alternating current circuit fundamentals

7/30/2019 Alternating Current Circuit Fundamentals

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Alternating Current Circuit Fundamentals

A practical approach

Last revision:19 July 2006

Copy Right:

Department of Electrical, Electronic and Computer Engineering

University of Pretoria

Author:

Werner Badenhorst

Contact details:

Tel: +27 12 420 2587

Fax: +27 12 362 5000

e-mail: [email protected]


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Copy Right: Dept EEC Engineering, University of Pretoria 2

INDEX

1. The Sinusoidal waveform 3

2. AC Voltage, Current and Power for a resistive load 6

3. The Root Mean Square (RMS) value 9

4. Capacitors and Inductors: A practical explanation 124.1 Inductors 124.2 Capacitors 13

5. Inductor characteristics in DC and AC circuits 155.1 DC Circuit characteristics 155.2 AC Circuit characteristics 17

6. Capacitor characteristics in DC and AC circuits 256.1 DC Circuit characteristics 256.2 AC Circuit characteristics 28

7. Resistance, Reactance and Impedance 32

8. Basic single phase AC circuit analysis 34

9. Power in AC circuits 359.1 Solving the voltages and current 359.2 Instantaneous and average power 369.3 Complex power 389.4 All is revealed 40

10. Balanced three phase AC circuit analysis and power 4210.1 What does a balanced 3-phase

AC circuit look like 4210.2 Analysis 4210.3 Three phase power 4510.4 Transforming from Delta to Wye 46

AC CIRCUIT ANALYSIS AND POWER FORMULAS 47


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1. The Sinusoidal waveform

The equation for a sinusoidal waveform should be well known and familiar:

+=+=+=

T

tAftAtAtf

2sin)2sin()sin()(

where:

= radian frequency = 2f

f = frequency in hertz [Hz]= /2

T = signals period

A = Amplitude

t = time

= phase angle

Figure 1.1 gives a graphical representation of the following functions:

)1801002sin(3)(

)45502sin(4)(

)0502sin(2)(

+=

+=

+=

tth

ttg

ttf

Figure 1.1: Three different sinusoidal waveforms illustrating differences in

amplitude, frequency and phase shift.


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The basic sinusoidal waveform is represented by f(t) having no phase shift and an

amplitude of 2. Besides g(t) having double the amplitude of f(t), it also seems as if g(t)

has been shifted 45 to the left of with respect to f(t) due to the 45 phase angle. This can

clearly be seen when comparing the positions of the peaks. Thinking in a time frame this

makes perfectly sense. 45s equates back to 2.5ms on a 50Hz time scale and hence the

added 45 angle causes g(t) to reach its peak 2.5 ms earlier than f(t). It can also be thought

of as g(t) having been given a 2.5ms head start to f(t). In terms of an equation we can

relate g(t) and f(t) as follows:

)5.2(2)( mstftg +=

A comparison of f(t) and h(t) provides us with a few other interesting observations. Firstwe observe that where f(t) completes only one cycle (period) in 20 ms, h(t) completes two

cycles. This is due to the 100Hz frequency of h(t) being double that of f(t)s 50Hz. The

second observation is that where f(t) starts out positive, h(t) starts out negative. This is

due to the 180 phase shift of h(t). Mathematically f(t) and h(t) has the following

relationship:

)102(5.1)2(5.1)1802(5.1)( mstftftfth +==+=

As seen in equation above, a 180 phase shift can also be substituted with a negative sign.Hence the following statement will read true:

)10()180()( mstftftf ==

The phase shift angle plays a crucial role in AC circuit analysis since capacitors and

inductors causes current waveforms to be out of phase with the voltage waveforms. This

will however be discussed in detail later.

Where does the term Alternating Current come from?

The voltage generated by big power stations is sinusoidal at a frequency of 50Hz in South

Africa. As a result all loads connected to this sinusoidal voltage supply have sinusoidal

currents flowing through them. As shown in figure 1.1 the value of the sinusoidal

waveform is positive for one half cycle and negative for the other half cycle, hence

sinusoidal voltages and currents will also alternate between a positive and negative value


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each cycle. It is this alternation between a positive and negative value that led to the term

Alternating Current.

Why is electricity primarily transmitted and distributed as AC instead of DC?

AC provided the only means of changing between different voltage levels prior to the

development of effective power electronic devices capable of operating at high voltages

and currents. AC allows the use of power transformers to step up voltages at the power

stations for High Voltage Transmission over long distances and step down voltages for

Distribution between customers. How transformers operate is beyond the scope of this

publication.


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2. Instantaneous AC Voltage, Current and Power for a resistive

load

Since South Africa operates at a frequency of 50Hz all signals used henceforth will have a

frequency of 50Hz setting = 2f = 250 = 314.16 radians. Assume the voltage applied

to a 5 resistor R in figure 2.1 is )sin(10)( ttv = V. Knowing we are working at 50 Hz

we can write v(t) in polar format: = 010V V.

