current and static electricity

5

Current and Static ElectricityUNIT 1 CURRENT AND STATIC

ELECTRICITY

Structure 1.1 Introduction

Objectives

1.2 DC Circuit and Steady Current and Voltage 1.3 Basic Definitions Related to Circuit Analysis 1.4 Ohm’s Law and its Limitations 1.5 Resistance

1.5.1 Classification of Substances 1.5.2 Resistance Law 1.5.3 Effect of Temperature on Resistance 1.5.4 Temperature Coefficient of Resistance 1.5.5 Resistance in Series 1.5.6 Resistance in Parallel 1.5.7 Classification of Resistors

1.6 Kirchhoff’s Laws 1.6.1 Kirchhoff’s Voltage Law 1.6.2 Kirchhoff’s Current Law 1.6.3 Application of Kirchhoff’s Laws

1.7 Source Transformations

1.8 Heating Effect of Electric Current

1.9 Summary

1.10 Answers to SAQs

1.1 INTRODUCTION

The development of electricity and its application has a great history. In 600 BC, Thales, a Greek Mathematician, philosopher could produce a feel of electricity. He found that when amber is rubbed against wool or fur it gets “electrified” and attracts small particles of straw and produces spark. Static positive and negative charges can be created by rubbing amber with a piece of wool or fur. Some of the valence electrons from the one material become free electrons and are transferred to another material. The charges on the amber and the wool tend to remain stationary, thus the name static electricity.

The names like electricity, electrons and electronics originated from the word ‘elektron’ which is the Greek word of amber.

Above was the first experiment in the field of static electricity. After a long time in 1785, Coulomb, in France, first established the mathematical relation for such experiment. He defined the charge and force between two charges. In 1800, Galvani and Volta produced the electricity by chemical means. They made the first DC source known as battery. In 1825-26, Ohm established the relation between voltage and current which became the basis for Kirchhoff’s Voltage and Current law in 1845.

Most of the useful application of static electricity do not rely on static discharge which ionize the air and occurs when the electric force field between a positive charge and a negative charge becomes too strong. Rather they make use of the force of attraction between unlike charges or the force of repulsion between like charges. These forces are used to move charged particles to desired location.

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In this unit, we will learn basic definitions of the terms which are related to electricity, such as voltage, current, power and energy. Circuit element resistance will be discussed in detail.

Electricity

The relationship between current (I), voltage (V) and resistance (R) was discovered by a German scientist named Georg Ohm. In this honor this relationship is named Ohm’s law. Here, in this unit, we will discuss Ohm’s law with its limitations and two basic laws of the behaviour of current and voltage, which are known as Kirchhoff’s Laws. These laws enable scientists to understand and, therefore, evaluate the behaviour of electrical networks. How to convert a voltage source into a current source and vice-versa will also be discussed under source transformation technique.

Objectives After studying this unit, you should be able to

• explain the concept of DC circuit and steady voltage, current, power and energy,

• define Ohm’s law and its limitations, • explain the resistance law and its dependency on temperature, • find the equivalent resistance for parallel and series circuits, • write the voltage and current equations using Kirchhoff’s voltage and

current laws, • understand source transformation technique, and • calculate the power loss and energy dissipated in electric circuit.

1.2 DC CIRCUIT AND STEADY CURRENT AND VOLTAGE

Figure 1.1 shows a simple DC circuit which consists of DC source and passive element like resistance. Different types of DC sources are cell, battery, DC generator etc. In DC Analysis voltage and current remain constant with respect to time, while alternating current (or AC voltage) changes direction and magnitude with respect to time. Voltage and current waveforms for given simple DC circuit are shown in Figures 1.2(a) and (b).

