT and branch currents.

ANSWER KEY

Lec

11

Lec

12

Lec

13

Lec

14

Lec

15

Lec

16

Lec

17

Lec

18

Lec

19

Lec

20

Lec

21

Q1. d c a b b b a a c a b

Q2. b b b b b a a a a b d

Q3. b b a c b c b a c a a

Q4. b a a a b b b b b b b

Q5. c b a c a a a b c a b

Q6. c c b b a -- d a -- c --

Q7. d a b b d -- c -- -- -- --

Q8. b c b a c -- -- -- -- -- --

Q9. a d a a c -- -- -- -- -- --

Q10. -- b c d c -- -- -- -- -- --

223

MODULE 3

THREE PHASE CIRCUITS

224

LECTURE 22

FUNDAMENTALS OF THREE PHASE CIRCUITS

OBJECTIVES

• To introduce the concepts and principles of the three-phase electrical supply and the

corresponding circuits

• To describe the generation of the three-phase supply and distinguish between star (3 and

4-wire) and delta connections.

22.1 Generation of three phase supply

In order to understand the reasons for, and the method of generating a three-phase supply, let us

firstly consider the generation of a single phase supply. Alternating voltage is provided by an a.c.

generator, more commonly called an alternator. It was shown that when a coil of wire, wound on

to a rectangular former, is rotated in a magnetic field, an alternating (sinusoidal) voltage is

induced into the coil. You should also be aware that for electromagnetic induction to take place,

it is the relative movement between conductor and magnetic flux that matters. Thus, it matters

not whether the field is static and the conductor moves, or vice versa.

For a practical alternator it is found to be more convenient to rotate the magnetic field, and to

keep the conductors (coil or winding) stationary. In any rotating a.c. machine, the rotating part is

called the rotor, and the stationary part is called the stator. Thus, in an alternator, the field system

is contained in the rotor. The winding in which the emf is generated is contained in the stator.

The reasons for this are as follows:

In this context, the term field refers to the magnetic field. This field is normally produced by

passing d.c. current through the rotor winding. Since the winding is rotated, the current is passed

to it via copper slip-rings on the shaft. The external d.c. supply is connected to the slip-rings by a

pair of carbon brushes.

(a) When large voltages are generated, heavy insulation is necessary. If this extra mass has to be

rotated, the driving device has to develop extra power. This will then reduce the overall

efficiency of the machine. Incorporating the winding in the stator allows the insulation to be as

heavy as necessary, without adversely affecting the efficiency.

(b) The contact resistance between the brushes and slip-rings is very small. However, if the

alternator provided high current output (in hundreds of ampere), the I 2 R power loss would be

significant. The d.c. current (excitation current) for the field system is normally only a few amps

225

or tens of amps. Thus, supplying the field current via the slip-rings produces minimal power loss.

The stator winding is simply connected to terminals on the outside of the stator casing.

(c) For very small alternators, the rotor would contain permanent magnets to provide the rotating

field system. This then altogether eliminates the need for any slip-rings. This arrangement is

referred to as a brushless machine.

The basic construction for a single-phase alternator is illustrated in Fig. 22.1.

Fig 22.1

The conductors of the stator winding are placed in slots around the inner periphery of the stator.

The two ends of this winding are then led out to a terminal block on the casing. The rotor

winding is also mounted in slots, around the circumference of the rotor. This figure is used to

illustrate the principle. A practical machine would have many more conductors and slots.

Now in a three phase system there are three identical single-phase alternators contained in the

one machine. The three stator windings are brought out to their own separate pairs of terminals

on the stator casing. These stator windings are referred to as phase windings, or phases. They are

identified by the colors red, yellow and blue. Thus we have the red, yellow and blue phases. The

circuit representation for the stator winding of such a machine is shown in Fig. 22.2. In this

figure, the three phase windings are shown connected, each one to its own separate load. This

arrangement is known as a three-phase, six-wire system. However, three-phase alternators are

rarely connected in this way.

226

Fig 22.2

Since the three generated voltages are sine waves of the same frequency, mutually out-of-phase

by 120°, then they may be represented both on a waveform diagram using the same angular or

time axis, and as phasors. The corresponding waveform and phasor diagrams are shown in Figs.

22.3 and 22.4 respectively. In either case,

Fig 22.3

227

Fig 22.4

we need to select a reference phasor. By convention, the reference is always taken to be the red

phase voltage. The yellow phase lags the red by 120°, and the blue lags the red by 240° (or, if

you prefer, leads the red by 120°). The windings are arranged so that when the rotor is driven in

the chosen direction, the phase sequence is red, yellow, and blue. If, for any reason, the rotor was

driven in the opposite direction, then the phase sequence would be reversed, i.e. red, blue, and

yellow. We shall assume that the normal sequence of R, Y, and B applies at all times. It may be

seen from the waveform diagram that at any point along the horizontal axis, the sum of the three

voltages is zero. This fact becomes even more apparent if the phasor diagram is redrawn as in

Fig. 22.5. In this diagram, the three phasors have been treated as any other vector quantity. The

sum of the vectors may be determined by drawing them to scale, as in Fig. 22.5, and the resultant

found by measuring the distance and angle from the beginning point of the first vector to the

arrowhead of the last one. If, as in Fig. 22.5, the first and last vectors meet in a closed figure, the

resultant must be zero.

Fig 22.5

228

22.2 Three Phase four wire systems

It is not necessary to have six wires from the three phase windings to the three loads, provided

there is a common ‘return’ line. Each winding will have a ‘start’ (S) and a ‘finish’ (F) end. The

common connection mentioned above is achieved by connecting the corresponding ends of the

three phases together. For example, either the three ‘F’ ends or the three ‘S’ ends are commoned.

This form of connection is shown in Fig. 22.6, and is known as a star or Y connection. With the

resulting 4-wire system, the three loads also are connected in star configuration.

The three outer wires are called the lines, and the common wire in the centre is called the

neutral.

Fig 22.6

If the three loads were identical in every way (same impedance and phase angle), then the

currents flowing in the three lines would be identical. If the waveform and/or phasor diagrams

for these currents were drawn, they would be identical in form to Figs. 22.3 and 22.4. These

three currents meet at the star point of the load. The resultant current returning down the neutral

wire would therefore be zero. The load in this case is known as a balanced load, and the neutral

is not strictly necessary. However it is difficult, in practice, to ensure that each of the three loads

is exactly balanced. For this reason the neutral is left in place. Also, since it has to carry only the

relatively small ‘out-of balance’ current, it is made half the cross-sectional area of the lines.

Let us now consider one of the advantages of this system compared with both a single-phase

system, and the three-phase 6-wire system. Suppose that three identical loads are to be supplied

with 200 A each. The two lines from a single-phase alternator would have to carry the total 600

A required. If a 3-phase, 6-wire system was used, then each line would have to carry only 200 A.

Thus, the conductor cross sectional area would only need to be 1/3 that for the single-phase

system, but of course, being six lines would entail using the same total amount of conductor

material. If a 4-wire, 3-phase system is used there will be a saving on conductor costs in the ratio

of 3.5:6 (the 0.5 being due to the neutral).If the power has to be sent over long transmission lines,

such as the National Grid System, then the 3-phase, 4-wire system yields an enormous saving in

cable costs. This is one of the reasons why the power generating companies use three-phase, star-

connected generators to supply the grid system.

229


1. Compared to the single phase systems, the three phase power systems are:

a. Operational advantages b. efficient

c. Economical d. all of the above

2. In a two-phase system, the phase voltages differ by

a. 60° b. 120°

c. 90° d. 180°

3. In a three phase system, the phase sequence is used to indicate the

a. frequency of the voltages

b. magnitude of the voltages

c. angle between the voltages

d. order in which the phase voltages attain peak values

4. What is the meaning of positive phase sequence?

a. Phase voltage R leads the phase voltage Y by 120°

b. Phase voltage B leads the phase voltage R by 120°

c. Phase voltage Y leads the phase voltage B by 120°

d. All of the above

5. Which of the following represents a negative phase sequence?

a. B-Y-R b. R-B-C

c. Y-R-B d. All of the above

230

LECTURE 23

RELATIONSHIP BETWEEN LINE AND PHASE VOLTAGE

AND CURRENTS IN STAR CONNECTED SYSTEM

OBJECTIVES

• To explain and derive the relationship between line and phase quantities in a star

connected system.

• To explain delta and mesh connections.

23.1 Relationship between Line and Phase quantities in a star connected

system

Consider Fig. 23.1, which represents the stator of a 3-phase alternator connected to a 3-phase

balanced load. The voltage generated by each of the three phases is developed between the

appropriate line and the neutral. These are called the phase voltages, and may be referred to in

general terms as Vph or specifically as VRN, VYN and VBN respectively.

However, there will also be a difference of potential between any pair of lines. This is called a

line voltage, which may be generally referred to as VL, or specifically as VRY, VYB arid VBR

respectively.

Fig 23.1

A line voltage is the phasor difference between the appropriate pair of phase voltages. Thus, VRY

is the phasor difference between VRN and VYN. In terms of a phasor diagram, the simplest way to

231

subtract one phasor from another is to reverse one of them, and then find the resulting phasor

sum. This is, mathematically, the same process as saying that a - b = a + (- b). The corresponding

phasor diagram is shown in Fig. 23.2.

Fig 23.2

Note: If VYN is reversed, it is denoted either as -VYN or as VNY. We shall use the first of these.

The phasor difference between VRN and VYN is simply the phasor sum of VRN + (-VYN).

Geometrically this is obtained by completing the parallelogram as shown in Fig.23.2. This

parallelogram consists of two isosceles triangles, such as OCA. Another property of a

parallelogram is that its diagonals bisect each other at right angles. Thus, triangle OCA consists

of two equal right-angled triangles, OAB and ABC. This is illustrated in Fig. 23.3. Since triangle

OAB is a 30°, 60°, 90° triangle, then the ratios of its sides AB:OA:OB will be 1:2:√3

respectively.

Fig 23.3

232

Using the same technique, it can be shown that:

……….. (23.1)

The complete phasor diagram for the line and phase voltages for a star connection is shown in

Fig. 23.4. Also, considering the circuit diagram of Fig. 23.1, the line and phase currents must be

the same.

Hence, in star configuration, IL = Iph ………….. (23.2)

Fig 23.4

We now have another advantage of a 3-phase system compared with single-phase. The star-

connected system provides two alternative voltage outputs from a single machine. For this

reason, the stators of all alternators used in electricity power stations are connected in star

233

configuration. These machines normally generate a line voltage of about 25 kV. By means of

transformers, this voltage is stepped up to 400 kV for long distance transmission over the

National Grid. For more localized distribution, transformers are used to step down the line

voltage to 132 kV, 33 kV, 11 kV, and 415 V. The last three of these voltages are supplied to

various industrial users. The phase voltage derived from the 415 V lines is 240 V, and is used to

supply both commercial premises and households.


Q1. A 415 V, 50 Hz, 3-phase supply is connected to a star-connected balanced load. Each phase

of the load consists of a resistance of 25Ω and inductance 0.1H, connected in series.

Calculate

(a) Phase voltage,

(b) The line current drawn from the supply, and

(c) The power dissipated.

Solution:

Fig 23.5

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23.2 Delta or Mesh Connection

If the start end of one winding is connected to the finish end of the next, and so on until all three

windings are interconnected, the result is the delta or mesh connection. This connection is shown

in Fig.23.6. The delta connection is not reserved for machine windings only, since a 3-phase load

may also be connected in this way.

235

Fig 23.6


1. Which of the following are the necessary conditions for an entire three-phase system to be

balanced?

a. Line voltages are equal in magnitude

b. Impedance in each phase must be identical

c. Phase difference between successive line voltages are equal

d. All of the above

2. Which of the following is true related with a three phase delta-connected circuit?

a. Line currents are equal to phase currents

b. Line voltage is equal to phase voltage

c. Line currents are 90 degrees apart

d. Line voltage is 1.732 times the phase voltage

3. Which of the following is not true related to a three phase power systems?

a. The instantaneous sum of power delivered is always constant

b. A non-zero starting torque for the three phase motors

236

c. Output pf a three phase machine is less than that of a single phase machine for the same

frame size

d. For the same voltage the weight of a conductor is 25 % less than in a single phase

4. How delta connection is formed?

a. The starting end of one coil is connected to the finishing end of the other

b. All the starting ends of the coils are connected together

c. The starting and the finishing end points of each coil are left unconnected

d. None of the above

5. What is the expression for the neutral current of a three phase four wire system?

a. IN= IR+ IY + IB b. IN = 0

c. IN = IR+ IY d. IN= IR+IB

237

LECTURE 24

RELATIONSHIP BETWEEN LINE AND PHASE VOLTAGE

AND CURRENTS IN STAR CONNECTED SYSTEM

OBJECTIVES

• To explain and derive the relationship between line and phase quantities in a delta

connected system.

• To derive relation in the power dissipation in star and delta connected load

24.1 Relationship between Line and Phase Quantities in a Delta-connected

system

It is apparent from Fig. 23.12 that each pair of lines is connected across a phase winding. Thus,

for the delta connection:

VL=Vph ………………………………… (24.1)

IL=√3Iph ……………………………… (24.2)

Fig 24.1

The phasor diagram for the phase and line currents in delta connection is as in Fig. 24.2. Note

that the provision of a neutral wire is not applicable with a delta connection. However, provided

that the load

238

Fig 24.2

is balanced, there is no requirement for one. Under balanced load conditions the three phase

currents will be equal, as will be the three line currents. If the load is unbalanced, then these

equalities do not exist, and each phase or line current would have to be calculated separately.