Figure 2.1: AC voltage source connected to a pure resistive load

Using the same power equation as in DC circuits, the instantaneous power dissipated in

and the current flowing through the resistor R is:

( )( ))2cos(110

5

)sin(10)()()()(

222

tt

R

tvtitvtp

==== [W]

whereR

tvti

)()( = [A] and { })2cos(1

2

1)sin( 2 xx =

The energy consumed by the resistor at time t1 is given by:

+

=

10

0

)()(

tt

t

dttpte [J]

meaning that the energy consumed at a particular point in time ( t1) is the area underneath

the power waveform up to that particular time (t1) relative to a starting time (t0). The total

energy consumption over one period T is calculated by substituting t1 with T:

++

==

Tt

t

Tt

t

dttvR

dttpE0

0

0

0

2)(1

)( [J]

In this exampleEequates to 200 mJ.

Dividing the energy consumed during a period by the period results in an average rate of

energy consumption, also known as the average power:

v(t) R

i(t)


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++

===

Tt

t

Tt

t

dttvRT

dttpTT

EP

0

0

0

0

2)(1

)(1

This can be interpreted as the area underneath v(t)2 over a single period divided by the

constant being the resistance R multiplied by the period T.

Figure 2.2 provides a graphical interpretation of the above equations.

Figure 2.2: Instantaneous voltage, current, power and energy (divided by factor 10)

waveforms.

From figure 2.2 the following graphical observations can be made:

i. The voltage and current are positive and negative at the same time, i.e. they are in

phase.

ii. Multiplying v(t) and i(t) in real time equalsp(t) which is observed to be:

a. Always positive (since v(t) and i(t) is in phase and v(t)2 is always positive)

b. Double the frequency ofv(t) and i(t)

c. Not constant with time as with a constant DC source

Both a and b are confirmed by the trigonometric identity { })2cos(12

1)sin( 2 xx = .

iii. Regarding the instantaneous power and average power:

a. The instantaneous power is never negative

b. The average powers value equals the amplitude of the instantaneous power.


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c. The area underneathp(t) equals the area underneath P as illustrated in figure

2.3

Figure 2.3: Graphical illustration of area underneath real time and average

power being the same

iv. The energy consumption does not increase linearly with time as with a constant

DC source. Remember that the energy waveform was divided by factor 10 to

allow proportional representation on the same vertical axis.

v. Graphically or Mathematically, calculating instantaneous power and energy for an

AC circuit is much more complex than calculating DC power and energy.

Isnt there an easier way?!

The last observation calls for a way of simplifying AC circuit analysis such that power

and energy can be calculated with the same ease as in DC circuit analysis. In order to

achieve this one simply has to transform the AC source into its equivalent DC source.

The core behind the transformation lies in the energy consumed by the resistor, which

must be the same at any point in time using either the AC instantaneous analysis or the

equivalent DC analysis. The equivalent DC value for the AC source that adheres to the

above energy requirement is known as the Root Mean Square value of the AC source.


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3. The Root Mean Square (RMS) value

An analytical approach

As mentioned the RMS value of an AC source is the equivalent DC source value that will

result in the same energy dissipation in a resistor at any point in time. Figure 3.1 gives a

graphical representation of the transformation:

Figure 3.1: Graphical representation of the RMS transformation

When connecting a constant DC voltage source to the resistor, the energy dissipated in the

resistor over a period T is given by.

TR

VE dc

2

=

Continuing with the previous example, simply substituting Vdc with the amplitude Vmax ofthe sinusoidal waveform results in an energy consumption of

mJmsE 400205

10 2==

This is double the correct amount of energy. So what must Vdc be if not the amplitude

Vmax? Surely there must be a simple conversion factor that relates Vdc with Vmax such that:

maxkVVdc =

As mentioned the key lies in the amount of energy consumed by the resistor having to be

the same for both circuits in figure 3.1. It has already been shown that the energy in an

AC circuit equals +

=

Tt

t

dttvR

E0

0

2)(1

and in a DC circuit the energy equals TR

VE dc

2

= .

Hence we can state that

+

=

Tt

t

dc dttv

R

T

R

V 0

0

2

2

)(1

v(t) R

i(t)

Vdc R

Idc


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This leads us to an equation that provides the RMS value for any periodic signal:

+

==

Tt

t

dcrms dttvT

VV0

0

2)(1

In the case of a sinusoidal waveform we can use the existing example to determine the

conversion factor k. Substituting the known energy consumption of 200 mJ over a period

of 20ms into TR

VE dc

2

= , provides us with:

VT

ERVdc 071.750

02.0

52.0=

==

Substituting this result into maxkVVdc = leads us to:

5.0100

50

10

50

max==== V

V

kdc

or 2

1

=k

The last result leads us to the conventional RMS equation for a sinusoidal waveform:

2

maxVVrms =

Using this conversion in our example we find that:

VV

Vrms

0711.7502

10

2

max===

Hence

2.002.05

502

=== TR

VE rms [J] = 200 mJ

In the same manner the rms current and average power can be calculated as:

R

VI rmsrms =

and

RIR

VRI

R

VIVP rms

rmsrmsrms

2

max

2

max22

2

1

2=====

Then of course the energy is calculated as:

PTE= .