RV

I

Figure 1.1 : Simple DC Circuit

V

0 t

I

0 t

VR

(a) (b)

Figure 1.2 : Waveforms (a) Direct Voltage; and (b) Direct Current

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Current and Static Electricity1.3 BASIC DEFINITIONS RELATED TO CIRCUIT

ANALYSIS

In general, we will use uppercase letters V, I, P to indicate constant voltage, current and power respectively and lower case letters v, i, p to indicate instantaneous values of time varying signals. Let us know about some basic definition such as voltage, current, power and energy which are commonly used in electrical. Voltage (or Potential)

Voltage is the electric pressure that causes current to flow. Voltage is also known as electromotive force (e.m.f.), or potential difference. All these terms refer to the same thing, that is, the force that sets charges in motion. Potential difference is actually a potential energy difference that exists between two points. An electric charge possesses potential energy (energy at rest). Potential energy is capable of doing work when we provide the right conditions to convert it from its stored form into another form (potential to kinetic energy). Voltage is the work done for moving unit positive charge from one point to another in electric field.

Energy ( )Voltage ( )Charge ( )

WVQ

=

Potential difference can also be defined as the work done for bringing a unit positive charge from infinity to any point in electric field. Instantaneous value of voltage can be given as

joules/coulombor volts.dwvdq

=

Current Electric current is the moment of charged particles in a specified direction. The charged particle is often referred to as a current carrier. In a solid, such as copper wire the charged particle (current carrier) is the electron. However, in both liquids and gases, the ions are free to move about and become current carriers. In semiconductors, the charge is carried by electron and holes, the holes behaving like positive charges. In conducting materials, a large number of free electrons are available which move from one atom to other at random when a potential difference is applied between two points of the conducting material and the current starts flowing. In electricity the amount of current is specified in terms of the charge and time required to move the charge past a given point. The amount of electric current is, therefore, specified in Coulombs per second (ampere). When we think about current, keep two points in mind. First, the effect of current is almost instantaneous. Current in a wire travels at nearly the speed of light, i.e. 186,000 miles per second (3 × 108 meters per second). Second an individual electron moves much more slowly. It may take minutes for an electron to travel a few feet in the wire. The rate of flow of charges through any cross-section of conductor is called a current and is denoted as i.

Amperesdqidt

=

where i is the instantaneous value of the current (value at any particular instant).

8

Hence one ampere is the current, which flows when a charge of one coulomb moves across the cross-section of a conductor in one second.

Electricity

The steady current ‘I’ is given as

AmperesQIt

=

where Q is the uniform flow of the charges through the cross-section of the conductor in time ‘t’.

Otherwise CoulombsQ i dt= ∫The term static electricity comes into the picture when flow of electron is steady or the flow of electron is at constant drift velocity.

Static electricity includes the study of current, potential and power related to DC circuits.

Electrical Power

Power is defined as the rate of doing work and it is expressed in Joules per second

Joules/secondWPt

=

When one coulomb of electrical charge moves through a potential difference of one volt in one second the work done is one Joule/sec and in electrical engineering it is expressed as Watt.

So, power supplied WattsP V I= ×

By Ohm’s law ( ) WattsP I R I= × ×

2 WattsI R=

or, 2

WattsV VP VR R

⎛ ⎞= × =⎜ ⎟⎝ ⎠

For AC circuit or circuits with unidirectional source, electrical power is expressed as

WaP v i tt= ×

2i R=

where, p, v, i all stands for instantaneous values.

Energy

Electrical energy is the total amount of work done and it is expressed in Joules or in Watt-Second in electrical engineering.

In DC circuit, energy is dissipated in the form of heat in resistor for a time t seconds and is given by

Watt-sec.W P t= ×

V I t= × ×

2 Watt-sec.W I Rt=

or 2

Watt-sec.VW tR

= ×

For AC circuits, energy can be expressed as

9

Current and Static Electricityc. Watt-seW p d t= ∫

Watt-sec.W v i d t= ×∫where, p, v, i are instantaneous values.

Energy expressed in terms of Kilo-Watt-hour (kWh) or units, sine Watt-sec. is a small unit,

1 unit of energy = 1 kWh

Electrical energy supplied to the consumer is charged by Power Distribution Companies in terms of Standard Energy Units known as Kilo-Watt-Hours.

1.4 OHM’S LAW AND ITS LIMITATIONS

Ohm’s law is a central concept to most electrical engineering theories. In 1825-26, Ohm gave the relation between electric current and potential. This relation is known as Ohm’s law. According to Ohm’s law potential difference across a conductor is directly proportional to the current flowing through the conductor, the temperature of the conductor remains constant.

i.e. V I ∝

V R I=

where R is the constant of proportionality is known as resistance,

V RI= (resistance in ohms)

Electrical resistance is the hindrance to the flow of electrons in a given material.