This technique is beyond the scope of the syllabus you are now studying.

24.2 Power Dissipation in Star and Delta connected load

We have seen in Example 24.1 that the power in a 3-phase balanced load is obtained by

multiplying the power in one phase by 3. In many practical situations, it is more convenient to

work with line quantities.

239

Substituting these values into eqn [1] will yield the same result. Thus, the equation for

determining power dissipation, in both star and delta-connected loads is exactly the same.

However, the value of power dissipated by a given load when connected in star is not the same

as when it is connected in delta. This is demonstrated in the following example.


Q1. A balanced load of phase impedance 120Ω is connected in delta. When this load is

connected to a 600 V, 50 Hz, 3-phase supply, determine (a) the phase current, and (b) the

line current drawn.

Solution:

240

Q2. A balanced load of phase impedance 100Ω and power factor 0.8 is connected (a) in star, and

(b) in delta, to a 400 V, 3-phase supply. Calculate the power dissipation in each case.

Solution:

Fig 24.4

241

Fig 24.5

Comparing the two answers for the power dissipation in the above example, it may be seen that:

Power in a delta-connected load is three times that when connected in star configuration.

242


1. The sum of the instantaneous voltages around a delta-connected supply is equal to

a. The maximum line voltage b. Twice the maximum line voltage c. √2 times the rms value of the phase voltage d. Zero

2. What is the expression for the real power in a three-phase system?

a. Vph Iph cosΦ b. 3 Vph Iph cosΦ

c. 3VL IL cosΦ d. (b) and (c)

3. A 415- V three-phase star connected system is connected to a delta connected balanced load

of 100 ohm each. How much is the line current?

a. 2.866 A b. 4.15A c. 7.18A d. 2.39A

4. Which of the following system is a four wire system?

a. Star b. Delta c. (a) and (b) d. Star with neutral

5. The power in a three phase system is given by √3VL IL cosΦ , where Φ is the phase angle

between

a. line voltage and line current

b. line voltage and phase current

c. phase voltage and phase current

d. phase voltage and line current

243

LECTURE 25

MEASUREMENT OF THREE PHASE POWER

OBJECTIVES

• To learn measurement of three phase power is explained


Q1. A balanced star-connected load is fed from a 400 V, 50 Hz, three-phase supply. The

resistance in each phase of the load is 10Ω and the load draws a total power of 15 kW.

Calculate (a) the line current drawn, (b) the load power factor, and (c) the load inductance.

Solution:

245

Q2. A balanced delta-connected load takes a phase current of 15 A at a power factor of 0.7

lagging when connected to a 115 V, 50 Hz, three-phase supply. Calculate (a) the power drawn

from the supply, and (b) the resistance in each phase of the load.

Solution:

25.2 Measurement of three phase Power

In an a.c. circuit the true power may only be measured directly by means of a wattmeter. The

instrument has a fixed coil through which the load current flows, and a moving voltage coil (or

pressure coil) connected in parallel with the load. The deflection of the pointer, carried by the

246

moving coil, automatically takes into account the phase angle (or power factor) of the load. Thus

the wattmeter reading indicates the true power,

P = VI cos φ watt.

If a three-phase load is balanced, then it is necessary only to measure the power taken by one

phase. The total power of the load is then obtained by multiplying this figure by three. This

technique can be very simply applied to a balanced, star-connected system, where the star point

and/or the neutral line are easily accessible. This is illustrated in Fig. 25.1.

Fig 25.4

In the situation where the star point is not accessible, then an artificial star point needs to be

created. This is illustrated in Fig. 25.5, where the value of the two additional resistors is equal to

the resistance of the wattmeter voltage coil.

In the case of an unbalanced star-connected load, one or other of the above procedures would

have to be repeated for each phase in turn.

The total power P = P 1 + P 2 + P 3, where P 1, P 2 and P 3 represent the three separate readings.

Fig 25.5

247

For a delta-connected load, the procedure is not quite so simple. The reason is that the phase

current is not the same as the line current. Thus, if possible, one of the phases must be

disconnected to allow the connection of the wattmeter current coil. This is shown in Fig. 25.6.

Again, if the load was unbalanced, this process would have to be repeated for each phase.

Fig 25.6


1. In a three phase star connected system, ERY lags behind EB by how many electrical degrees?

a. 90 b. 60 c. 30 d. 120

2. In a three phase star connected load the neutral current is equal to a. IL b. IP c. 1 d. 0

3. The relationship between the line and phase voltage of a delta connected circuit is given by

a. VL= VP b. VL=VP/3 c. VL=√3VP d. VL=2VP

4. For what type of load, the one wattmeter method will be used to measure power in a three phase circuit?

a. Balanced b. Balanced star connected load c. Balanced star connected with the neutral accessible load d. Balanced delta connected load

5. What is the expression for the reactive power in a balanced three phase system?

a. √3VL IL cosΦ b. √3VL IL sinΦ c. √3VL IL d. 3VL IL

248

6. To which of the following circuits the two wattmeter method is not suitable to measure

power in a three phase circuit?

a. Unbalanced star connected loads with the neutral inaccessible.

b. Balanced star or delta connected loads

c. Unbalanced delta connected loads

d. None of these

249

LECTURE 26

MEASUREMENT OF THREE PHASE POWER

OBJECTIVES

• To study the measurement of three phase power using two wattmeter method

• To study advantages of two wattmeter method.

26.1 Two-wattmeter method

The measurement of three-phase power using the above methods can be very awkward and time-

consuming. In practical circuits, the power is usually measured by using two wattmeters

simultaneously, as shown in Fig. 26.1

The advantages of this method are:

(a) Access to the star point is not required.

(b) The power dissipated in both balanced and unbalanced loads is obtained, without any

modification to the connections.

(c) For balanced loads, the power factor may be determined.

Fig 26.1

250

Consider now the phasor diagram for a resistive-inductive balanced load, with the two

wattmeters connected as in Fig. 26.1. This phasor diagram appears as Fig. 26.2, below.

251

Fig 26.2

From these results, and using Fig. 26.3 , the following points should be noted:

Fig 26.3

1] If the load p.f. > 0.5 (i.e. φ < 60°); both meters will give a positive reading.

252

2] If the load p.f. = 0.5 (i.e. φ = 60°); W 1 indicates the total power, and W 2 indicates zero.

3] If the load p.f. < 0.5 (i.e. φ > 60°); W 2 attempts to indicate a negative reading. In this case,

the connections to the voltage coil of W 2 need to be reversed, and the resulting reading recorded

as a negative value. Under these circumstances, the total load power will be P = P 1 - P 2.

4] The load power factor may be determined from the two wattmeter readings from the equation:


1. A three phase star connected symmetrical load consumes P watts of power from a balanced

supply. If the same load is connected in delta to the same supply, the power consumption will be a. 3P b. P c. √3P d. 2P

2. W1 and W2 are the readings of two wattmeters used to measure power of a three phase balanced load. The reactive power drawn by the load is a. W1+ W2 b. W1-W2 c. √3(W1-W2) d. √3(W1+W2)

3. In the measurement of the three phase power by the two wattmeter method, if the two wattmeter readings are equal, the power factor of the load is a. 1 b. 0 c. 0.8 lagging d. 0.8 leading

4. In a two wattmeter method measurement of the three phase power, one of the wattmeter is

showing zero reading. Find the power factor of the circuit.

a. 1 b. 0.5 c. 25 d. 0

5. In a delta connected load, if one of the resistors is removed what will be the power?

a. power will be two third of the original

b. power will be zero

c. power will be double

d. power will be infinite

253

LECTURE 27

NUMERICAL SOLVING ON THREE PHASE CIRCUITS

OBJECTIVE

• To solve numericals based on three phase ac circuits


Q1. The power in a 3-phase balanced load was measured, using the two-wattmeter method. The

recorded readings were 3.2 kW and 5 kW respectively. Determine the load power and power

factor.

Solution:

Q2. A 3-phase balanced load takes a line current of 24 A at a lagging power factor of 0.42, when

connected to a 415 V, 50 Hz supply. If the power dissipation is measured using the two-

wattmeter method, determine the two wattmeter readings, and the value of power dissipated.

Comment on the results.

Solution:

254

Q3. A delta-connected load has a phase impedance of 100 _ at a phase angle of 55°, and is

connected to a 415 V three-phase supply. The total power consumed is measured using the

two-wattmeter method. Determine the readings on the two meters and hence calculate the

power consumed.

Solution:

Fig 27.1

255

27.2 Advantages of a three phase system

In previous sections, some of the advantages of three-phase systems, compared with single-phase

systems, have been outlined. These advantages, together with others not yet described, are listed

below. The advantages may be split into two distinct groups: those concerned with the

generation and distribution of power, and those concerning a.c. motors.

1] The whole of the stator of a three-phase machine is utilized. A single phase alternator utilizes

only two thirds of the stator slots.

2] A three-phase alternator has a better distribution factor.

3] A three-phase, four-wire system provides considerable savings in cable costs, for the

distribution of an equivalent amount of power.

4] A three-phase, four-wire system provides alternative voltages for industrial and domestic

users.

5] For a given machine frame size, the power output from a three-phase machine is greater than

that from a single-phase machine.

6] A three-phase supply produces a rotating magnetic field; whereas a single-phase supply

produces only a pulsating field.

7] Three-phase motors are inherently self-starting; whereas single-phase motors are not.

8] The torque produced by a three-phase motor is smooth; whereas that produced by a single-

phase motor is pulsating.

256


1. What is the phase angle between the emfs in a three phase system? a. 100 degree b. 90 degree c. 120 degree d. 60 degree

2. A circuit draws 2A when it is connected to a %)-V supply. The wattmeter connected in the circuit reads 80W. What is the power factor of the circuit? a. 1 b. 0.8 c. 1.5 d. 0

3. A three phase star connected symmetrical load consumes P watts of power from a balanced supply. If the same load is connected in delta to the same supply, what will be the power consumption? a. 3P b. 2P c. √3P d. P

4. A three phase 500V motor load has a power factor of 0.707. Two wattmetters are connected to measure the power that shows an input of 30W. What is the reactive power? a. 0 VAR b. 15 VAR c. 10 VAR d. 30 VAR

5. What will happen if the phase sequence at the three phase load reverses? a. Phase current changes in phase angle but not in magnitude

b. Phase current as well as magnitude both changes

c. Phase current remains same but magnitude changes


257

EXERCISES

1) A three-phase load is connected in star to a 400 V, 50 Hz supply. Each phase of the load

consists of a coil, having inductance 0.2 H and resistance 40Ω. Calculate the line current.

2) If the load specified in Question 1 above is now connected in delta, determine the values for

phase and line currents.

3) A star-connected alternator stator generates 300 V in each of its stator windings. (a) Sketch

the waveform and phasor diagrams for the phase voltages, and (b) calculate the p.d. between

any pair of lines.

4) A balanced three-phase, delta-connected load consists of the stator windings of an a.c.

motor. Each winding has a resistance of 3.5 Ω and inductance 0.015 H. If this machine is

connected to a 415 V, 50 Hz supply, calculate (a) the stator phase current, (b) the line current

drawn from the supply, and (c) the total power dissipated.

5) Repeat the calculations for Question 4, when the stator windings are connected in star confi

guration.

6 )Three identical coils, connected in star, take a total power of 1.8 kW at a power factor of

0.35, from a 415 V, 50 Hz supply. Determine the resistance and inductance of each coil.

7) Three inductors, each of resistance 12 Ω and inductance 0.02 H, are connected in delta to a

400 V, 50 Hz, three-phase supply. Calculate (a) the line current, (b) the power factor, and (c)

the power consumed.

8) The star-connected secondary of a three-phase transformer supplies a delta-connected motor,

which takes a power of 80 kW, at a lagging power factor of 0.85. If the line voltage is 600 V,

calculate (a) the current in the transformer secondary windings, and (b) the current in the motor

windings.

9)A star-connected load, each phase of which has an inductive reactance of 40 Ω and

resistance of 25 Ω , is fed from the secondary of a three-phase, delta-connected transformer. If

the transformer phase voltage is 600 V, calculate (a) the p.d. across each phase of the load, (b)

the load phase current, (c) the current in the transformer secondary windings, and (d) the power

and power factor.

10) Three coils are connected in delta to a 415 V, 50 Hz, three-phase supply, and take a line

current of 4.8 A at a lagging power factor of 0.9. Determine (a) the resistance and inductance

of each coil, and (b) the power consumed.

258

11) The power taken by a three-phase motor was measured using the two-wattmeter method.

The readings were 850 W and 260 W respectively. Determine (a) the power consumption and,

(b) the power factor of the motor.

12) Two wattmeters, connected to measure the power in a three-phase system, supplying a

balanced load, indicate 10.6 kW and - 2.4 kW respectively. Calculate (a) the total power

consumed, and (b) the load phase angle and power factor. State the significance of the negative

reading recorded.

13) Using the two-wattmeter method, the power in a three-phase system was measured. The

meter readings were 120 W and 60 W respectively. Calculate (a) the power, and (b) the power

factor.

14) Each branch of a three-phase, star-connected load, consists of a coil of resistance 4 _ and

reactance 5 Ω . This load is connected to a 400 V, 50 Hz supply. The power consumed is

measured using the two-wattmeter method. Sketch a circuit diagram showing the wattmeter

connections, and calculate the reading indicated by each meter.