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A Graphical example

Figure 3.2 provides a graphical illustration showing what the equations look like in time

in calculating the rms value for the function ttf sin2)( = . Since the amplitude is 2 the

rms value will be 2

Step 1: Calculate ( )ttf 2cos12)( 2 =

Step 2: Calculate the area underneath the square: dttfArea

Tt

t

+

=

0

0

2)(

Step 3: Calculate the mean of the area underneath the square: 2)(1 0

0

2==

+

dttfT

ms

Tt

t

Step 4: Finally calculate the root of the mean of the area underneath the square:

2

22)(

1 0

0

2===

+

dttfT

F

Tt

t

rms

Figure 3.2: Graphical illustration of the steps in calculating the rms value for a

sinusoidal waveform.


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4. Capacitors and Inductors: A practical explanation

Up to now we are able to convert an AC source into its equivalent DC value, using the

rms value calculated from the amplitude of the AC source. We then utilise the rms value

in analysing AC circuits with purely resistive loads in exactly the same manner as DC

circuits with resistive loads. (Un)fortunately purely resistive loads are almost non-existent

in AC networks. AC network loads consist primarily out of inductive and to lesser extent

capacitive loads, both in combination with resistive loads. It is therefore VERY important

to understand the characteristics of capacitors and inductors in an AC environment.

The objective is not to give a detail mathematical analysis of the transient responses of

capacitors and inductors.

In essence inductors and capacitors are energy storage elements. The difference lies in

how and the type of energy that is stored by each.

4.1 Inductors

Inductors basically consist of a magnetic metal with a conductor wound around it. When

current flows through the coil, a magnetic field is created in and around the coils of the

conductor according to Faradays laws.

Figure 4.1: Illustration of magnetic field created by a current in a conducting coil

around a magnetic metal

This magnetic field flows through and around the coils and metal as illustrated in figure

4.1. Part of the energy provided by the current source is stored in the magnetic field.

From this it is clear that an inductor stores energy in the form of a magnetic field

i


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created by current flowing through the inductor coil. The stronger the current or

higher the number of coils (higher inductance), the stronger the magnetic field and the

greater the stored amount of energy will be. The inductor does however have a limit as

to the amount of energy it can store and the rate at which it can store the energy, but more

about this later.

It is important to realise that it takes energy out of the source to set up the magnetic field

in the inductor just as a resistor takes energy out of a source and dissipates it in the form

of heat. There is however a crucial difference! The energy is STORED in the magnetic

field of an inductor, implying that it can be retrieved. In contrast energy in a resistor is

DISSIPATED, implying it cannot be retrieved. This characteristic of an inductor will be

put to good and important use in future discussions.

A detailed study of self-inductance falls beyond the scope of this work. For the purposes

of this publication suffice to say that Inductance is represented by the letter L and

measured in Henrys [H] as Resistance is represented by the letter R and measured in

Ohms [].

4.2 Capacitors

A capacitor fundamentally consists of two conductive plates separated by an insulating,

dielectric material such as air or Teflon making it impossible for DC current to flow

through the circuit shown in figure 4.2. Though current cannot flow, the capacitor plates

are charged by the voltage source such that the plate connected to the positive terminal is

charged positively and vice versa. Hence an electrical field with a potential difference v

is created between the capacitor plates.

Figure 4.2: Illustration of the electrical field created by a voltage source between

two conductive plates separated by an insulator

v

+ + + + + +


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Should the voltage source be disconnected from the capacitor, the energy stored within

the electrical field will maintain the potential difference between the two plates for a

certain period of time. The time depends on factors such as the dielectric insulation used,

distance between the plates, voltage of the voltage source and the area of the capacitor

plates. From the above it is clear that a Capacitor stores energy in the form of anelectrical field set-up by potential difference across the capacitor plates. The higher

the voltage or the bigger the capacitor (higher capacitance) the stronger the electrical field

and the more energy is stored within the capacitor. Like the inductor, a capacitor has a

limit to the amount of energy it can store.

Again it takes energy out of the voltage source to set-up the electrical field. However, as

with the inductor this energy is STORED within the electrical field and can hence also beretrieved. Capacitance is represented by the letter C and is measured in Farad [F].

Summary

i. Inductors store energy in the form of a magnetic field created by current flowing

through a conductive coil wound around a magnetic material.

ii. Inductance is represented by the letter L and measured in Henrys [H]

iii. Capacitors store energy in the form of an electrical field set-up by the potential

difference over the two conductive plates separated by a dielectric insulator.

iv. Capacitance is represented by the letter C and is measured in Farad [F].