V and I represent constant value of current and voltage.

Above relation is shown in the form of resistance curve in Figure 1.3, which is linear in nature.

(a) Resistance Curve for Ohm’s Law (b) Circuit Represents Ohm’s Law

Figure 1.3

Ohm’s law can also be applied to AC circuits

volv Ri= ts

where v and i are instantaneous values of voltage and current respectively.

Limitations of Ohm’s Law

(a) Ohm’s law is valid only if the physical conditions like temperature, pressure remain constant.

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(b) Ohm’s law is not applicable to non-metallic conductors. For example, for silicon carbide, the relationship is given by , where k and n are constants. This relation is not linear.

nV k I=

Electricity

(c) Since the resistance also depends on the length and area of cross section of conductor, so for the application of Ohm’s law the dimensions of conductor should remain constant.

Example 1.1

A lamp load of 1000 Ω resistance is connected across the DC supply of 25 V. What is power absorbed in lamp and what amount of heat will be released in 10 sec.

Solution The current taken by the lamp load

25 0.025 Amp100

I = =

25 m Amp=

Power loss 2I R=

2(0.025) 1000= ×

3625 10 Watts−= ×

625 mW=

Heat released or energy consumed in 10 sec.

2W I Rt=

Pt=

3625 10 10−= × ×

2625 10−= ×

6.25 joules= .

SAQ 1 An electric iron operates from a 230 volt outlet and draws 8 amperes of current. At Rs. 4/ kWh, how much does it cost to operate the iron for 2 hours?

1.5 RESISTANCE

It is the property of any substance due to which it opposes the flow of current through it. It has the same role in electric circuit as that of friction in mechanical system. This opposition is basically due to the molecular structure of the substance. When electrons flow through any substance then they collide with the other molecules or atoms of the substance. In each collision, some energy is dissipated in the form of heat. So we can say due to resistance some energy is wasted in the form of heat (which is given by I2 Rt). The resistance is defined as the ratio of voltage and current in any circuit and its unit is ohm (Ω).

ohm ( )VRI

= Ω

11

Current and Static Electricity

Reciprocal of resistance is known as the conductance (G) and its unit is mho (Ω).

∴ 1 mhoIGR V

= = .

1.5.1 Classification of Substances On the basis of their resistance, substances may be classified as good conductor, semiconductor and bad conductor. Good Conductor

Materials with low resistance and high conductance are known as the good conductors of electricity. Example

Metals (like copper, aluminum, silver etc.), acids and electrolytes. Semiconductor

Materials which are bad conductors at low temperature and good conductors at high temperature are classified as the semiconductors. Such materials are partially conducting, but also has properties of an insulator. The amount of current conduction that can be supported can be varied by “doping” the material with appropriate materials, which results in the increased presence of free electrons for current flow. They have medium resistance (between good and bad conductors) at room temperature. Example

Germanium, Silicon, GaAs. Bad Conductors

Materials which offer very high resistance to flow of electricity are known as the bad conductors of electricity. They are normally used as the insulator in electrical machines. Example

Mica, Glass, Paper, Rubber, Wood, Bakelite etc.

1.5.2 Resistance Law Resistance of any material is directly proportional to the length of material and inversely proportional to its area of cross-section. i.e. R l∝

1RA

∝

∴ lRA

∝

or lRA

= ρ

where l = length in metre, A = area of cross section in metre2, and

resistivity of material (specific resistance). =ρ

It is defined as the resistance between the opposite faces of a metre cube of any material.

ARl

ρ =

2metre= ohm ×

metre

= Ohm-metre

12

Reciprocal of resistivity is known as conductivity or the specific conductance. It is denoted by ‘σ’ and its unit is Ω/metre.

Electricity

1σ =

ρ

. mho/metrlGA

= e

1.5.3 Effect of Temperature on Resistance In ideal conditions, resistance is the constant element of the circuit. But as the current flows, heat is produced, and temperature is increased. With increase in temperature following effects are observed :

(a) Resistance of the metal conductors increases with increase in temperature. The Resistance-temperature graph is a straight line and shown in Figure 1.4.