15) A three-phase, 415 V, 50 Hz supply is connected to an unbalanced, star-connected load,

having a power factor of 0.8 lagging in each phase. The currents are 40 A in the red phase, 55

A in the yellow phase, and 62 A in the blue phase. Determine (a) the value of the neutral

current, and (b) the total power dissipated.


1) In a 3ф power measurement by two wattmeter method, both the wattmeter read the

same value. What is the power factor of the load? Justify your answer.

2) Deduce the relationship between phase and line quantities (voltage, current, power)

in the circuit of a three phase delta connected system. Also draw neat diagrams.

3) Each of delta connected load consists of a 50 Ω resistor in series with a 50µ F

capacitor. The supply is 440 V 3phase, 50 Hz. Determine phase current, line current,

Power. Draw phasor diagram showing the voltage and currents.

4) Explain with phasor how two wattmeters can be used to measure power in system?

Also explain the variation in wattemeter readings with load power factors.

5) Three identical coils each having a reactance of 20 Ω and resistance of 10 Ω are

connected in (a) star and (b) delta across a 440 V 3ф line. Calculate for each method

of connection the line current and readings on each of the two wattmeters connected

to measure power.

259

6) Three similar coils, connected in star, take a total power of 1.5 kW ata P.F. of 0.2

lagging from a three phase, 440V, 50 Hz supply. Calculate the resistance and

inductance of each coil.

7) Explain merits of two wattmeter method for power measurement, giving circuit

diagram and phasor.

8) A balanced 3 phase load connected in delta, draws a power of 10kW at 440V at a

power of 0.6 lead, find the values of circuit elements and reactive voltampere drawn.

9) Write briefly what is power triangle, name the sides with units.

10) Enlist the advantages of three phase system.

ANSWER KEY

Lec

22

Lec

23

Lec

24

Lec

25

Lec

26

Lec

27

Q1. d d d a a c

Q2. c b d d c b

Q3. c c c a a a

Q4. d a d d a a

Q5. d a c a a a

260

MODULE 4

SINGLE PHASE

TRANSFORMER

261

LECTURE-28

CONSTRUCTION AND PRINCIPLE OF OPERATION OF A

SINGLE PHASE TRANSFORMER

OBJECTIVES

• To introduce single phase transformer in terms of principle of operation

• To explain construction of single phase transformer

• To dicuss basic types of single phase transformer.

28.1 Introduction

A transformer is an electrical device having no moving parts. It is used for transferring electric

power from one circuit to another at same frequency, at same or different voltages with a

corresponding decrease or increase in the current. Transformer converts generated voltage of

about 11 KV (determined by generator design limitations) to higher values of 132KV, 220KV,

400KV etc. Such an increase in voltage is necessary for economical transmission of electrical

power. Due to this, the costs of transmission lines as well as power losses are tremendously

reduced. Apart from this, transformers are used for distribution of electrical power,

communication circuits, radio and TV circuits, telephone circuits etc.

28.2 Principle of operation

In its simplest form, it consists of two windings wound on an iron core. One of the windings is

connected to an alternating source of supply frequency f. This source supplies a current to this

winding (called primary winding) which in turn produce an alternating flux in the iron core. The

alternating flux φalternates at same frequency (f). N1 is the primary winding and N2 is secondary

winding as shown in Fig.28.1.The secondary winding which feeds the load is also linked by this

flux. Emfs are induced in the two windings due to this alternating flux in the core by Faraday’s

laws of electromagnetic induction. If the secondary circuit is closed, a current flows in the

secondary. Thus transformer operates on the principle of mutual induction between two coils.

262

Fig.28.1Basic Principle

In brief, we can say the following:

• The transformer is a static device. • It transfers electrical power from one circuit to another. • During transfer of power, there is no change of frequency. • It uses electromagnetic induction to transfer electrical power. • The two electrical circuits are in mutual inductive influence of each other.

Power transformer is used for the transmission purpose at heavy load, voltage greater than 33 KV & high efficiency. It also having a big in size as compare to distribution transformer, it used in generating station and Transmission substation. The distribution transformer is used for the distribution of electrical energy at low voltage as less than 33KV in industrial purpose and 440V-220V in domestic purpose. It work at low efficiency at 50-70%, small size, easy in installation, having low magnetic losses & it is not always fully loaded.

28.3 Construction (Reference: “Electrical Technology” by M.S Naidu & Kamakshaiah)

Fig 28.2 Constructional details of a distribution transformer

263

28.3.1 Main Parts

1.Laminated core 2.Windings 3.Tank 4.Conservator 5.Bushing 6.Breather 7.Radiators 8.Winding leads(input) 9.Winding leads (output) 10.Transformer Oil

Laminated iron core - Provides a magnetic path of low reluctance for the flux (CRGOSS)

Windings – Primary and secondary winding made of high grade copper. Winding turns are insulated from each other. Bare Cu wires are given enamel coating also.

Container/Tank – To house the core and windings, made of sheet metal. Oil – Medium for insulation and cooling. Conservator – Provides the means for the oil to settle down by expanding under heavy

loads. Otherwise high pressures will be developed inside which will lead to bursting of tank

Breather – prevents air coming in to the contact with oil. Radiators - Cooling Bushings – Insulating and bringing out winding terminals from the tank.

28.4 Types of Transformers There are two types, based on the type of core used as shown in Fig.28.3

Fig.28.3 Core and shell type transformer

Core type Shell type

1) Winding surround a considerable

part of the core

Core surrounds a considerable portion

of windings.

2) More space for insulation, hence

Transformers. preferred for high voltages

More economical for low voltage

3) Leakage flux is more Leakage flux is less.

264


1. If a transformer is continuously operated , the maximum temperature rise occurs in

a. Core b. Windings c. Tank d. None of these

2. Core type transformer suitable for high voltages because

a. More number of turns can be wound b. More space for insulation c. More flux available for induction d. None of the above

3. Special feature of the transformer that a. It is static device b. It transfer electrical power from one circuit to another circuit c. It does not change the frequency d. All of the above

4. Basic function of laminated core is a. To provide mechanical support to the winding b. To provide magnetic path of low reluctance c. To provide cooling to the winding d. None of the above

5. _________ prevents air coming in to the contact with oil. a. Conservator tank b. Bushing c. Breather d. Radiators

265

LECTURE 29

IDEAL TRANSFORMER UNDER NO LOAD

OBJECTIVES

• To explain emf equation of transformer

• To develop phasor diagram of Ideal transformer on no load condition

29.1 Derivation of emf equation

Let φ be the instantaneous value of the magnetic flux, V1 be the applied voltage, φm be the peak

value of flux and ω is the angular frequency. Then φ = φm sinωt

Induced emf, dt

tSindN

dt

dNe m )( ωφ Φ

−=−= = ωNφm(-Cosωt)=ωNφmSin(ωt-π/2)

Hence peak value of the induced emf is ωNφm = 2πfNφm.

∴Emax = 2πfNφm

∴Erms=Emax/√2=2π/√2 fNφm=4.44fNφφφφm

E1=4. 44fN1φm&E2=4. 44fN2φm

∴E1/E2=N1/N2=K, where K is called transformation ratio or turns ratio.

Neglecting the drops, E1≅ V1 E2≅ V2

∴V1/V2=N1/N2 For an ideal transformer i.e. having no losses mmf is same.

∴ N1I1=N2I2 ,whereI1is theprimary current and I2is the secondary current

∴V1////V2=N1////N2= I2////I1

If N2>N1, V2>V1,the device is known as a step down transformer.

29.2 Solved Problems Q1. A 25KVA transformer has 500 turns on primary and 40 turns on the secondary winding.The

primary winding is connected to a 3KV,50Hz source. Calculate a) secondary voltage b)I1and I2 c) Maximum flux

Solution:

V1/V2=N1/N2

266

3000/V2=500/40∴V2=3000x40/500=240V

O/P KVA = 25 KVA∴I2=25x103/240=104.17A

I2/I1= N2/N1∴I1=40/500 x 104.17=8.33 A

V1≅ E1=4.44fN1φm∴φm=3000/4.44x50x500=0.027wb

Q2. A250 KVA,11000/400V.50Hz 1φ transformer has 80 turns on secondary. Calculate a) approx. values of I1 and I2 b) approx.. N c) Max.value of the flux Solution:

V1I1= 250 x 103∴11000 X I1≅ 250 X 103 ∴I1=250X103/11000=22.7A

V2I2= 250 x 103∴I2=250x103/400=625A

V1/V2=N1/N2∴N1=V1/V2XN2 = 80x11000/400=2200

V1≅4.44fφm N1 ∴φm=11000/4.44x50x2200 =22.5mwb

Q3. A 1φ transformer has 350 primary and 1050 secondary turns.The net cross sectional area of

the core is 55cm2.If the primary winding is connected to a 400V,50 Hz, 1φ supply. Calculate i) Bmax in the core ii)induced voltage in secondary Solution:

V1≅ E1=400V

E1=4. 44fφm N1 asφm =Bm.Ac , Therefore Bm=0.93wb/m2

E1/E2=N1/N2∴ E2=1200V

29.3 Ideal Transformer⇒ no R in winding, no leakage flux

On no load ⇒ secondary is open, No secondary current. When a transformer is on no load, the input current at the primary has to supply the losses in the core and also magnetise the core.

Fig.29.1 Ideal transformer on no load

267

29.4 Phasor diagram on no load

Flux is common to both primary and secondary is taken as reference phasor to draw the phasor diagram as shown in Fig.29.2.V1is the supply voltage , Iw = active component of no load current Io required to meet losses in the core, Im= magnetising component of no load current Io required

to produce flux in the core,φ0= no load pf angle.

Fig.29.2 phasor diagram on no load

Im=I0Sinφ0, Iω=IoCosφo

Io=22

wm II +

Cosφo=Iw/Io

Iw is usually small and Cosφoisalso small.Iois3 to 5% of full load primary current. Hence copper losses due to Iocan be neglected. This means that no load primary input is practically equal to the iron losses in the transformer core.

29.5 Equivalent circuit on no load

No load current Io and the transformer core loss are generally a function of supply voltage and therefore the shunt branch is connected across the supply.From the equivalent circuit shown in Fig.29.3, Xm=V1/Im , Ro=V1/Iw

Fig.29.3 Equivalent circuit on no load

268


1. A transformer has a 6:1 voltage ratio. Find the current in the secondary if the current in

the primary is 200 mA.

a. 0.833A b. 1.2A c. 1kA d. .0333A

2. There are 400 turns of wire in an iron-cored coil. If this coil is to be used as the primary of

a transformer, how many turns must be wound on the coil to form the secondary

winding of the transformer to have a secondary voltage of one volt if the primary

voltage is five volts?

a. 80 b. 400 c. 2000 d. 5

3. A sinusoidal flux 0.02Wb (maximum) links with 55 turns of a transformer secondary

coil operating at 50 Hz. The rms value of the induced emf in the secondary is

a. 144.2 V b. 244.2 V c. 344.6 V d. 100 V

4. A Step up transformer is a transformer in which a. N2 < N1 b. N2 = N1 c. N2 > N1

5. Significance of the relation 1

2

2

1

I

I

V

V= is

a. KVA rating of transformer remains same in both primary and secondary

b. As the voltage increases on primary side, current in primary also increases

c. As the current increases on secondary side, voltage at secondary decreases


6. Basic principle of operation of transformer is based on

a. Fleming’s left hand rule

b. Faraday’s law of electromagnetic induction

c. Lenz’s law

d. Kirchhoff’s law

269

LECTURE 30

IDEAL TRANSFORMER ON LOAD

OBJECTIVES

• To explain phasor diagram of ideal transformer on load condition

• To discuss about the parameters do be considered for practical transformer

30.1 Ideal transformer on load

When a load is connected in the secondary, the load current that flows through the secondary turns will produce a load component of mmf, thereby flux, which according to Lenz’s law is in

such a direction as to oppose the flux φm which is producing it, consequently the flux and emf induced in the secondary is momentarily reduced. This change increases the difference between

V1and induced in primary (E1) and will make the primary draw more current so as to maintain φm

constant. Hence flux in a transformer remains constant.

I2 in the secondary causes a current ′2I to flow in the primary side such that, ′

2I = I2*

N1/N2.′

2I represent the component of primary current to neutralize the demagnetizing effect of

the secondary current. In addition to ′2I , the no load current I0 also flows in the primary. So I1= I0

+ ′2I . Primary side pf= cosφ1, where φ1 = angle between V1 and I1.However as far as the

magnetization of the core is considered, I0 is same as on no load. Consider a load on the

secondary such that pf is lagging. So I2lag V2 by φ2as shown in Fig.30.1.

Fig.30.1 Phasor diagram of ideal loaded transformer

270

30.2 Resistance, Leakage flux and leakage reactance

In ideal transformer, the only flux that exists is φm that links with both primary and secondary and winding resistances were considered negligible. But in practice, not all of the flux produced by the primary winding links the secondary winding and both primary and secondary winding has resistance. The difference between the total flux linking the primary windings and the mutual flux linking both winding is called the primary leakage flux. i.e. some part of the primary flux, completes its path by passing through air rather than the core. Leakage flux always involves appreciable air paths and although some iron is included in the closed path, the reluctance experienced by the leakage flux is practically that of the air.

Fig.30.2 Practical transformer on load

This leakage flux induces an emf in primary winding i.e. self induced emf. The inductance is the property of the circuit by which flux is established whenever current is flowing through it. Thus each leakage flux will have a corresponding inductance that is responsible for its production. The effect of inductances is to produce the respective reactance X1 and X2 in the primary and secondary windings. Hence it is equivalent to as small inductive coil in series with each winding such that voltage drop in each series coil is equal to that produced by leakage flux. The quantity X1 is called the primary leakage reactance. It is a fictitious quantity introduced to represent primary leakage flux. Similarly when a transformer is loaded, secondary leakage flux is also there. This is represented by X2. R1, R2 are the resistances of primary and secondary windings respectively.