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5. Inductor characteristics in DC and AC circuits

The explanation of the characteristics of inductors in this section and capacitors in the

next will not be entirely textbook style. The aim is for the reader to gain insight into what

happens inside the circuit thinking in engineering laymens terms and reference. The

mathematical explanations will be limited to the utmost necessary as will be required in

single and three phase AC circuit analysis.

Figure 5.1 displays an inductor L and resistor R connected in series to a voltage source v,

also referred to as a RL circuit. Assume there is no energy stored in the inductor when

the switch is closed.

5.1 DC circuit characteristics

The switch is closed

When the switch is closed, the potential difference provided by the energy stored in

the DC voltage source is put across the series combination of the resistor and inductor.

The potential difference causes a current to start flowing, at which moment a magnetic

field is created through the inductor. However, creating this magnetic field requires

energy as does creating the flow of current through the resistor. Hence the energy in

the voltage source is divided between creating the magnetic field and creating a

current flow through the resistor and inductor.

As mentioned earlier, the inductor has a limit to the amount of energy it can store and

given enough time will reach a maximum. How much time? That depends on the

amount of current that is allowed through the circuit, which is in turn governed by

Lv

R

i

+

vL

_

Figure 5.1: RL circuit with the switch still open.


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Ohms law (I = V/R). The larger the current, the faster the inductor will reach its

maximum.

As the energy level increases towards its maximum, the inductor consumes less and

less energy from the voltage source. This allows for more and more energy from thevoltage source to increase the current flow through the circuit. As long as the inductor

is consuming energy there is a potential difference, a voltage drop, across the inductor

in the direction of the current flow (vL > 0).

The inductor is fully charged

Once the inductor is fully charged, meaning the magnetic field / energy level has

reached its maximum, all the energy in the voltage source is consumed by the resistorvia the flow of current. When fully charged the inductor acts as a short circuit, since it

is not consuming any energy from the source, allowing the current to flow freely

through it. Since the potential difference across a short circuit is zero, so the potential

difference across the inductor is zero (vL = 0) once the energy level and current

remains constant, thereby not consuming any energy. This is illustrated in figure 5.2.

The voltage source is switched off and short circuited

Next the voltage source is switched off and short-circuited, providing a closed path for

the current to continue flowing. With no voltage source to supply energy to the

circuit, the magnetic energy stored in the inductor is used to maintain the current flow

through the circuit thereby supplying the resistor with energy. As the resistor

dissipates the energy stored in the magnetic field, the current flow decreases over time

until all the energy in the magnetic field has been dissipated.

Figure 5.2: RL circuit with inductorfully charged, acting as a short circuit.

Figure 5.3: RL circuit with voltagesource switched off, and short-circuited.

LV

R

i

+

vL=0

_

LV

R

i

+

vL


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As long as the inductor is supplying the resistor with energy, the inductor is a source

of energy instead of a consumer of energy. Again a potential difference appears

across the inductor because of the decreasing energy level. This time however, the

polarity is opposite to when the energy level increased (vL < 0), even though the

direction of the current has not changed. This is depicted in figure 5.3.

What it looks like in real time

Figure 5.4 illustrates what the inductor voltage vL and the current i would look like if

measured with an oscilloscope.

Figure 5.4: Real time illustration of the circuit current and inductor voltage.

In figure 5.4 the switch was closed at t = 0 and at t = 25 the voltage source was

switched off and short-circuited.

5.2 AC circuit characteristics

Figure 5.5 illustrates the same RL circuit, but this time with a sinusoidal AC voltage

source connected. Remember, a sinusoidal AC sources polarity changes between

positive and negative once every cycle, or once per period. The polarity indicated on

the voltage source serves only as a starting reference point.


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This way then that way

In a DC circuit the frequency is 0 Hz whilst per definition, an AC sources frequency

will always be greater than 0 Hz. Because the frequency is not zero, meaning thepolarity of the voltage across the resistor and inductor changes the whole time, the

current through the circuit will change direction the whole time. This in turn causes

the magnetic field of the inductor to change polarity the whole time (refer back to

figure 4.1).

In other words, because the current changes direction the whole time (alternating

current remember!!), the inductor is charged in one direction during the first half cycle

and charged in another direction during the second half. BUT THERE IS A CATCH!!

The catch lies in momentum

To explain the catch, let us first make use of an analogy, one that hopefully all can

relate with: pushing a car that just wont start (for whatever reason). We pick up the

action at the time when you are pushing forward at your hardest.

i. The car is still accelerating, but it is approaching its maximum velocity due the

weight of the car, the friction on the road and the fact that you cant push any

harder

ii. You start growing tired and pushing less, but still the car accelerates because the

energy you are putting in is still enough to overcome the road friction with extra to

put into the momentum of the car.