Figure 1.4 : Resistance-Temperature Curve for Metals

(b) Resistance of alloy also increases with increase in temperature. But this increment is relatively slow and irregular.

(c) Resistance of semiconductors decreases with increase in temperature. At very low temperature, they acts like insulators but at high temperature they show the property of conductors.

(d) Resistance of electrolytes and insulators (bad conductors) decreases with increase in temperature.

1.5.4 Temperature Coefficient of Resistance Let R0 is resistance at any initial temperature t0 and Rt is the resistance at higher temperature t. Then increment in resistance (Rt – R0) is directly proportional to initial value of resistance R0 and increment in temperature (t – t0). i.e. 0 0 0 0( )tR R R t t− = α −

where α0 is the temperature coefficient of resistance referred to temperature t0

00

0 0( )tR R

R t t−

α =−

If initial temperature t0 = 0oC then 0

0

tR RR t−

α =×

. Unit of temperature coefficient is

per oC. Resistance at any temperature is defined as

0 (1 )tR R t= + α (here t0 = 0oC)

13


The value of temperature coefficient of resistance (α) is not constant. Its value depends upon the initial temperature on which the increment of resistance is based. If the value of α at t1

oC is α1 then its value at t2oC will be

2

2 11

11 ( )t t

α =+ −

α

The temperature coefficient of resistance ‘α’ is positive for metals and alloys. α is negative for semiconductors, electrolytes and insulators. Example 1.2

A platinum coil has a resistance of 3.146 Ω at 40oC and 3.767 Ω at 100oC. Find the resistance at 0oC and the temperature-coefficient of resistance at 40oC.

Solution

0 0 0 0( )tR R R t= + α − t

0

Take t0 = initial temperature = 0o.

0 0tR R t= + α R

0 0(1 )R t= + α

100 0 0(1 100 )R R= + α at t = 100oC . . . (i)

and 40 0 0(1 40 )R R= + α at t = 40oC . . . (ii) From Eqs. (i) and (ii)

o00

0

1 1003.767 0.00379 per C3.146 1 40

+ α= ⇒ α =

+ α

From Eq. (i)

03.767 (1 100 0.00379)R= + ×

∴ 0 2.732,R =

o40

2 10

1 1 1 per C1 1 304( ) (40 0)0.00379

t tα = = =

+ − + −α

1.5.5 Resistance in Series The equivalent resistance of series combination of resistances is the summation of all resistances connected in series.

eq 1 2 3R R R R= + +

R1 R3

I

V

V2V1 V3

R2

Figure 1.5 : Resistances in Series

In series circuit, current remains same through all the resistances while the voltage is divided in all resistances.

1 2V V V V= + + 3

3

1 2IR IR IR= + +

14

Electricity q 1 2 3 e( )I R R R I R= + + =

Voltage Division Rule Since in series circuit, shown in Figure 1.5, current is given by equation

eq

VIR

=

then 11 1 .

eq

RV IR VR

= =

22 2 .

eq

RV IR VR

= =

33 3 .

eq

RV IR VR

= =

1.5.6 Resistance in Parallel Figure 1.6 shows the resistance connected is parallel.

I

V

R1

R3

R2

I1

I3

I2

Figure 1.6 : Parallel Resistors

In parallel circuits, current is divided while the voltage remains same across all parallel branches.

eq 1 2 3

1 1 1 1R R R R

= + +

2 3 1 3 1 2

1 2 3

R R R R R RR R R+ +

=

∴ 1 2 3eq

1 2 2 3 3 1

R R RR

R R R R R R=

+ +

It is better to represent the circuit in conductance form.

i.e. eq 1 2 3G G G G= + +

Current Division Rule

Current, 11

VIR

=

But, . eqV I R=

∴ eq 2 31

1 1 2 2 3 3. .

R R RI I I

1R R R R R R R= =

+ +

Similarly 1 32

1 2 2 3 3 1.