Q1. A 1φ transformer has 1000 turns on the primary and 200 turns on the secondary. The no load current is 3A at a p.f. of 0.2 lagging and the secondary current is 280A at a p.f. of 0.8 lagging. Calculate the primary current and p.f. Assume the voltage drop in the windings to be negligible. Also find Iom and Iol

Solution:

Io= 3A φo=Cos 1− 0.2 =78.46 Iom=IoSinφo

I2= 280A φ2=Cos 1− 0.8 =36.86 Iol=IoCosφo = 3x0.2 =0.6 A

271

′2I /I2 = N2/N1∴

′2I = I2. N2/N1 = 56A

3 -φ0+56 -φ2=58.3-38.50 =58.28-38.80A


1. The emf induced in the primary

a. Is in phase with the flux b. Lags behind the flux by 900 c. Leads the flux by 900 d. Is in phase opposition to that of flux

2. The leakage flux in the transformer is represented in equivalent circuit as

a. Resistance connected in parallel with both primary and secondary winding

b. Inductance connected in parallel with both primary and secondary winding

c. Resistance connected in series with both primary and secondary winding

d. Inductance connected in series with both primary and secondary winding

3. The rated secondary voltage of 220V for a transformer means the terminal voltage is 220V on

a. No load b. Rated full load c. Half full load d. ¾ full load

4. As the load current (I2) increases the primary current of transformer also increases

a. To increase the mmf component of secondary current and hence increase the flux in the

core

b. To nullify the mmf component of increased secondary current and hence to maintain

constant flux in the core

c. To supply the loss component of increased load current


272

LECTURE 31

EQUIVALENT CIRCUIT OF A TRANSFORMER

OBJECTIVES

• To explain equivalent circuit of transformer

31.1 Development of Equivalent Circuit

R1, R2 = resistance of primary & secondary windings of the actual transformer

X1, X2 = reactance of the windings due to leakage flux in the actual transformer

Ro= value is such that it takes a current equal to the core loss of the actual transformer

Io= current which the transformer takes on no load

Xm=Magnetizing reactance

Fig.31.1 Equivalent circuit of practical transformer

For ease of analysis equivalent circuit parameters are transferred from secondary side to primary

side or vice versa. The effect of circuit parameters shouldn’t be changed while transferring the

parameters from one side to another side. We can replace R2 of the secondary by inserting

additional resistance R2’ in the primary circuit such that the power absorbed in R2

’ when carrying

the primary current is equal to that in R2 due to the secondary current i.e.

I21 R2

’ = I22 R2 therefore R2

’=(I2/I1)2 R2= (N1/N2)2 R2 =K2 R2

Similarly, since the inductance of a coil is proportional to the square of the no. of turns, the

secondary leakage reactance X2 can be replaced by an equivalent. Reactance X2’ in the primary

circuit, such that.

X2’=X2(N1/N2)

2 ≡ X2(V1/V2)2 =K2X2

273

Fig.31.2 Transferring parameters to primary

If Ro1 is a single resistance in the primary circuit equivalent to the primary and secondary

resistance of the actual transformer, then, Ro1= R1 + R2’ = R1 + R2 (V1 / V2)

2. If X01is the single

reactance in the primary circuit equivalent to X1 and X2 of the actual transformer, then

X01= X1 + X2’ = X1+X2 (V1/V2)

2.Equivalent circuit referred to primary is shown in Fig.31.3

Fig.31.3 Equivalent circuit referred to primary

Since the no load current of transformer is only about 3-5 % of the full-load primary current, we

can omit the parallel circuits R0 and Xm without introducing an appreciable error, when we are

considering the behavior of the transformer on full load and can be drawn as shown in

Fig31.4.This is known as approximate equivalent circuit.

274

Fig.31.4 Approximate equivalent circuit


1. The equivalent resistance of the primary of the transformer having K = 5, R1 = 0.1 ohm

when referred to secondary becomes ____ Ohm

a. 0.5

b. 0.02

c. 0.004

d. 2.5

2. The no load current in a transformer is ________ % of full load current

a. 10- 15 b. 1-3 c. 3 – 10 d. None of these

3. Iron loss component in transformer is represented as an equivalent resistance in the equivalent circuit

a. Connected in series with secondary winding b. Connected in parallel across the primary winding c. Connected in parallel across secondary winding d. Connected in series with primary winding

4. Total copper loss of a single phase transformer can be found out as

a. I12R1+I2

2R2 b. I1

2R01 c. I2

2R02

275

d. All of the above

5. Equivalent resistance of transformer referred to secondary is given by the relation

a. 2

12

k

RR +

b. 2

21

k

RR +

c. 2

12 kRR +

d. 2

21 kRR +

276

LECTURE-32

PRACTICAL TRANSFORMER

UNDER DIFFERENT LOAD CONDITIONS

OBJECTIVES

• To explain phasor diagram of practical transformer under different load

conditions.

32.1 Phasor Diagram

From the equivalent circuit of a practical transformer we can write, V1 = E1 + I1R1 + jI1X1 and

E2=V2+I1R1 + jI1X1 .Phasor diagrams for resistive load, lagging load and leading load are shown

in Fig.32.1,32.2 & 32.3 respectively.

Resistive load

Fig.32.1Resistive Load

Lagging pf Load

277

Fig.32.2Lagging Load

Leading pf load

Fig.32.3Leading Load

278

LECTURE-33

LOSSES AND EFFICIENCY

OBJECTIVES

• To explain different types of losses in a transformer

• To define efficiencyof a transformer

• To derive condition for maximum efficiency

• To discuss voltage regulation

33.1 Losses in a transformer

1) Copper losses in primary and secondary windings = I12R1 + I2

2R2 = I12R01 or I2

2R02

2) Iron Losses = Hysteresis loss + Eddy current losses

Hysteresis Loss

The magnetic core of transformer is made of ′Cold Rolled Grain Oriented Silicon Steel′. Steel is very good ferromagnetic material. Ferromagnetic substances have numbers of domains in their structure. These domains are arranged inside the material structure in such a random manner, that net resultant magnetic field of the said material is zero.

Whenever external magnetic field or mmf is is applied to that substance, these randomly directed domains are arranged themselves in parallel to the axis of applied mmf. The mmf applied in the transformer core is alternating. For every cycle, due to the domain reversal there will be extra work done. For this reason, there will be a consumption of electrical energy which is known as Hysteresis loss of transformer.

Eddy Current Losses

The alternating flux of transformer may also link with other conducting parts like steel core or iron body of transformer etc. As alternating flux links with these parts of transformer, there would be a locally induced emf. Due to these emfs there would be currents which will circulate locally at that parts of the transformer. These circulating current will not contribute in output of the transformer and dissipated as heat. This type of energy loss is called eddy current loss.

Hysteresis loss and eddy current loss, both depend upon magnetic properties of the materials used to construct the core of transformer and its design. So these losses in transformer are fixed and do not depend upon the load current. So core losses in transformer which is alternatively known as iron loss in transformer and can be considered as constant for all range of load.

279

Therefore, Total losses = I12R1 + I2

2R2 + Iron losses (Pi).No rotational losses since it is a static

device.

33.2 Efficiency of a transformer

280

Note: Transformer is rated in KVA because the angle between V & I depends on the load

connected to it.

33.3 Voltage Regulation

Power system network operate at standard voltages. The equipment connected as a load operates

at standard voltage within certain agreed tolerance limits. The input to this equipment is supplied

from secondary side of the transformer. There are additional drops inside the transformer due to

the load currents. While input voltage is the responsibility of the utility, the voltage at the load is

important for satisfactory operation of load. If excessive voltage drop is permitted to occur inside

the transformer the load voltage becomes too low and affects its performance. It is therefore

necessary to enumerate the drop that takes place inside a transformer when certain load current,

281

at any power factor, is drawn from its output leads. This drop is termed as the voltage regulation

and is expressed as a ratio of the terminal voltage.

It can be defined in two ways –

1) Down regulation 2) Up regulation

Down regulation: This is defined as ” the change in terminal voltage when a load current at any power factor is applied, expressed as a fraction of the no-load terminal voltage”. Expressed in symbolic form we have,

Down regulationnl

lnl

V

VV −=

Where,

Vnl is the no-load terminal voltage Vl is load terminal voltage

This term is more popularly used. In case of the transformers, the no-load voltage being the one given by the utility, no-load voltage is taken as the reference. Regulation up: This is defined as the ratio of the change in the terminal voltage when a load at a given power factor is thrown off, and the on load voltage. This definition if expressed in symbolic form results in

Up regulationl

lnl

V

VV −=

Where, Vnl is the no-load terminal voltage Vl is load terminal voltage

Normally full load regulation is of importance compared to part load regulation In the case of transformers both type of regulations will have not much difference since the transformer impedance is very low and the power factor of operation is quite high. Hence a convenient starting point is the load voltage. The full load output voltage is specified on the name plate. Hence regulation up has some advantage when it comes to its application. Fig. shows the phasor diagram of operation of the transformer under loaded condition. The no-load current I0 is neglected in view of the large magnitude of I2. Then I1 = I2’ Total voltage drop = E2-V2 =OC – OA = OM –OA = AM = AN + NM Since AN is very small,

Approximate voltage drop ≅ AN = AD + DN = AD + BL

= I2R02 cosΦ + I2X02 sinΦ Thus,

282

% voltage regulation = 100sincos

2

022022 ×±

E

XIRI φφ

Where, ‘+’ sign is used for lagging load and ‘-’ sign is used for leading load

D N M

L B

C

A

φ

φ I2R02

I2X02

O

E2

V2

I2

(b)

Fig.33.1 Phasor diagram


1. Which loss is not common between a transformer and rotating machines?

a. Eddy current loss

b. Copper loss

c. Windage loss

d. Hysteresis loss

I2’

283

2. The full load copper loss of a transformer is 1600W. The loss at half load will be

a. 1600W

b. 6400

c. 800

d. 400

3. The purpose of providing an iron core in a transformer is to

a. Provide support to windings

b. Reduce hysteresis loss

c. decrease the reluctance of the magnetic path

d. reduce eddy current losses

4. In a given transformer, for a given voltage, the losses which remain constant irrespective of

load changes are

a. Frictional and windage losses

b. Copper losses

c. Hysteresis and eddy current loasses


5. The efficiency of a transformer will be maximum when

a. Copper losses = Eddy current losses

b. Hystersis loss = Eddy current losses

c. Copper losses = Iron losses

d. Hystersis loss = Copper losses

284

LECTURE- 34

NUMERICAL SOLVING ON TRANSFORMER EFFICIENCY

AND MAXIMUM EFFICIENCY

OBJECTIVES

• To solve numericals based on transformer efficiency and maximum efficiency


Q1. 150KVA, 50Hz, 1phase transformer is having voltage ratio 5000/250 volts. Find a) No. of

turns on each winding. b) Full load efficiency at 0.8 pf lagging and half load efficiency

c) Load at which maximum efficiency occurs. Maximum core flux is 0.06wb. Full load

Copper losses are 1800W and iron losses are 1500W. d) Maximum efficiency.

Solution:

V1/V2 = N1/N2

5000/250 = 376/N2

Therefore N2 = 19

V1 = 4.44fφmN1 .Therefore N1 = 5000/4.44*50*0.06= 376

(x =1 for Full load)

Full load efficiency = x* Rated KVA *pf*103* 100/(x* Rated KVA *pf* 103 + Pcu + Pi)

= 150 x 103 x 0.8 x 100/150 x 103 x 0.8 + 1800w + 1500 = 97.32%

Load at which maximum efficiency occurs = F.L KVA . = 150 x = 136.93KVA

Max. efficiency = = 97.33%

Half Load Efficiency = = 96.85%

Q2. Find the maximum efficiency at F.L of a 100KVA, 1Phase transformer at upf.If the iron

loss at half full load is 2KW.

Solution:

285

Pi = 2000W, For Max. efficiency ,Pcu = Pi = 2000W.

Max efficiency at full load = = 96.15%

Q3. The primary and secondary windings of a 500KVA transformer have resistances 0.42ohm

and 0.0019 ohm respectively. The primary and secondary voltages are 11000V and 400V

respectively, and the core loss is 2.9Kw, assuming the pf of the load to be 0.8. Calculate the

ƞ on full load and half load. Find the O/P at which efficiency of the transformer is max.and

calculate it’s value.