Figure 5.5: RL circuit with asinusoidal AC voltage source.

YOU CAR ROAD

L = 6.14 mH

R=2.65

i(t)

+

vL

_

v(t)

+ vR -


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iii. Your input keeps dropping until you finally stop pushing. At some point however

the friction became strong enough to start slowing the car down. But even when

you stop pushing the car keeps going forward due to the kinetic energy stored in

the momentum of the car.

iv. While the car is still moving forward you realise you were suppose to go in theopposite direction and so you run to the

front of the car and slowly start pushing

harder and harder in the opposite direction.

v. Does the car start moving backwards, in the opposite direction, the moment you

start pushing in the opposite direction? No way!! The kinetic energy stored in the

cars momentum keeps pushing forward even though both you and the roadsfriction are pushing in the opposite direction. The car does however keep on

losing speed and at some point comes to a standstill.

vi. Since you are still pushing backwards the car now starts accelerating backwards

with the roads friction now pushing against you.

vii. You keep increasing your energy input into pushing until again you are pushing at

your hardest in a backward direction. And so the cycle can continue if you decide

to push forwards, then backwards, then forwards, etc

How does this translate to and inductor in and AC circuit??

Table 5.1 provides the key to the analogy:

Table 5.1: Key to Car and Inductor analogy

You and the car The voltage source and the inductorYour effort put into pushing The potential difference of the voltage source

Speed of the car Current flowing through the circuit

Kinetic energy Potential magnetic energy

Momentum Magnetic field

Friction of the road Resistance

Weight of the car Inductance

YOUCAR

ROAD

YOUCAR

ROAD


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Again we pick up the action where the voltage is at its maximum:

i. The current is still increasing in a positive direction, but it is approaching its peak

value due to the inductance, the resistance and the fact the voltage source has

reached its maximum potential difference.

ii. The voltage starts dropping, but still the current increases because the energysupplied by the voltage source is enough to overcome the resistance with extra to

put into the magnetic field of the inductor.

iii. The voltage keeps dropping and finally becomes zero. At some point however the

resistance became strong enough to start decreasing the current. But even when

the voltage becomes zero the current keeps on flowing in the positive direction due

to the magnetic energy stored in the magnetic field of the inductor.

iv. While the current is still flowing in a positive direction and decreasing, the voltagestarts increasing in a negative direction.

v. Does the current start flowing in the negative direction the moment the voltage

becomes negative? No way!! The magnetic energy stored in the inductors

magnetic field keeps the current flowing in a positive direction even though the

resistance is consuming the magnetic energy and the voltage source is pushing

in the negative direction. The current does however keep on decreasing and at

some point becomes zero.

vi. Since the voltage is still increasing in the negative direction, the current now starts

increasing in the negative direction with the resistance now again consuming

energy from the voltage source.

vii. The voltage source keeps increasing until it reaches its maximum negative value.

And so the cycle will continue between positive and negative as long as the

voltage source is on.

The important thing in the analogy you need to remember is this:

You cant change the direction of a moving car instantaneously by pushing in the

opposite direction due to the kinetic energy stored in the momentum of the car.

Just so the voltage source cant change the direction of current flow instantaneously by

changing polarity due to the magnetic energy stored in the magnetic field of the

inductor.


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Figure 5.6 provides a graphical illustration of the relationship between the voltage and

the current described in the section above.

Observe in figure 5.6 that the current waveforms peak comes after the voltage

waveforms peak in time, 2ms to be exact. The terminology used for this phenomenon

where the current waveform lags behind in time with respect to the voltage waveform

is as follows: The current lags the voltage by 36 degrees (2ms).

The full picture in real time

Figure 5.7 looks at the entire circuit, including the voltage across the resistor and theinductor. Taking the voltage source as our point of reference it has the phasor

equation of:

rmspS VVV == 093.2310328

where Vp stands for peak voltage and Vrms stands for the rms voltage.

Figure 5.6: Graphical illustration of the relationship between thevoltage source and the current through the inductor in time.


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In electrical engineering AC distribution systems the convention is to give voltage and

current values in RMS values. Henceforth all values will be given in RMS values and

not peak/amplitude values.

The current equals AI = 3671.70 since it is lagging the voltage by 36 degrees.

The voltage across the resistor is still governed by Ohms law and since the resistance

is a constant, the angle of the voltage across the resistor is the same as the

currents flowing through it. The phasor equation is VRIVR == 3663.187 if

the resistance is 2.65.

Figure 5.7: Instantaneous voltage and current in the AC RL circuit of figure 5.5

The voltage across the inductor deserves some special attention. To understand what

happens, we need to go back to the DC circuit and figure 5.4. There it became evident

that when the current through the inductor is increasing the voltage drop across the

inductor was positive, and when the current was decreasing the voltage drop became

negative. The same principle applies in AC circuits.