R RI I

R R R R R R=

+

15

Current and Static ElectricityAnd, 1 2

31 2 2 3 3 1

. R RI IR R R R R R

=+ +

Voltage and Current Division Rules for Combination of Two Resistances Voltage Division Rule

1 21 2

1 2 1 2. , .R RV V V V

R R R= =

+ + R

I

V

R1 R2

V2V1

Figure 1.7 : Series Resistances

Current Division Rule

2 11 2

1 2 1 2. , .R RI I I I

R R R= =

+ + R

R1

R2

I

V

I1

I2

Figure 1.8 : Parallel Resistances

1.5.7 Classification of Resistors On the basis of construction :

(a) Carbon composition resistor

(b) Wire-wound resistor on ceramic or plastic forms (example : rheostat)

(c) Deposit carbon resistor

(d) Metal film resistor

(e) Printed, painted or earthed circuit resistance

On the Basis of Value

Fixed Resistors

Their values remain constant. Fixed resistors are commonly made of carbon composition.

R

Figure 1.9 : Fixed Resistor

Tapped Resistors There are tapping on the resistor so that any desired value can be achieved.

16

Electricity

Figure 1.10 : Tapped Resistor

Variable Resistor

If the value of resistor can be changed with the help of movable contact then it is known as the variable resistor (like rheostat).

Figure 1.11 : Variable Resistor

Example 1.3

Find the equivalent resistance for the circuit shown below.

Figure for Example 1.3

Solutions

After adding the series resistances, the circuit becomes :

5 4 6 15seR = + + = Ω

Now solve the parallel branch

parallel 2 || 2R =

2 2 12 2×

= = Ω+

∴ eq se parallelR R R= +

= 15 + 1 = 16 Ω.

Example 1.4

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Find the current through each element of the given network and also find potential difference across 15Ω resistor.

15Ω

5Ω

100V 10Ω

10Ω


Solution

The circuit can be reduced in the following form :

Total current i in the circuit is obtained by

100 100 4.76 Amp5 10 6 21

ViR

= = = =+ +

So the current through 5Ω and 10Ω resistor will be i = 4.76 Amp.

Now, current through 10 resistor in parallel branch is determined by current division rule :

Ω

21

1 2

ri ir r

= ×+

154.76 2.85 Amp10 15

= × =+

Similarly, current through 15Ω resistor is

21 2

10i ir r

= ×+

104.76 1.9 Amp10 15

= × =+

∴Potential difference across 15 resistor will be Ω

2 15V i= × 1.9 15 28.5 volts= × =

Example 1.5

18

Find the equivalent resistance of network across the source terminals and determine the current drawn from the source.

Electricity

4Ω100V 6Ω

2Ω

2Ω

2Ω4Ω


Solution

After solving both the parallel branches, we get

eq 4 0.67 2.9 7.067R = + + = Ω

∴ The current drawn from the source

eq

100 14.16 Amp7.067

VIR

= = = .

SAQ 2

(a) A tungsten filament has a temperature of 2,050oC and a resistance of 500 Ω when taking normal working current, calculate the resistance of the filament when it has a temperature of 25oC. Temperature coefficient at 0oC is 0.005/oC.

(b) Find equivalent resistance for the following passive network.

Figure for SAQ 2(b)

1.6 KIRCHHOFF’S LAWS

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In 1845, Kirchhoff, at the age of 23, published his paper regarding two basic law for solving the electrical circuits. These laws became basis for the development of the network analysis or circuit theory.

1.6.1 Kirchhoff’s Voltage Law (KVL) This law is based on the law of conservation of energy. According to this, electrical energy supplied to any circuit (closed loop) is equal to the energy consumed by its passive elements. Since this law relates the voltage in a closed circuit of an electrical network it is known as Kirchhoff’s Voltage Law (KVL) or Kirchhoff’s Mesh Law. Kirchhoff’s voltage law states that in a closed electric circuit the algebraic sum of emfs and voltage drops is zero. By convention, the emfs or voltage rises are taken to be positive and voltage drops are taken to be negative.

I1

R1 R2

V1 V2

E2E1

A

B C

D Figure 1.12 : Closed Loop

In the closed circuit ABCDA, shown in Figure 1.12, applying Kirchhoff’s voltage law, we have 1 1 2 2 0E V V E− − − =

1 1 2 2 0E IR IR E− − − =

1 2 1 2E E IR IR− = +

Sum of voltage rises = Sum of voltage drop or Energy supplied = Energy consumed.