Solution:

V2I2 = 500x103 therefore I2 = 500x103/400 = 1250A

I1 = 500x103/11000 = 45.5A

I22R2 = 2969w and I1

2R1 = 870w

Total I2 R loss = 3.84 Kw

Total loss = 3.84Kw + 2.9Kw= 6.74Kw

O/P Power on full load = 500*0.8=400Kw

I/P Power on full load = (400+6.74) Kw =406.74 Kw

Efficiency full load = 400Kw/406.74 *100 = 98.3%

Efficiency half full load = (1/2)*400*100/(1/2)*500*0.8+(1/2)2 *3.84+2.9=98.1%

x = √Pi/Pc = 0.867

Kilovolts-amps at efficiency max = 433 KVA

Efficiency max = 0.867*400*100/0.867*400 + (0.867)2*3.84+ 2.9 = 98.4%

Q.5) A 20 KVA Transformer has iron loss of 450W and full load copper loss of 900W. Assume

pf of 0.8 lagging. Find i) Full load and half load ƞ ii) Load at which ƞ max occurs. iii) Value of

max ƞ

Full load efficiency = x * rated KVA * 103*pf*100/x*rated KVA*103*pf + x2 *FL Cu loss+ iron

loss

=20*103*0.8*100/20*103*0.8+900+450 = 92.22%

Load at which efficiency max occur = F.L KVA .√Pi/PCu= 20*√450/900 = 14.14 KVA

x=0.707 ,Max efficiency = 92.63%

286

LECTURE 35

O.C. AND S.C. TEST

OBJECTIVES

• To explain O.C. and S.C. test conducted on transformer

• To explain procedure for estimation of circuit parameters from these tests results

35.1 No Load Tests on Transformer

The exact equivalent Circuit Of the transformer has a total of six parameters. A knowledge of

these parameters allows us to compute the performance of the Transformer under all Operational

conditions. O.C. and S.C. tests on a transformer enable the η and the voltage regulation to be

calculated without actually loading the Transformer and with a reasonable accuracy. Also power

required to carry out these tests is very small compared with the full load output of the

Transformer.

O.C. test

H.V. winding is left open and L.V. winding is connected to the supply, normal voltage and

frequency. Hence there is no Secondary Current and no Secondary copper Loss. Since we are not

loading the no load current is very small. (which is measured by a/m). Hence Cu loss in Primary

Winding Is negligibly small and nil in Secondary (open circuit). Hence w/m reading represents

practically the core loss under no load condition, which is the same for all loads.

Fig.35.1OC test set up

A W/M, V/M and A/M are connected in the l.v. side of Primary Winding as shown in Fig.35.1

With normal voltage applied to the Primary, normal flux will be set up in the core, hence normal

iron losses will occur, which are recorded by w/m. If meters are on the l.v. side, low range

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meters can be used and is safe to operate on lv side. Hence OC test is normally conducted on lv

side.

Readings - Vo , Io, Wo

Wo = Vo Io Cosφo ∴

Iom = Io Sin φo

Iw = Io Cos φo ∴Xm = Ro =

Hence, Ro ,Xmcan be found

Short Circuit Test:

Low voltage winding is shorted. A variable voltage is applied to the other end, so that rated

current is passing through the winding. In this test, the applied voltage is a small percentage of

the normal voltage, the mutual flux produced is also a small % of its normal value. Hence, core

losses are very small. So w/m reading represents the full load Cu loss for the whole Transformer.

Fig.35.2 SC test set up

Readings - Vsc, Isc, Wsc,

Z01 =

Wsc= Isc2 R01

∴ R01 =

X01 =

R01 = R1 + R2’

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If R1 can be measured, R2’ can be found.

Hence all equivalent circuit parameters are obtained by OC and SC test. Efficiency at any load

can be predetermined from these equivalent circuit parameters.


Q1. O.C. and S.C. test readings of a 4KVA, 200V/400V, 50Hz, 1Q Transformer are

O.C. test: 200 V, 0.7 A, 70 W = on l.v side

S.C. test: 15 V, 10 A, 85 W = on h.v. side

Find the i) equivalent circuit constants ii) ηfor 80% lagging pf load iii) η At half full load

and at 0.8 pf load .Draw the equivalent circuit referred to Primary

Solution:

O.C. test

VoIo Cosφo = Wo ∴ = 0.5

Iom = Io Sin φo = 0.61 A

Iol = Io Cos φo = 0.35 A

Ro = = 571.4 Ω

Xm = = 330 Ω

S.C. test on Sec. Side

Z02= = = 1.5 Ω Z01= = 0.375 Ω

Isc2 R02 = Wsc ∴R02= = 0.85 Ω

R01 (pri.) = = = 0.21 Ω

X01 (pri.) = = 0.31 Ω

F.L. sec Current =

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F.L. o/p at 0.8 pf = 40 x x 0.8 = 3200 W

F.L. losses = 70 + 85 = 155 W

∴F.L.i/p = 3200 + 155 = 3355

Half Load

Q2 . A 20 KVA, 2000 / 200 V, Transformer Pri. Resistance and reactance of 2.3 Ω and 4.2 Ω

resp. Corresponding Sec. values are 0.025Ω and 0.04Ω. Open circuit loss is 200 W.

Determine (i) equivalent Resist. And react. Ref. to primary and sec. ii) η at 0.8 pf lagging.

Solution:

R1 = 2.3 Ω, X1 = 4.2 Ω, R2 = 0.025 Ω, X2 = 0.04 Ω

290

Q3 . Calculate the (a) full load η at 0.8 pf lag (b) Terminal voltage when supplying full load at

0.8 pf (lag) for an input voltage of 500V. The result of O.C. and S.C. test on 5 KVA, 500 /

250 V, 50 Hz, 1 φ Transformer are as follows.

Solution:

O.C.: 500 V, 1 A, 50 W

S. C.: 15 V, 6 A, 21.6 W

(a)

Cu loss at 6 A = 21.6 W

Q4. O.C. and S.C. tests on a 5 KVA, 200 / 400 V, 50 Hz Transformer gave the following results.

O.C: 200 V, 1 A, 100 W (lv side), S. C.: 15 V, 10 A, 85 W (hv side).Draw the equivalent

circuit referred to primary and put all values.

Solution:

From OC test:

V1 = Vo = 200V , Io = 1 A, Wo = 100 W

Wo = Vo IoCos φo∴ Cos φo = 0.5

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Sin φo = 0.866

,

SC test:

, ,


1. During short circuit test, the iron losses are negligible because

a. The current on the secondary side is negligible

b. The voltage on the secondary side does not vary

c. The voltage applied on the primary side is low

d. The full load current is not supplied to the transformer

2. No load test on a transformer is carried out to find

a. Copper loss

b. magnetizing current

c. Magnetizing current and loss

d. Efficiency of the transformer

3. During testing of a transformer

a. Both SC and OC tests are performed at rated current

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b. Both SC and OC tests are performed at rated voltage

c. SC test is performed at rated voltage and OC test is performed at rated current

d. SC test is performed at rated current and OC test is performed at rated voltage

4. At relatively light loads, transformer efficiency is low because

a. secondary output is low

b. transformer losses are high

c. fixed loss is high in proportion to the output

d. Cu loss is small.

5. The short circuit test on a transformer gives

a. Equivalent resistance and leakage reactance b. Magnetizing current and core loss at rated voltage c. Copper loss d. Both A and C

EXERCISES

1. A 50kVA, single phase transformer has 600 turns on the primary winding and 40 turns on the

secondary winding. The primary winding is connected to a 2.2kV, 50Hz supply, Determine

1. Secondary voltage at no laod

2. Primary and secondary currents at full load

2. A single phase transformer has a primary voltage of 230V. No load primary current is 5A, no

load pf is 0.25. Number of primary turns are 200 and frequency is 50Hz.

Calculate i) Maximum value of flux in the core ii) core loss iii) Magnetizing current

3. A 30kVA, 2400/ 120V 50Hz transformer has high voltage winding resistance of 0.1Ω and

leakage reactance of 0.22Ω. The low voltage winding resistance is 0.035Ω and leakage

reactance is 0.012Ω. Calculate

1. Equivalent reactance as referred to both primary and secondary

2. Equivalent impedance as referred to both primary and secondary

3. Copper loss at full load

4. Copper loss at half load

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4. A resistance connected across the secondary of an ideal transformer has a value of 800Ω as

referred to the primary. The same resistance when connected across the primary has a value

of 3.125Ω as referred to the secondary. Find the turns ratio of transformer.

5. In a 25 KVA, 2000/200V, transformer, the F.L iron and copper losses are 350W and 400W

respectively. Calculate the at upf a) at F.L b) H.L c) Determine the load KVA for

efficiency max d)value of efficiency max and Pcu and Pi in this case.

Ans: a) efficiency fl = 97.08 b) efficiency h.F.L= 96.52 c) 23.38KVA

d) 97.09%, Pcu = Pi = 350W

6. Obtain an equivalent circuit of a 200/400V, 50Hz, 1φ transformer from the following test

data

O.C. Test: 200V, 0.7A, 70W (on L.V. side)

S.C. Test: 15V, 10A, 85W (on H.V. side)

6. A 50kVA, 2200/220V, 50Hz, 1φ transformer gave the following test result,

O.C. Test: 2200V, 0.5A, 500W (on H.V. side)

S.C. Test: 100V, 20A, 500W (on H.V. side)

Calculate

i) Efficiency at full load 0.8 lag p.f.

ii) Maximum efficiency at 0.8 p.f. and load at maximum efficiency

7. An open circuit and short circuit test on a 5kVA, 200/400V, 50Hz, 1φ transformer gave the

following test result

O.C. Test: 200V, 1A, 100W (on L.V. side)

S.C. Test: 15V, 10A, 85W (with primary S.C.)

Draw the equivalent circuit diagram referred to primary

8. 250kVA, 50Hz, 1φ transformer has 98.135% efficiency at full load 0.8 lag p.f. The efficiency

at full load, 0.8lag p.f. is 97.751%. Calculate the iron loss and full load copper loss

9. Draw a phasor diagram for single phase transformer for lagging and leading load condition

10. Explain how to get approximate equivalent circuit of a single phase transformer by open

circuit and short circuit test.

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11. State the characteristics of an ideal transformer

12. What are the losses in transformer? explain why rating of transformer is in kVA not in kW?

13. Derive the condition of maximum efficiency of transformer

14. Derive EMF equation of transformer

15. Explain construction and working principle of single phase transformer


1. Draw a phasor diagram for single phase transformer for lagging and leading load condition

2. Explain how to get approximate equivalent circuit of a single phase transformer by open

circuit and short circuit test.

3. State the characteristics of an ideal transformer

4. What are the losses in transformer? explain why rating of transformer is in kVA not in kW?

5. Derive the condition of maximum efficiency of transformer

6. Derive EMF equation of transformer

7. Explain construction and working principle of single phase transformer

8. Draw phasor diagram of a single phase transformer for lagging load

9. Draw phasor diagram of a single phase transformer for upf load

10.Draw phasor diagram of a single phase transformer for lagging, leading and upf load

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ANSWER KEY

Lec

28

Lec

29

Lec

30

Lec

31

Lec

33

Lec

35

Q1. b d c d c c

Q2. b a d d d c

Q3. d b a b c d

Q4. b c b d d c

Q5. c a -- c c d

Q6. -- b -- -- -- --

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MODULE 5

ELECTRONICS

297

LECTURE-36

PN JUNCTION DIODE

OBJECTIVES

• To learn about semiconductor materials and their properties

• To explain PN junction diode ; its construction and working

• To understand V-I characteristics of pn junction diode

36.1 Semiconductor Basics

Semiconductors materials such as silicon (Si), germanium (Ge) and gallium arsenide (GaAs), have electrical properties between those of a "conductor" and an "insulator". They are not good conductors nor good insulators (hence their name "semi"-conductors). They have very few "free electrons" because their atoms are closely grouped together in a crystalline pattern called a "crystal lattice". However, their ability to conduct electricity can be greatly improved by adding certain "impurities" to this crystalline structure thereby, producing more free electrons than holes or vice versa.

By controlling the amount of impurities added to the semiconductor material it is possible to control its conductivity. These impurities are called donors or acceptors depending on whether they produce electrons or holes respectively. This process of adding impurity atoms to semiconductor atoms (the order of 1 impurity atom per 10 million (or more) atoms of the semiconductor) is called Doping.

The most commonly used semiconductor basic material by far is silicon. Silicon has four valence electrons in its outermost shell which it shares with its neighbouring silicon atoms to form full orbital of eight electrons. The structure of the bond between the two silicon atoms is such that each atom shares one electron with its neighbour making the bond very stable.

As there are very few free electrons available to move around the silicon crystal, crystals of pure silicon (or germanium) are therefore good insulators, or at the very least very high value resistors.

Silicon atoms are arranged in a definite symmetrical pattern making them a crystalline solid structure. A crystal of pure silica (silicon dioxide or glass) is generally said to be an intrinsic crystal (it has no impurities) and therefore has no free electrons.

A Silicon Atom Structure

298

Fig. 36.1

Fig 36.1 shows the structure and lattice of a 'normal' pure crystal of Silicon.

N-type Semiconductor

In order for silicon crystal to conduct electricity, we need to introduce an impurity atom such as Arsenic, Antimony or Phosphorus into the crystalline structure making it extrinsic (impurities are added). These atoms have five outer electrons in their outermost orbital to share with neighbouring atoms and are commonly called "Pentavalent" impurities.

This allows four out of the five orbital electrons to bond with its neighbouring silicon atoms leaving one "free electron" to become mobile when an electrical voltage is applied (electron flow). As each impurity atom "donates" one electron, pentavalent atoms are generally known as "donors".

Antimony (Sb) or Phosphorus (P), are frequently used as a pentavalent additive to the silicon. The resulting semiconductor basics material has an excess of current-carrying electrons, each with a negative charge, and is therefore referred to as an "N-type" material with the electrons called "Majority Carriers" while the resulting holes are called "Minority Carriers".

Fig 36.2

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Fig. 36.2 shows the structure and lattice of the donor impurity atom Antimony.

P-Type Semiconductor

We can also introduce a "Trivalent" (3-electron) impurity into the crystalline structure, such as Aluminium, Boron or Indium, which have only three valence electrons available in their outermost orbital, the fourth closed bond cannot be formed. Therefore, a complete connection is not possible, giving the semiconductor material an abundance of positively charged carriers known as "holes" in the structure of the crystal where electrons are effectively missing.