Looking at figure 5.7 and comparing vL(t) and i(t) you will notice that when i(t) is

increasing, it causes vL(t) to be positive and vice versa. It is important to note that it is

the current that induces the voltage across the inductor and not the other way around!!


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This effect causes vL(t) to be leading i(t) by 90, or conventionally stated: The

current flowing through an inductor lags the inductor voltage by 90. Putting the

inductor voltage into phasor format gives:

VVVVRSL === 5433.1363663.187093.231

Inductors and Ohms law

The question now arises: Is there a way to calculate the voltage across an inductor

using Ohms law like we do for a Resistor? The answer is yes! Assigning the

symbolXL to the resistance of the inductor we find XL using Ohms law:

==

= L

L jXI

V9093.1

3671.70

5433.136

Looking at the answer we notice a 90 angle represented byj in terms of a complexnumber. This is the same 90 angle we found when we discussed the waveforms in

figure 5.7. The next component to explain is the relationship between the magnitude

of 1.93 and the inductance L measured in Henry.

kLXL =

Before presenting what kis lets first use another analogy to understand the

relationship. The analogy I want to use is that of moving your hand to and fro through

water. If you do it slowly, it is easy to move your hand. But the faster you want to go,i.e. the higher your frequency, the more difficult it becomes to move your hand

through the water.

The same principle applies to inductors. The higher the frequency of the voltage and

current, the more resistance it will have because you want to charge and discharge the

inductor a higher rate. So now we have established that one part ofk is the frequency

f. Given that the inductance used is 6.14 mH we findk to be:

01.3141014.6

928.13=

==

L

Xk L

and since fkk '= :

228.650

' ==k

k

meaning that XL equals:

LfLXL ==

2


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XL is called the reactance of the inductor instead of the resistance because inductors

consumes/stores energy for half a cycle and delivers that energy during the next half

cycle back to the circuit, hencereacting with the circuit. This is unlike resistors that

just dissipate energy in the form of heat.

Reactance is however only a scalar, a magnitude. We found using Ohms law that:

= LL jX

I

V

So the true resistance of an inductor is not only the reactance but also includes the

90 angle. The term assigned to the vector combination of reactance and the angle is

called the impedance. Note that the impedance of a resistor is RR =0 .

Having an equation with which to calculate the reactance of the inductor, we are now

able to write Ohms law for an inductor:

=== 90IXIjXILjV LLL

The road we followed to get to these VERY important equations is not exactly

acceptable in mathematical terms, but it does show and explain where it comes from in

a more practical way.


25/47


6. Capacitor characteristics in DC and AC circuits

Fortunately Capacitors and Inductors have a lot in common. In principle the role that

current plays in inductors is the role voltage plays in capacitors. And the role that voltage

plays in inductors is the role current plays in capacitors. So current and voltage basically

swaps around. But of course there are a few subtle differences. Again the explanations

will be somewhat crude and practical, but hopefully insightful.

Figure 6.1 displays the RC circuit that will be used in the capacitor discussions to follow.

Again assume that there is no energy stored in the capacitor when the switch is closed.

Just to recap quickly, remember that a capacitor consists of two conductive plates

separated by an insulating material called a di-electricum. This literally means that the

two plates are isolated from each other, which implies that current cannot flow from one

plate to the other. Think for a moment. You should come up with more or less the

following question: If the current cant flow through the capacitor, then how can there

be a current flowing through the circuit? The key lies in opposites attract: positive

charge will attract negative charge and negative charge attracts positive charge. I alsostrongly advice you read through par 4.2 again before continuing.

6.1 DC circuit characteristics

The switch is closed

When the switch is closed, the potential difference provided by the energy stored in

the DC voltage source is put across the series combination of the resistor and

Figure 6.1: RC circuit with the switch still open.

Cv

R

i

+

vC

_


26/47


capacitor. Since there is no energy stored in the capacitor it means there is no

electrical field and no potential difference across the capacitor, vC= 0 V. So how does

the capacitor charge if the current cant flow?? Well

What (kind of) happens is this: The positive charge on the voltage sources positiveterminal attracts electrons from the top capacitor plate leaving the top plate with more

protons (positively charged) than electrons (negatively charged). Hence the top plate

becomes positive. At the same time the negative charge on the voltage sources

negative terminal attracts the protons on the bottom plate of the capacitor likewise

leaving the bottom plate negatively charged.

Since the definition of current is the flow of positive charge and opposite that ofnegative charge, this movement of protons and electrons from the capacitor plates to

the voltage source actually results in the flow of current through the circuit. It also

creates the illusion of current flowing through the capacitor because of the separation

of charge on the capacitor plates leaving the one positive and the other negative.

The energy of the voltage source is therefore used to attract the charges from the

capacitor and to move this charge through the resistor. The energy used to move the

charge through the resistor is dissipated in the form of heat. However the energy used

to separate the charge from the capacitor plates is converted into an electric field,

which can again be extracted at a later stage.