1.6.2 Kirchhoff’s Current Law (KCL) Kirchhoff’s current law states that algebraic sum of all the current meeting at a point or junction is zero. Since this law relates the current flowing through the circuit, it is known as Kirchhoff’s Current Law (KCL). In Figure 1.13, applying KCL to the junction, we have

1 3 4 2 5 0I I I I I+ + − − =

or, 1 3 4 2I I I I I+ + = + 5

I5I4

I3

I2

I1

O

Figure 1.13 : KCL at Junction

∑ Current entering the junction = Current leaving the junction ∑

1.6.3 Application of Kirchhoff’s Laws Application of Kirchhoff’s Voltage Law

20

Kirchhoff’s voltage law is used to determine the value of current in the multi loop circuit. For this purpose, Maxwell’s loop method is used as explained by an example given below.

Electricity

Figure 1.14 shows the circuit having two loops and one voltage source in first loop.

I1

R4

I2

V

R1 R3

R2

Figure 1.14 : Maxwell’s Loop Method

Assume, two loop currents flowing in the directions shown in Figure 1.14. Now we can write the KVL equations for each loop.

KVL Equation for First Loop

∑ Voltage rise = ∑ Voltage drops

1 1 2 1 2( )V R I R I I= + −

1 2 1 2( ) 2R R I R I= + − . . . (i)

KVL Equation in Second Loop

2 2 1 3 2 4 20 ( )R I I R I R I= − + +

2 1 2 3 4 2( )R I R R R I= − + + + . . . (ii)

After solving equations, we can find two loop currents I1 and I2.

Voltage division rule is also one of the applications of KVL.

Application of Kirchhoff’s Current Law

Kirchhoff’s current law is used to determine the voltages at different nodes or junctions of the circuit. For this purpose, Maxwell’s node method is used. This method is being explained with the help of an example given below.

Again consider the same circuit of Figure 1.14.

For Applying Maxwell’s Node Method

Assume voltage V1 at node as shown in Figure 1.15 and one reference node with zero voltage.

I1

R4

Reference Node (Voltage = 0)

I3

V

R1 R3V1

I2

R2

Figure 1.15 : Maxwell’s Node Method

Assume three currents I1, I2 and I3 leaving the node V1, then apply KCL : Algebraic of currents at any node is zero.

21

Current and Static Electricity1 2 3 0I I I+ + =

1 1 1

1 2 3 40V V V V

R R R R−

+ + =+

11 2 3 4

1 1 1 VV1R R R R R

⎡ ⎤+ + =⎢ ⎥+⎣ ⎦

From this equation node, voltage V1 can be obtained and then current passing through different elements can be determined. For example, current through R1 is

11

1

V VIR−

=

If value of I1 comes negative, this means actual direction of current I1 will be opposite to the assumed direction as shown in Figure 1.15. Current division rule is also one of the applications of KCL.

Example 1.6 Find currents I1 and I2 in the given circuits by applying KVL.


Solution We apply KVL for first loop :

1 110 1 1 ( )I I I= + − 2

2

110 2 I I= − . . . (i)

KVL for second loop :

2 1 21 ( ) 4 0I I I− + =

2 15 0I I− = . . . (ii)

∴ 12 0.2

5II = = 1I

1I

. . . (iii)

Put the value of I2 in Eq. (i)

1 110 2 0.2 1.8I I= − =

⇒ 1 5.55 AmpI =

From Eq. (iii), . 2 1.11 AmpI =

Example 1.7 Find the node voltage V and then current passing through each element by using KCL.

22

Electricity


Solution

Apply KCL at the node whose voltage is V.

Algebraic summation of all the current meeting at node is zero.

10 01 1 4

V V V−+ + =

11 1 104

V ⎡ ⎤+ + =⎢ ⎥⎣ ⎦

2.25 V = 10

V = 4.44 volt.

Now we have to find current through each element.

10 10 4.44 5.56 Amp1

VI −= = − =

1 4.44 Amp1VI = =

2 1.11 Amp4VI = = .

SAQ 3

(a) In the given DC circuit, determine the currents I1 and I2 by using KVL.