As there is now a hole in the silicon crystal, a neighbouring electron is attracted to it and will try to move into the hole to fill it. However, the electron filling the hole leaves another hole behind it as it moves. This in turn attracts another electron which in turn creates another hole behind it, and so forth giving the appearance that the holes are moving as a positive charge through the crystal structure (conventional current flow). This movement of holes results in a shortage of electrons in the silicon turning the entire doped crystal into a positive pole. As each impurity atom generates a hole, trivalent impurities are generally known as "Acceptors" as they are continually "accepting" extra or free electrons.

Boron (symbol B) is commonly used as a trivalent additive as it has only five electrons arranged in three shells around its nucleus with the outermost orbital having only three electrons. The doping of Boron atoms causes conduction to consist mainly of positive charge carriers resulting in a "P-type" material with the positive holes being called "Majority Carriers" while the free electrons are called "Minority Carriers". Then a semiconductor basics material is classed as P-type when its acceptor density is greater than its donor density. Therefore, a P-type semiconductor has more holes than electrons.

Fig 36.3

Fig. 36.3 shows the structure and lattice of the acceptor impurity atom Boron.

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36.2 PN junction

N and P-type materials do very little on their own as they are electrically neutral, but when we join (or fuse) them together these two materials behave in a very different way producing what is generally known as a PN Junction.

When the N and P-type semiconductor materials are first joined together a very large density gradient exists between both sides of the junction so some of the free electrons from the donor impurity atoms begin to migrate across this newly formed junction to fill up the holes in the P-type material producing negative ions.

However, because the electrons have moved across the junction from the N-type silicon to the P-type silicon, they leave behind positively charged donor ions (ND) on the negative side and now the holes from the acceptor impurity migrate across the junction in the opposite direction into the region where there are large numbers of free electrons. As a result, the charge density of the P-type along the junction is filled with negatively charged acceptor ions (NA), and the charge density of the N-type along the junction becomes positive. This charge transfer of electrons and holes across the junction is known as diffusion.

This process continues back and forth until the number of electrons which have crossed the junction have a large enough electrical charge to repel or prevent any more charge carriers from crossing over the junction. Eventually a state of equilibrium (electrically neutral situation) will occur producing a "potential barrier" zone around the area of the junction as the donor atoms repel the holes and the acceptor atoms repel the electrons.

Since no free charge carriers can rest in a position where there is a potential barrier, the regions on either sides of the junction no become completely depleted of any more free carriers in comparison to the N and P type materials further away from the junction. This area around the junction is now called the Depletion Layer.

The PN junction

Fig. 36.4

301

The total charge on each side of the junction must be equal and opposite to maintain a neutral charge condition around the junction. If the depletion layer region has a distance D, it therefore must therefore penetrate into the silicon by a distance of Dp for the positive side, and a distance of Dn for the negative side giving a relationship between the two of Dp.NA = Dn.ND in order to maintain charge neutrality also called equilibrium.

PN junction Distance

Fig. 36.5

As the N-type material has lost electrons and the P-type has lost holes, the N-type material has become positive with respect to the P-type. Then the presence of impurity ions on both sides of the junction causes an electric field to be established across this region with the N-side at a positive voltage relative to the P-side. The problem now is that a free charge requires some extra energy to overcome the barrier that now exists for it to be able to cross the depletion region junction.

A suitable positive voltage (forward bias) applied between the two ends of the PN junction can supply the free electrons and holes with the extra energy. The external voltage required to overcome this potential barrier that now exists is very much dependent upon the type of semiconductor material used and its actual temperature. Typically at room temperature the voltage across the depletion layer for silicon is about 0.6 - 0.7 volts and for germanium is about 0.3 - 0.35 volts. This potential barrier will always exist even if the device is not connected to any external power source.

The significance of this built-in potential across the junction is that it opposes both the flow of holes and electrons across the junction and is why it is called the potential barrier. In practice, a PN junction is formed within a single crystal of material rather than just simply joining or fusing together two separate pieces. Electrical contacts are also fused onto either side of the crystal to enable an electrical connection to be made to an external circuit. Then the resulting device that has been made is called a PN junction Diode.

The Junction Diode

If electrical connections is made at the ends of both the N-type and the P-type materials and then connected to a battery source, an additional energy source exists to overcome the barrier

302

resulting in free charges being able to cross the depletion region from one side to the other. The behaviour of the PN junction with regards to the potential barrier width produces an asymmetrical conducting two terminal device, better known as the Junction Diode.

A diode is one of the simplest semiconductor devices, which has the characteristic of passing current in one direction only. However, unlike a resistor, a diode does not behave linearly with respect to the applied voltage as the diode has an exponential V-I relationship and therefore we can not described its operation by simply using an equation such as Ohm's law.

If a suitable positive voltage (forward bias) is applied between the two ends of the PN junction, it can supply free electrons and holes with the extra energy they require to cross the junction as the width of the depletion layer around the PN junction is decreased. By applying a negative voltage (reverse bias) results in the free charges being pulled away from the junction resulting in the depletion layer width being increased. This has the effect of increasing or decreasing the effective resistance of the junction itself allowing or blocking current flow through the diode.

Then the depletion layer widens with an increase in the application of a reverse voltage and narrows with an increase in the application of a forward voltage. This is due to the differences in the electrical properties on the two sides of the PN junction resulting in physical changes taking place

Before we can use PN junction as a practical device or as a rectifying device we need to firstly bias the junction, ie connect a voltage potential across it. On the voltage axis above, "Reverse Bias" refers to an external voltage potential which increases the potential barrier. An external voltage which decreases the potential barrier is said to act in the "Forward Bias" direction.

There are two operating regions and three possible "biasing" conditions for the standard Junction Diode and these are:

• Zero Bias - No external voltage potential is applied to the PN-junction. • Reverse Bias - The voltage potential is connected negative, (-ve) to the P-type material

and positive, (+ve) to the N-type material across the diode which has the effect of increasing the PN-junction width.

• Forward Bias - The voltage potential is connected positive, (+ve) to the P-type material and negative, (-ve) to the N-type material across the diode which has the effect of decreasing the PN-junction width.

Zero Biased Junction Diode

When a diode is connected in a Zero Bias condition, no external potential energy is applied to the PN junction. However if the diodes terminals are shorted together, a few holes (majority carriers) in the P-type material with enough energy to overcome the potential barrier will move across the junction against this barrier potential. This is known as the "Forward Current" and is referenced as IF

303

Likewise, holes generated in the N-type material (minority carriers), find this situation favourable and move across the junction in the opposite direction. This is known as the "Reverse Current" and is referenced as IR. This transfer of electrons and holes back and forth across the PN junction is known as diffusion, as shown below.

Fig. 36.6

The potential barrier that now exists discourages the diffusion of any more majority carriers across the junction. However, the potential barrier helps minority carriers (few free electrons in the P-region and few holes in the N-region) to drift across the junction. Then an "Equilibrium" or balance will be established when the majority carriers are equal and both moving in opposite directions, so that the net result is zero current flowing in the circuit. When this occurs the junction is said to be in a state of "Dynamic Equilibrium".

The minority carriers are constantly generated due to thermal energy so this state of equilibrium can be broken by raising the temperature of the PN junction causing an increase in the generation of minority carriers, thereby resulting in an increase in leakage current but an electric current cannot flow since no circuit has been connected to the PN junction.

Reverse Biased Junction Diode

When a diode is connected in a Reverse Bias condition, a positive voltage is applied to the N-type material and a negative voltage is applied to the P-type material. The positive voltage applied to the N-type material attracts electrons towards the positive electrode and away from the junction, while the holes in the P-type end are also attracted away from the junction towards the negative electrode.

The net result is that the depletion layer grows wider due to a lack of electrons and holes and presents a high impedance path, almost an insulator. The result is that a high potential barrier is created thus preventing current from flowing through the semiconductor material.

304

Reverse Biased Junction Diode showing an Increase in the Depletion Layer

Fig. 36.7

This condition represents a high resistance value to the PN junction and practically zero current flows through the junction diode with an increase in bias voltage. However, a very small leakage current does flow through the junction which can be measured in microamperes, (µA). One final point, if the reverse bias voltage Vr applied to the diode is increased to a sufficiently high enough value, it will cause the PN junction to overheat and fail due to the avalanche effect around the junction. This may cause the diode to become shorted and will result in the flow of maximum circuit current and this shown as a step downward slope in the reverse static characteristics curve below.

Reverse Characteristics Curve for a Junction Diode

Fig. 36.8

Sometimes this avalanche effect has practical applications in voltage stabilising circuits where a series limiting resistor is used with the diode to limit this reverse breakdown current to a preset maximum value thereby producing a fixed voltage output across the diode. These types of diodes are commonly known as Zener Diodes.

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Forward Biased Junction Diode

When a diode is connected in a Forward Bias condition, a negative voltage is applied to the N-type material and a positive voltage is applied to the P-type material. If this external voltage becomes greater than the value of the potential barrier, approx. 0.7 volts for silicon and 0.3 volts for germanium, the potential barriers opposition will be overcome and current will start to flow.

This is because the negative voltage pushes or repels electrons towards the junction giving them the energy to cross over and combine with the holes being pushed in the opposite direction towards the junction by the positive voltage. This results in a characteristics curve of zero current flowing up to this voltage point, called the "knee" on the static curves and then a high current flow through the diode with little increase in the external voltage as shown below.

Forward Characteristics Curve for a Junction Diode

Fig. 36.9

The application of a forward biasing voltage on the junction diode results in the depletion layer becoming very thin and narrow which represents a low impedance path through the junction thereby allowing high currents to flow. The point at which this sudden increase in current takes place is represented on the static V-I characteristics curve above as the "knee" point.

Forward Biased Junction Diode showing a Reduction in the Depletion Layer

Fig. 36.10

306

This condition represents the low resistance path through the PN junction allowing very large currents to flow through the diode with only a small increase in bias voltage. The actual potential difference across the junction or diode is kept constant by the action of the depletion layer at approximately 0.3V for germanium and approximately 0.7 V for silicon junction diodes.

Since the diode can conduct "infinite" current above this knee point as it effectively becomes a short circuit, therefore resistors are used in series with the diode to limit its current flow. Exceeding its maximum forward current specification causes the device to dissipate more power in the form of heat than it was designed for resulting in a very quick failure of the device.

Junction Diode Symbol and Static V-I Characteristics

Fig. 36.11


1. The potential barrier at a PN junction is due to the charges on either side of the junction.

These charges are

a. Minority carriers

b. Majority carriers

c. Both majority and minority carriers

d. Fixed donor and acceptor ions

2. In a PN junction diode, holes diffuse from the P region to the N region because

307

a. The free electrons in the N region attract them.

b. They are swept across the junction by the potential difference.

c. There is greater concentration of holes in the P region as compared to N region


3. In a semiconductor diode the barrier potential offers opposition to only

a) Majority carriers in both regions.

b) Minority carriers in both regions

c) Free electrons in the N region

d) Holes in the P region

4. When forward bias is applied to a junction diode, it

a) Increases the potential barrier

b) Decreases the potential barrier

c) Reduces the majority-carrier current to zero

d) Reduces the minority carrier current to zero

5. When we apply reverse bias to a junction diode, it

a) Lowers the potential barrier

b) Raises the potential barrier

c) Greatly increases the minority-carrier current

d) Greatly increases the majority-carrier current

6. The number of minority carriers crossing a junction depends on the

a) Concentration of doping impurities

b) Magnitude of potential barrier

c) Magnitude of forward bias voltage

d) Rate of thermal generation of electron-hole pair

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LECTURE-37

HALF WAVE RECTIFIER

OBJECTIVES

• To learn about rectifiers and its types

• To explain working of half wave rectifier

• To study the performance parameters of half wave rectifier

37.1 Rectifiers

We know that a semiconductor signal diode will only conduct current in one direction from its anode to its cathode (forward direction), but not in the reverse direction. A widely used application of this feature is in the conversion of an alternating voltage (AC) into a continuous voltage (DC). In other words, Rectification.

There are many possible ways to construct rectifier circuits using diodes. The three basic types of rectifier circuits are:

• The Half Wave Rectifier

• The Full Wave Rectifier

37.2 Half wave Rectifier – Its working

A simple half-wave rectifier is shown in Figure 37.1. It consists of a resistive load; a rectifying element i.e., p-n junction diode and the source of a.c. voltage, all connected is series. The a.c. voltage is applied to the rectifier circuit using step-down transformer.

Fig. 37.1

The input to the rectifier circuit is V = Vm sin ωt, where Vm is the peak value of secondary a.c. voltage. For the positive half-cycle of input a.c. voltage, the diode is forward biased and hence it conducts. Now a current flows in the circuit and there is a voltage drop across RL. The waveform of the diode current (or) load current is shown in figure 37.2.

309

For the negative half-cycle of input, the diode D is reverse biased and hence it does not conduct. Now no current flows in the circuit i.e., i=0 and Vo=0. Thus for the negative half-cycle no power is delivered to the load.

Fig. 37.2

37.3 Performance parameters

For a half wave rectifier, the following parameters are to be analyzed. i) DC output current ii) DC output voltage iii) RMS current iv) RMS voltage

v) Rectifier Efficiency (η)

vi) Ripple Factor (γ) vii) Transformer Utilization Factor (TUF) viii) Peak Factor

Let a sinusoidal voltage Vi be applied to the input of the rectifier.

Then Vi = Vm sin ωt

where Vm is the maximum value of the secondary voltage.