The capacitor becomes fully charged

Like the inductor, the capacitor has a limit to the amount of energy it can store in the

electrical field. Current always flow from a high potential difference to a lower

potential difference. Initially the capacitor voltage is 0 V, meaning a big potential

difference between the capacitor and source resulting in a big positive current flowing

towards the capacitor. As time passes the electric field becomes stronger increasing

the voltage across the capacitor plates. This causes the current to decrease as the

potential difference between the capacitor and voltage source decreases. Finally the

current becomes zero when the voltage across the capacitor equals the voltage across

the voltage source.


27/47


Since the current through the circuit is now zero, the capacitor is said to act as an open

circuit similarly to the inductor that acted as a short circuit. This state of the RC

circuit is depicted in figure 6.2.

The voltage source is switched off and short circuitedAgain the voltage source is switched off and short-circuited, putting the voltage of the

capacitor across the resistor and providing a closed path for the current to flow

(negative charge attracts positive charge through the resistor). Because of the

potential difference across the capacitor there will be a current flowing through the

resistor where the energy stored in the electrical field is dissipated as heat.

HOWEVER!!! The polarity of the voltage across the capacitor wont change because

the direction of the electrical field doesnt change. Hence the current flowing through

the resistor is in a negative direction, opposite to when the capacitor was charging.

As time passes the resistor dissipates more and more of the energy stored in the

electrical field causing the potential difference across the capacitor plates to drop.

This drop in voltage again causes the current magnitude to drop until finally the

voltage and current both become zero.

What it looks like in real time

Figure 6.4 illustrates what the capacitor voltage vC and the current i would look like if

measured with an oscilloscope.

CV

R

i=0

+vC=V

_

CV

R

i


28/47


Figure 6.4: Real time illustration of the circuit current and capacitor voltage.

Again switch was closed at t = 0 and at t = 25 the voltage source was switched off and

short-circuited.

6.2 AC circuit characteristics

Figure 6.5 illustrates the same RC circuit, but this time with a sinusoidal AC voltage

source connected.

Im definitely not going to give an elaborate explanation of what happens in the RC

circuit like I did in the RL circuit. Instead I will highlight the differences in table 6.1.

Figure 6.5: RC circuit with a

sinusoidal AC voltage source.

C = 1.65 mFR=2.65

i(t)

+

vC

_

v(t)

+ vR -


29/47


Table 6.1 comparison of Inductor and Capacitor AC characteristics

Inductor Capacitor

Current lags voltage Voltage lags current or, current leadvoltage

Current direction cannot changeinstantaneously

Voltage polarity cannot changeinstantaneously

Inductor current lags inductor voltage by90

Capacitor voltage lags capacitor currentby 90

Change in current induces the Inductorvoltage

Voltage difference between source andcapacitor creates the current

Very important to remember now is that it is the difference between the capacitor and

source voltages that causes the flow of current. With the inductor the change in

current induced the inductor voltage. Based on this we can again use the car example

we used with the inductor but with on VERY important difference:The speed of the car is now the voltage across the capacitor instead of the current

through the circuit.

You can now work through the analogy yourself.

The full picture in real time

Figure 6.6 is the equivalent of figure 5.7 showing the real time voltage and current

waveforms in the AC RC circuit diagram in figure 6.5.

Figure 6.6: Real time voltage and current waveforms in the AC RC circuit of figure 6.5


30/47


Again we take the voltage source as our reference.

rmsS VV = 093.231

The circuit component values ware again chosen to give a current that equals

AI = 3671.70

This time however it is leading the voltage by 36 degrees instead of lagging as

indicated by the +36. This means that the angle of the voltage across the 2.65

resistor also becomes positive remaining in phase with the current:

VRIVR == 3663.187 .

Finally the voltage across the capacitor can be calculated from the above as:

VVVV RSC === 5433.1363663.187093.231

Regarding the current in the circuit there are three observations I want to highlight:

a. The current is positive whenever the voltage across the capacitor is increasing and

vice versa. This underlines the fact that the current flows when the voltage across

the capacitor changes due the extraction of protons and electrons.

b. The current is positive whenever the supply voltage exceeds the capacitor voltage.

This underlines the fact that current flows from a high potential difference towards

a low potential difference.

c. The current through the capacitor lags the voltage across the capacitor by 90.