Figure for SAQ 3(a)

(b) In the following circuit, find the node voltage V and then determine the current passing through each element (use KCL).

23


Figure for SAQ 3(b)

(c) Find the current passing through 2Ω resistor by using Maxwell loop method.

10V

5Ω

25V 5Ω

2Ω 1Ω

Figure for SAQ 3(c)

(d) Find the current passing through 1Ω resistor by using KVL (by Maxwell loop method)

Figure for SAQ 3(d)

(e) Solve the above problem (SAQ 3(d)) by using KCL (apply node method).

1.7 SOURCE TRANSFORMATIONS

The voltage and current sources are mutually transferrable. Any practical voltage source (or simply, a voltage source) consists of an ideal voltage source in series with an internal resistance (for ideal source, this impedance is being zero and the output becomes independent of the load current). Any practical current source is demonstrated by ideal current source in parallel with their internal resistance. Let Figures 1.16(a) and (b) demonstrates their equivalence.

(a) A Practical Voltage Source (b) A Practical Current Source Figure 1.16

Assume some load resistance rL connected at terminal a-b of both voltage and current source.

aL

a L

VIr r

=+

b

bL

b L

rI I

r r=

+

24

Two sources to become identical they should deliver same current to load Electricity

b

a L b L

rV Ir r r r

=+ +

However, for the current source, the terminal voltage at a-b would be I rb, a-b being open

i.e. bV I r=

Now, we finally get

a L br r r rL+ = +

i.e. a br r=

So, for any practical voltage source, if ideal voltage be V and internal resistance be ra, the voltage source can be replaced by a current source I with the internal resistance in parallel to the current source.

Example 1.8

Convert the following current source to equivalent voltage source.


Solution

For equivalent voltage source V is given by

V I r= ×

10 10A= × Ω

100 V=

Series resistance for voltage source will be same as parallel resistance in case of current source (that is 10 Ω as given).

So, equivalent voltage source is

Figure

1.8 HEATING EFFECT OF ELECTRIC CURRENT

We know that the resistance opposes the flow of current in a conducting material. This is due to atomic structure of the material. When current flows in a conductor (flow of free electrons) then electrons collide with the immobile charges of the material. In each collision, kinetic energy of electron is converted into heat energy. This is the heating effect of electric current. Material which has the minimum value of resistance show the minimum heating effect.

Power dissipated in the resistor = I2 R watts

25


∴ Energy dissipated in the form of heat

= power × time

= I2 Rt Joules.

This is the expression for total heat developed in resistor of R Ω in time t sec and this is directly proportional to square of current I flows in the conductor.

Example 1.9

Find total power lost in the circuit. Also find the energy dissipated in 5 sec.


Solution

The given circuit can be redrawn as :

Current I is given by

250 10 Amp25

I = =

So, power loss = I2 × R

= (10)2 × 25

= 2500 watts

Energy dissipated = I2 Rt = 2500 × 5 = 12500 joules

= 12.5 k joules.

SAQ 4 Find the power loss in 10 Ω resistor and also find energy dissipated in 10 seconds.

Figure for SAQ 4

26

Electricity 1.9 SUMMARY

In this unit, we have discussed basic terms related to electricity like current, emf, power, energy etc. Concept of three basic laws of electrical engineering – Ohms law, Kirchhoff’s Voltage Law and Kirchhoff’s Current Law – with their applications have also been discussed. This unit provides the platform to understand the concept of complicated electrical problems which will be discussed in Unit 2. Effect of temperature on resistance and the heating effect of electric current have also been presented here. Solved examples along with SAQs related to each topic have been given for better understanding.

1.10 ANSWERS TO SAQs

SAQ 1

Given V = 230 V

I = 8 Amp

t = 2 h

Rate = Rs. 4 per kWh (Rs. 4 for one unit)

Cost = ?

W = Pt, P = V I

P = 8 Amp × 230 V

= 1840 Watt

Energy consumed, W = 1840 Watt × 2 h

= 3,680 Wh

= 3.68 kWh

Cost = Rs. 4 per kWh × 3.68 kWh

= Rs. 14.72

The cost is Rs. 14.72.