Let the diode be idealized to piece-wise linear approximation with resistance Rf in the forward direction i.e., in the ON state and Rr (=∞) in the reverse direction i.e., in the OFF state.

Now the current ‘i’ in the diode (or) in the load resistance RL is given by

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i) Average (or) DC Output Current (Iav or Idc)

The average dc current Idc is given by

ii) Average (or) DC Output Voltage (Vav or Vdc)

The average dc voltage is given by

311

iii) RMS Output Current (Irms)

The value of the rms current is given by

iv) RMS Output Voltage (Vrms)

RMS voltage across the load is given by

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The rectifier efficiency is defined as the ratio of dc output power to the ac input power ie.

Theoretically the maximum value of rectifier efficiency of a half wave rectifier is 40.6%

313

vi) Ripple Factor (γ)

The ripple factor, γ is given by

vii) Transformer Utilization Factor

The d.c. power to be delivered to the load in a rectifier circuit decides the rating of the

transformer used in the circuit. So, transformer utilization factor is defined as

The factor which indicates how much is the utilization of the transformer in the circuit is called Transformer Utilization Factor (TUF). The a.c. power rating of transformer = Vrms Irms

The secondary voltage is purely sinusoidal hence its rms value is 2

1times maximum while the

current is half sinusoidal hence its rms value is 2

1of the maximum.

The dc power delivered to the load,

314

viii) Peak Inverse Voltage (PIV)

It is defined as the maximum reverse voltage that a diode can withstand without destroying the junction. The peak inverse voltage across the diode is the peak of the negative half cycle. For a half wave rectifier, PIV = Vm

ix) Form factor (F)

The Form factor F is defined as

F = rms value / average value

x) Peak Factor (P)

The Peak factor P is defined as

P = Peak value / rms value

37.4 Disadvantages of Half-Wave Rectifier

1. The ripple factor is high. 2. The efficiency is low. 3. The Transformer Utilization Factor is low. Because of all these disadvantages, the half-wave rectifier circuit is normally not used as a power rectifier circuit.

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1. In a half wave rectifier, the load current flows for

a. The complete cycle of the input signal

b. Only for one half cycle of the input signal

c. Less than half cycle of the input signal

d. More than half cycle but less than the complete cycle of the input signal

2. If Vm is the peak voltage across the secondary of the transformer in a half-wave rectifier

(without any filter circuit), then the maximum voltage on the reverse –biased diode is

a. Vm b. 2Vm c. 0.5 Vm d. None of the above

3. The ratio of rms value of ac component to the dc component in the output is called

a. Average value b. TUF c. Ripple factor d. Efficiency

4. The ratio of the dc output power to ac input power

a. Average value b. TUF c. Ripple factor d. Efficiency

5. This parameter can be used to determine the rating of a transformer secondary. a. Average value b. TUF c. Ripple factor d. Efficiency

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LECTURE - 38

FULL WAVE RECTIFIERS

OBJECTIVES

• To study full wave rectifiers

• To discuss types of full wave rectifiers

• To analyse performance parameters of full wave rectifiers

38.1 Full Wave Rectifier

A full-wave rectifier converts an ac voltage into a pulsating dc voltage using both half cycles of the applied ac voltage. In order to rectify both the half cycles of ac input, two or four diodes are used in this circuit. There are two types of full wave rectifiers:

1. Centre-tap full wave rectifier 2. Bridge type full wave rectifier

38.2 Center Tap Full Wave Rectifier Circuit

In order to rectify both the half cycles of ac input, two diodes are used in this circuit. The diodes feed a common load RL with the help of a center-tap transformer. A center-tap transformer is the one which produces two sinusoidal waveforms of same magnitude and frequency but out of phase with respect to the ground in the secondary winding of the transformer. The full wave rectifier is shown in the Fig 38.1

Fig. 38.1

When point A of the transformer is positive with respect to point C, diode D1 conducts in the forward direction as indicated by the arrows. When point B is positive (in the negative half of the cycle) with respect to point C, diode D2 conducts in the forward direction and the current flowing through resistor R is in the same direction for both half-cycles. As the output voltage across the

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resistor R is the phasor sum of the two waveforms combined, this type of full wave rectifier circuit is also known as a "bi-phase" circuit.

The waveforms are shown in Fig 38.2

Fig. 38.2

38.3 Full Wave Bridge Rectifier

Another type of circuit that produces the same output waveform as the full wave rectifier circuit above is that of the Full Wave Bridge Rectifier.

The bridge rectifier circuit has four diodes connected to form a bridge. The ac input voltage us applied to diagonally opposite ends of the bridge. The load resistance is connected between the other two ends of the bridge. The bridge rectifier circuits is shown in Fig 38.3.

Fig. 38.3

The four diodes labelled D1 to D4 are arranged in "series pairs" with only two diodes conducting current during each half cycle. During the positive half cycle of the supply, diodes D1 and D2 conduct in series while diodes D3 and D4 are reverse biased and the current flows through the load as shown in Fig. 38.4

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Fig. 38.4

During the negative half cycle of the supply, diodes D3 and D4 conduct in series, but diodes D1 and D2 switch "OFF" as they are now reverse biased. The current flowing through the load is the same direction as before, as shown in Fig 38.5

Fig. 38.5

Advantages of Bridge rectifier circuit:

1) No center-tapped transformer is required. 2) The TUF is considerably high. 3) PIV is reduced across the diode.

Disadvantages of Bridge rectifier circuit:

The only disadvantage of bridge rectifier is the use of four diodes as compared to two diodes for center-tapped FWR. This reduces the output voltage.

38.4 Performance parameters of Full wave rectifiers

Let a sinusoidal voltage Vi be applied to the input of the rectifier.

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Then Vi = Vm sin ωt

where Vm is the maximum value of the secondary voltage.

i) Average (or) DC Output Current (Iav or Idc)

The average dc current Idc is given by

Substituting the value of Im, we get,

This is double of that of half wave rectifier.

ii) Average (or) DC Output Voltage (Vav or Vdc)

The dc output voltage is given by

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If RL >>Rf, then

iii) RMS Output Current (Irms)

The value of the rms current is given by

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iv) RMS Output Voltage (Vrms)

RMS voltage across the load is given by

If RL >> Rf, then,


The rectifier efficiency is defined as the ratio of dc output power to the ac input power

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Theoretically, the maximum value of rectifier efficiency of a full wave rectifier is 81.2%

when 0=L

f

R

R.

Thus full wave rectifier has efficiency twice that of half wave rectifier.

vi) Ripple Factor (γ)

The ripple factor, γ is given by

vii) Transformer Utilization factor (TUF)

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The average TUF in full wave rectifying circuit is determined by consideribg primary and secondary winding separately. There are two secondaries. Each secondary has a TUF of 0.287.

TUF of primary = Pdc / Volt-Amp rating of primary

If RL >> Rf, then

viii) Peak Inverse Voltage (PIV)

Peak inverse voltage is the maximum possible voltage across a diode when it is reverse biased.

In case of centre-tap FWR, when D1 is forward biased, ie. conducting and D2 is reverse biased, ie. non-conducting, voltage Vm is developed across load resistor RL. Now, voltage across D2 is the sum of voltages across load resistor RL and voltage across the lower half of transformer secondary Vm. Hence PIV of diode D2 = Vm + Vm = 2Vm.

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Similarly, PIV of diode D2 is 2Vm.

For a bridge type FWR, consider Fig. 38.6.

Fig. 38.6

It shows the circuit conditions during the positive half cycle. The load and ground connections are removed because we are concerned with the diode conditions only. In this circuit, diodes D1and D3 are forward biased and act like closed switches. They can be replaced with wires. Diodes D2 and D4 are reverse biased and act like open switches. The circuit of figure 19 is redrawn below. We can see that both diodes are reverse biased, in parallel, and directly across the secondary winding. The peak inverse voltage is therefore equal to Vm.

Therefore, Peak inverse voltage = Vm

ix) Form Factor (F)

The Form Factor F is defined as

F = rms value / average value

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x) Peak Factor (P)

The Peak Factor P is defined as

P = peak value / rms value

38.5 Comparison between Centre-tap and Bridge type FWR

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38.6 Comparison of HWR AND FWR


1. In a full wave rectifier , the current in each of the diodes flows for

a. The complete cycle of the input signal

b. Half cycle of the input signal

c. Less than half cycle of the input signal

d. Zero time

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2. In a center-tap full wave rectifier, Vm is the peak voltage between the center tap and one

end of the secondary. The maximum voltage across the reverse - biased diode is


3. The efficiency of a full wave rectifier is

a. 81.2% b. 40.6% c. 63.4% d. 48.2%

4. If a peak voltage of Vm is applied at the primary of the transformer, the PIV rating for a full wave bridge rectifier is


5. The form factor of a full wave recified output signal is

a. 1.414 b. 2 c. 0.707 d. 1.11

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LECTURE 39

FILTERS & BJT CONSTRUCTION

OBJECTIVES

• To study filters

• To understand construction and operation of bipolar junction transistor (BJT)

39.1 Filters

A power supply must provide ripple free source of power from an A.C. line. But the output of a rectifier circuit contains ripple components in addition to a D.C. term. It is necessary to include a filter between the rectifier and the loads in order to eliminate these ripple components. Ripple components are high frequency A.C. Signals in the D.C output of the rectifier. These are not desirable, so they must be filtered. So filter circuits are used. Many types of passive filters are in use.

Two such basic filters are:

1. Inductor filter

2. Capacitor filter

Series Inductor filter

The working of series inductor filter depends on the inherent property of the inductor to oppose any variation in current intend to take place. Fig 39.1 shows a series inductor filter connected at the output of a FWR. Here the reactance of the inductor is more for ac components and it offers more opposition to them. At the same time it provides no impedance for d.c. component. Therefore the inductor blocks a.c. components in the output of the rectifier and allows only d.c. component to flow through RL.

Fig. 39.1

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Fig. 39.2

The action of an inductor depends upon the current through it and it requires current to flow at all time. Therefore filter circuits consisting inductors can only be used together with full wave rectifiers. In inductor filter an increase in load current will improve the filtering action and results in reduced ripple. Series inductor filters are used in equipment of high load currents. The ripple factor in series inductor filter

Advantages

• Sudden changes in current is smoothen out

• Improved filtering action at high load currents across the inductor.

Disadvantages

• Reduced output voltage due to the drop

• Bulky and large in size

• Note suite for HWR

Shunt Capacitor filter

This type of filter consists of large value of capacitor connected across the load resistor RL as shown in figure 39.3. This capacitor offers a low reactance to the a.c. components and very high impedance to d.c. so that the a.c. components in the rectifier output find low reactance path through capacitor and only a small part flows through RL, producing small ripple at the output as shown in figure. Here Xc (=1/2πfC, the impedance of capacitor) should be smaller than RL. Because, current should pass through C and C should get charged. If C value is very small, Xc will be large and hence current flows through RL only and no filtering action takes place. The capacitor C gets charged when the diode (in the rectifier) is conducting and gets discharged

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(when the diode is not conducting) through RL. When the input voltage = is greater than the capacitor voltage, C gets charged. When the input voltage is less than that of the capacitor voltage, C will discharge through RL. The stored energy in the capacitor maintains the

load voltage at a high value for a long period. The diode conducts only for a short interval of high current. The waveforms are as shown in figure 39.4. Capacitor opposes sudden fluctuations in voltage across it. So the ripple voltage is minimized.

Fig 39.3

Fig. 39.4

The discharging of the capacitor depends upon the time constant C.RL. Hence the smoothness and the magnitude of output voltage depend upon the value of capacitor C and RL. As the value of C increases the smoothness of the output also increases. But the maximum value of the capacitor is limited by the current rating of the diode. Also decrease in the value of RL increases the load current and makes the time constant smaller. These types of filters are used in circuits with small load current like transistor radio receivers, calculators, etc.

The ripple factor in shunt capacitor filter is

Advantages

• Low cost

• Small size and weight

• Good characteristics

• Can be connected for both HW and FW rectifiers

• Improved d.c. output

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Disadvantages

• Capacitor draws more current

39.2 Bipolar Junction Transistor

Construction

A BJT is formed by joining three sections of semiconductor material, each with a different doping concentration. The three sections can be either a thin n region sandwiched between p+ and p layers, or a p region between n and n+ layers, where the superscript plus indicates more heavily doped material. The resulting BJTs are called pnp and npn transistors, respectively. Figure 39.5 illustrates the approximate construction, symbols, and nomenclature for the two types of BJTs.

Fig. 39.5

Operation

The operation of the npn BJT may be explained by considering the transistor as consisting of two back-to-back pn junctions. The base-emitter (BE) junction acts very much as a diode when it is forward-biased; thus, one can picture the corresponding flow of hole and electron currents from base to emitter when the collector is open and the BE junction is forward-biased, as depicted in Figure 39.6. Note that the electron current has been shown larger than the hole current, because of the heavier doping of the n side of the junction. Some of the electron-hole pairs in the base will recombine; the remaining charge carriers will give rise to a net flow of current from base to emitter. It is also important to observe that the base is much narrower than the emitter section of the transistor.

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Fig 39.6

Imagine, now, reverse-biasing the base-collector (BC) junction. In this case, an interesting phenomenon takes place: the electrons “emitted” by the emitter with the BE junction forward-biased reach the very narrow base region, and after a few are lost to recombination in the base, most of these electrons are “collected” by the collector. Figure 39.7 illustrates how the reverse bias across the BC junction is in such a direction as to sweep the electrons from the emitter into the collector. This phenomenon can take place because the base region is kept particularly narrow. Since the base is narrow, there is a high probability that the electrons will have gathered enough momentum from the electric field to cross the reverse-biased collector-base junction and make it into the collector. The result is that there is a net flow of current from collector to emitter (opposite in direction to the flow of electrons), in addition to the hole current from base to emitter. The electron current flowing into the collector through the base is substantially larger than that which flows into the base from the external circuit.

Fig 39.7

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One can see from Figure 39.7 that if KCL is to be satisfied, we must have

IE = IB + IC

The most important property of the bipolar transistor is that the small base current controls the amount of the much larger collector current

IC = β IB

where β is a current amplification factor dependent on the physical properties of the transistor. Typical values of β range from 20 to 200. The operation of a pnp transistor is completely analogous to that of the npn device, with the roles of the charge carriers (and therefore the signs of the currents) reversed. The symbol for a pnp transistor is shown in Figure 39.5.


1. The component used to reduce the ripple content of the rectified output is

a. Power supply b. Zener diode c. Pn junction diode d. Filter

2. An inductor filter is always connected in

a. Shunt b. Series c. any of the two

3. In a transistor with normal bias, the emitter junction

a. has a high resistance

b. has a low resistance

c. is reverse biased

d. emits such carriers into the base junction which are in majority

4. For transistor action

a. the collector must be more heavily doped than the emitter region

b. the collector base junction must be forward biased

c. base region must be very narrow

d. base region must be n-type

5. In a pnp transistor, electrons flow

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a. out of the transistor at the collector and base leads

b. into the transistor at the emitter and base leads

c. into the transistor at the collector and base leads

d. out of the transistor at the emitter and base leads

6. The emitter region in the pnp transistor is more heavily doped than the base region so

that

a. the flow across the base region will be mainly because of electrons

b. the flow across the base region will be mainly because of holes

c. recombination will be increased in base region

d. base current will be high

7. The arrowhead on the transistor symbol always points in the direction of

a. hole flow in the emitter region

b. electron flow in the emitter region

c. minority carrier flow in the emitter region

d. majority carrier flow in the emitter region

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LECTURE-40

BJT CONFIGURATIONS & CE CHARACTERISTICS

OBJECTIVES

• To understand different BJT configurations

• To study Common Emitter configuration characteristics

40.1 Bipolar Transistor Configurations

As the Bipolar Transistor is a three terminal device, there are basically three possible ways to connect it within an electronic circuit with one terminal being common to both the input and output. Each method of connection responding differently to its input signal within a circuit as the static characteristics of the transistor vary with each circuit arrangement.

(i) Common Base Configuration (ii) Common Emitter Configuration (iii)Common Collector Configuration

The Common Base (CB) Configuration

As its name suggests, in the Common Base or grounded base configuration, the base connection is common to both the input signal and the output signal with the input signal being applied between the base and the emitter terminals. The corresponding output signal is taken from between the base and the collector terminals as shown with the base terminal grounded or connected to a fixed reference voltage point.

The input current flowing into the emitter is quite large as its the sum of both the base current and collector current respectively therefore, the collector current output is less than the emitter current input resulting in a current gain for this type of circuit of "1" (unity) or less, in other words the common base configuration "attenuates" the input signal.

Fig. 40.1

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This type of amplifier configuration is a non-inverting voltage amplifier circuit, in that the signal voltages Vin and Vout are "in-phase". This type of transistor arrangement is not very common due to its unusually high voltage gain characteristics. Its output characteristics represent that of a forward biased diode while the input characteristics represent that of an illuminated photo-diode.

Also this type of bipolar transistor configuration has a high ratio of output to input resistance or more importantly "load" resistance (RL) to "input" resistance (Rin) giving it a value of "Resistance Gain". Then the voltage gain (Av) for a common base configuration is therefore given as:

Common Base Voltage Gain

Where: Ic/Ie is the current gain, alpha ( α ) and RL/Rin is the resistance gain.

The common base circuit is generally only used in single stage amplifier circuits such as microphone pre-amplifier or radio frequency amplifiers due to its very good high frequency response.

The Common Emitter (CE) Configuration

In the Common Emitter or grounded emitter configuration, the input signal is applied between the base, while the output is taken from between the collector and the emitter as shown. This type of configuration is the most commonly used circuit for transistor based amplifiers and which represents the "normal" method of bipolar transistor connection.

The common emitter amplifier configuration produces the highest current and power gain of all the three bipolar transistor configurations. This is mainly because the input impedance is low as it is connected to a forward-biased PN-junction, while the output impedance is high as it is taken from a reverse-biased PN-junction.

Fig. 40.2

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In this type of configuration, the current flowing out of the transistor must be equal to the currents flowing into the transistor as the emitter current is given as Ie = Ic + Ib. Also, as the load resistance ( RL ) is connected in series with the collector, the current gain of the common emitter transistor configuration is quite large as it is the ratio of Ic/Ib and is given the Greek symbol of Beta, ( β ). As the emitter current for a common emitter configuration is defined as Ie = Ic + Ib, the ratio of Ic/Ie is called Alpha, given the Greek symbol of α. Note: that the value of Alpha will always be less than unity.

Since the electrical relationship between these three currents, Ib, Ic and Ie is determined by the physical construction of the transistor itself, any small change in the base current ( Ib ), will result in a much larger change in the collector current ( Ic ). Then, small changes in current flowing in the base will thus control the current in the emitter-collector circuit. Typically, Beta has a value between 20 and 200 for most general purpose transistors.

By combining the expressions for both α and β, the mathematical relationship between these parameters and therefore the current gain of the transistor can be given as:

Where: "Ic" is the current flowing into the collector terminal, "Ib" is the current flowing into the base terminal and "Ie" is the current flowing out of the emitter terminal.

Then to summarize, this type of bipolar transistor configuration has a greater input impedance, current and power gain than that of the common base configuration but its voltage gain is much lower. The common emitter configuration is an inverting amplifier circuit. This means that the resulting output signal is 180o "out-of-phase" with the input voltage signal.

The Common Collector (CC) Configuration

In the Common Collector or grounded collector configuration, the collector is now common through the supply. The input signal is connected directly to the base, while the output is taken from the emitter load as shown. This type of configuration is commonly known as a Voltage

338

Follower or Emitter Follower circuit. The emitter follower configuration is very useful for impedance matching applications because of the very high input impedance, in the region of hundreds of thousands of Ohms while having relatively low output impedance.

Fig. 40.3

The common emitter configuration has a current gain approximately equal to the β value of the transistor itself. In the common collector configuration the load resistance is situated in series with the emitter so its current is equal to that of the emitter current. As the emitter current is the combination of the collector and the base current combined, the load resistance in this type of transistor configuration also has both the collector current and the input current of the base flowing through it. Then the current gain of the circuit is given as:

The Common Collector Current Gain

This type of bipolar transistor configuration is a non-inverting circuit in that the signal voltages of Vin and Vout are "in-phase". It has a voltage gain that is always less than "1" (unity). The load resistance of the common collector transistor receives both the base and collector currents giving a large current gain (as with the common emitter configuration) therefore, providing good current amplification with very little voltage gain.

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Characteristic Common

Base

Common

Emitter

Common

Collector

Input Impedance Low Medium High

Output Impedance Very High High Low

Phase Angle 0o 180o 0o

Voltage Gain High Medium Low

Current Gain Low Medium High

Power Gain Low Very High Medium

Common emitter configuration as an amplifier is the most widely used configuration due to its flexibility and high gain.

40.2 Characteristics of BJT in CE configuration

In CE configuration the emitter is made common to the input and output. It is also referred to as grounded emitter configuration. It is most commonly used configuration. In this, base current and output voltages are taken as impendent parameters and input voltage and output current as dependent parameters

VBE = f1 ( IB, VCE )

IC = f2 ( IB, VCE )

The figure shows the experimental set-up to plot input and output characteristics in a CE configuration.

Fig. 40.4

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Input Characteristic:

It is a graph between IB and VBE for different values of VCE . Since the base emitter junction of a transistor is a diode, therefore the characteristic is similar to diode one. The input current IB increases as the input voltage VBE increases for constant value of VCE.

As VCE increases, the depletion region in the collector-base increases. Hence width of the base available for conduction decreases and the collector gathers slightly more electrons thus reducing the base current (Early effect). So, the graph shifts towards the x axis. When collector is shorted with emitter then the input characteristic is the characteristic of a forward biased diode when VBE is zero and IB is also zero.

Fig. 40.5

Output Characteristic:

It is a graph of VCE and IC for various values of IB. The output characteristics can be divided into three parts.

(1) Active Region:

In this region collector junction is reverse biased and emitter junction is forward biased. As IB is maintained constant, IC increases as reverse bias voltage VCE increases. It is the area to the right of VCE = 0.5 V and above IB= 0. In this region transistor current responds most sensitively to IB. If transistor is to be used as an amplifier, it must operate in this region.

(2) Cut Off:

In this region, both the junctions are reverse biased. When the emitter base junction is reverse biased, the current due to majority carriers, ie. IB is zero. Since collector base junction is reverse

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biased, current due to minority carriers flows from collector to emitter which is represented as ICEO.

Cut off in a transistor is given by IB = 0, IC= ICEO.

(3) Saturation Region:

In this region both the junctions are forward biased by at least cut in voltage. When VCE is reduced to a small value such as 0.2 V, collector base junction is actually forward biased. In this region, there is a large change in IC with a small change in VCE.

Fig. 40.6


1. The CE input V-I characteristics is a plot of

a. VCB versus IC for constant values of IE

b. VCE versus IC for constant values of IB

c. VCE versus IE for constant values of VEB

d. VBE versus IB for constant values of VCE

2. In CE configuration, the output V-I characteristics is a plot of

a. VCB versus IC for constant values of IE

b. VCE versus IC for constant values of IB

c. VCE versus IE for constant values of VEB

d. VBE versus IB for constant values of VCE

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3. A transistor connected in common base configuration has

a. a low input resistance and a high output resistance

b. a high input resistance and a low output resistance

c. a low input resistance and a low output resistance

d. a high input resistance and a high output resistance

4. Compared to a CB amplifier, CE amplifier has

a. Lower input resistance

b. Higher output resistance

c. Lower current amplification

d. Higher current amplification

5. A transistor connected in common emitter configuration has

a. a high input resistance and a low output resistance

b. a medium input resistance and a high output resistance

c. very low input resistance and a low output resistance

d. a high input resistance and a high output resistance

EXERCISE

1. Explain forward biasing and reverse biasing with respect to p-n junction diode

2. Explain the terms ‘barrier potential’ and ‘depletion region’ as referred to p-n junction

diode.

3. Explain the working of a pn junction diode with characteristics. What is cut- in voltage?

4. What is rectifier? Draw the circuits diagram of a half wave and explain its operation with

the help of waveforms.

5. Derive the expressions for the following parameters of the half –wave rectifier circuit: (a)

average dc current(Idc) (b) rms value of load current (Irms) (c) output voltage (Vdc) (d)

rectifier efficiency(η) (e) ripple factor

6. What are the disadvantages of a half wave rectifier circuit?

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7. Derive the expressions for the following parameters of the full–wave rectifier circuit: (a)

average dc current (Idc) (b) rms value of load current (Irms) (c) output voltage (Vdc)

(d) rectifier efficiency(η) ) (e) ripple factor(r)

8. Compare full wave and half wave rectifier circuits

9. Draw the circuit diagram of a bridge rectifier and explain its operation.

10. Compare full wave centre-tap and bridge rectifier circuits.

11. Draw the circuit diagram of a full wave rectifier with capacitor filter and explain its

operation.

12. What is a transistor? State the types of transistors? Draw symbolic representation of each.

13. Draw basic constructional diagram of a n-p-n transistor and explain its working.

14. Draw basic constructional diagram of a p-n-p transistor and explain its working.

15. Explain the following facts:-

(i) Emitter region is highly doped

(ii) Base region is narrow

(iii) Transistor is called ‘transfer resistance’

16. What are different types of transistors configurations? Draw the circuit diagram of each

configuration for an n-p-n transistor.

17. Explain the input and the output characteristics of a transistor in common emitter

configuration

18. Describe experimental set-up to obtain the output characteristics of a transistor in the

CE configuration. Draw the output characteristics and indicate various regions of

operation on the characteristics.

19. Why is the common emitter configuration most commonly used?


1. Draw complete V-I characteristics of a diode.

2. Explain half wave rectifier circuit. Derive expression for rms and average value of load

current, TUF and rectifier efficiency, ripple factor.

3. Explain with circuit diagram and waveforms working of center tap full wave rectifier.

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4. Explain center tap full wave rectifier. Find expression for rms and average value of load

current, TUF and rectifier efficiency.

5. Derive the expressions for the following parameters of the full–wave rectifier circuit: a)

average dc current(Idc) b) rms value of load current (Irms) c) output voltage(Vdc) d)

rectifier efficiency(η) e) ripple factor(r)

6. Explain the working of a capacitor filter using waveforms ,with reference to a full wave

rectifier circuit.

7. Explain input characteristics of a common emitter configuration

8. Draw experimental set-up to plot input-output characteristics of CE configuration of BJT

and draw and explain output characteristics

9. Explain the input and output characteristics of a transistor in common emitter

configuration.

ANSWER KEY

Lec

36

Lec

37

Lec

38

Lec

39

Lec

40

Q1. d b a d d

Q2. c a b b b

Q3. a c a b a

Q4. b d a c d

Q5. b b d c b

Q6. d --- --- b ---

Q7. --- --- --- a ---

bee

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