SAQ 2

(a) 0 0tR R t= + α R

At t = temperature = 2050oC

Rt = 500

∴ 500 = R0 (1 + 2050 α) . . . (i)

at t = 25oC

Rt = R0 (1 + 25 α) . . . (ii)

Now, dividing Eqs. (ii) by (i)

1 25 5001 2050tR + α

= ×+ α

But α = 0.005 per oC

27

Current and Static Electricity 1 25 0.005 500 50

1 2050 0.005tR + ×= × =

+ ×Ω

50tR = Ω

(b)

Figure for Answers to SAQ 2(b)

Req in the parallel combination of 2 and 15 Ω resistors.

eq2 15 302 15 17

R ×= = Ω

+.

SAQ 3

(a)

Figure for Answers to SAQ 3(a)

By Maxwell loop method Apply KVL for loop 1 :

1 15 3 (I I I= + − 2 )

2

2

or, . . . (i) 1 25 4 3I I= −

KVL for loop 2 :

2 1 20 3 ( ) 2 4I I I I= − + +

1 20 3 9I I= − + . . . (ii)

1 3I I= . . . (iii)

Put Eq. (iii) in Eq. (i) to find value of I2

2 25 4 (3 ) 3I I= −

2 212 3I I= −

29I=

So, 25 Amp 0.555 Amp9

I = =

From Eq. (iii)

15 53 A9 3

I ⎛ ⎞= =⎜ ⎟⎝ ⎠

mp

or, 1 1.667 AmpI =

28

(b) Electricity

Figure for Answers to SAQ 3(b)

KCL at node :

5 01 1 6

V V V−+ + =

11 1 56

V ⎛ ⎞+ + =⎜ ⎟⎝ ⎠

V = 3.33 volt

(c)

Figure for Answers to SAQ 3(c)

For loop 1 1 1 1 2 125 5 2 5( ) 25 12 5I I I I I 2I= + + − ⇒ = − . . . (i)

For loop 2 2 2 1 110 5( ) 10 5 6I I I I 2I− = + − ⇒ − = − + . . . (ii)

Solving Eqs. (i) and (ii)

2

25 510 6

2.128 Amp12 5

5 6

I

−−

= =−

−

(d) Assume loop currents I1, I2, I3 as shown in figure.

Figure for Answers to SAQ 3(d)

Here current passing through 1 Ω is I1

KVL equations :

Loop 1 : 1 215 6( )I I I1= + + . . . (i)

Loop 2 : 1 2 2 315 6( ) 10 6( )I I I I= + + + − . . . (ii)

Loop 3 : 3 25 6( ) 2I I I3− = − + . . . (iii)

29


After solving Eqs. (i), (ii) and (iii), we get

I1 = 6.364 Amp (current passing through 1 Ω resistor).

(e)

(a) (b)

Apply source transformation

Now, network after assuming node voltages V1, V2, V3

Here we have to use super node analysis V2 and V3 from the super node. Apply KCL at node 1 :

1 1 2 53 3

V V V−+ =

1 2 1 21 1 1 5 2 13 3 3

V V V V⎛ ⎞+ − = ⇒ − =⎜ ⎟⎝ ⎠

5 . . . (i)

KCL at node 2 and 3 (super node)

3 32 1 2 5 03 1 6 2

V VV V V −−+ + + =

1 2 30.333 1.333 0.666 2.5V V V− + + = . . . (ii)

And the voltage source is equal to V2 – V3

V2 – V3 = 10 V

V3 = V2 – 10 . . . (iii) After solving Eqs. (i), (ii) and (iii), we get

V1 = 10.682 V

V2 = 6.3648 V

V3 = − 3.635 V

Current through 1 Ω, 2 6.3648 Amp1

VI = =

SAQ 4

Figure for Answers to SAQ 4

30

Electricity We have to find I2 for power dissipation in 10 Ω. Calculation for I :

eq5 10 50 255 5

(5 10) 15 3R ×

= + = + =+

eq

100 12 AmpIR

= =

By current division rule :

25 112 4 Amp

5 10 3I I= × = × =

+

Power loss in 10 Ω resistor : 2 22 10 4 10 160 WattP I= × = × =

Energy dissipated :

. 2 160 10 1600 JouleW I Rt= = × =

current and static electricity

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