Capacitors and Ohms law

Similarly to finding an equation for an inductors reactance in terms of, it is also

possible to find an equation for the impedance of a capacitor. Again we start of with

the following:

===

= C

C jXjI

V93.19093.1

3671.70

5433.136

The only difference with respect toXL is the angle being -90 instead of +90. If we

try the same equation for calculatingXCas we didXL we find that:

51836.065.1502 === mFCXC

This result clearly does not equal the required -1.93. A hint to the solution lies in the

fact that current and voltage as swapped roles as mentioned earlier. So instead of

calculating V/I let us calculate I/V to get:


31/47


mhojV

I

C

518.090518.05433.136

3671.70==

=

Hence we found the way to 518.0==CV

IC meaning that 93.1

1==

CI

VC

. Finally,

remember the minus, for the impedance isj1.93. This brings us to the finalconclusion that the reactance of a capacitorXCequals:

CfCXC

1

2

1==

Having an equation with which to calculate the reactance of the capacitor, we are now

able to write Ohms law for a capacitor:

==== 90C

IIjXCj

IC

IjV CC

Again, the road travelled to the solutions presented above is not mathematically sound

and cannot be used as proof!! You can however find the mathematical solutions in

most AC circuit literature.


32/47


7. Resistance, Reactance and Impedance

The concept of Impedance has already been introduced where impedance is a vector

consisting of a magnitude and an angle, or a real and imaginary component. Impedance

is assigned the letterZ such that:

=+= ZjXRZ

This leads us to Ohms law (V=IR) written in vector format for AC circuits:

ZIV =

From this we find that the impedance of an inductor and capacitor equals:

LLL jXXZ == 90 or CCC jXXZ == 90

If we combine resistors with inductors or capacitors in a circuit, the impedance will

respectively be:

LjXRZ += or CjXRZ +=

Plotting these impedances on Real and Imaginary axes we get the following triangles

known as impedance triangles:

Figure 7.1: Impedance triangles for inductive and capacitive loads combined with

Resistors.

From figure 7.1 the following is found:

i. The magnitude of the impedance is calculated using

Pythagoras: 222 XRZ += .

ii. The impedance angle is calculated using trigonometry:

=

R

X1tan .

iii. The sign of the impedance angle depends on whether it is primarily an

inductive load (positive angle) or capacitive load (negative angle)

R

Im

Z

R

XC0


33/47


Example:

The objective is to calculate the total circuit impedance for each of the two circuits.

Inductive load:

R = 2.65 and == 93.1jLjZL

=

+=

+=+=

3628.3tan

93.165.2

122

R

XXR

jjXRZ

LL

LTot

Capacitive Load:

R = 2.65 and == 93.11

jCj

ZL

=

+=

=+=

3628.3tan

93.165.2

122

R

XXR

jjXRZ

CC

CTot

Impedances in series and parallel

Exactly the same rules apply as for resistors in series or parallel. The mathematics is just

a bit more tedious because now you are working with complex numbers.

Exercise:i. Calculate the impedances for the inductors and capacitors in the circuit at 50Hz

ii. Write downZ1, Z2andZ3 in both polar ( = ZZ ) and rectangular

jXRZ += format.

iii. Calculate the equivalent impedance of the circuit between nodesa andb without the

capacitor connected and draw the impedance triangle. Write all calculated

impedances in polar and rectangular format.[Ans: = 99.7366.17Z ]

iv. Recalculate the equivalent impedance with the capacitor connected. What do you

observe regarding the reactance and impedance angle of the new equivalent

impedance compared to (iii)?

R1=3L1 = 25 mH

R2=2

L2 = 30 mH

L3 = 900 mH

a

b

C = 173F


34/47


8. Basic single phase AC circuit analysis

Analysing a circuit is all about finding the voltages across and currents through various

components or parts of circuit. All the methods used in DC circuit analysis can be used in

AC circuit analysis. The only difference is that now you are working with complex

numbers, continuously changing between polar and rectangular format.

Exercise

Let us use the same AC circuit used in the last exercise only this time we connect a

voltage source to terminals ab where VVS = 0230 .

i. Calculate all the currents (excluding IC) without the capacitor connected, giving the

results in Polar format. You might have to revisit your 2nd

year circuits

ii. Recalculate all the currents with the capacitor connected again giving the results in

polar format. What do you notice regardingIS?

R1=3L1 = 25 mH

R2=2

L2 = 30 mH

L3 = 900 mH

a

b

C = 173F

I3

I1 I2IC

IS

Vs


35/47


9. Power in single phase AC circuits

Section 2 already discussed and illustrated the concept behind real time or instantaneous

power and average power (figure 2.2). This section will explore these concepts for

inductors and capacitors based on the RLC circuit diagram in figure 9.1.

Figure 9.1: RLC circuit

9.1 Solving the voltages and current

Solving that circuit will yield the following results in RMS values:

VVVV

VVAIZ

CL

RTot

==

===

87.1261513.5330

87.362087.36587.365

Figure 9.2 provides a real time graphical illustration of the solution.

Figure 9.2: Real time voltage and currents found in the RLC circuit.

R= 4

i(t)

ZL= j6

ZC= -j3

+vC

_

+ vR - + vL -

VVS = 025Z=5

R=4

X=3=36.87


36/47


Note the phase difference of the voltages relative to the current! (5ms = 90). Also note

that the inductor and capacitor voltages are 180s out of phase, when the one is positive

the other is negative.

Because of the real time sinusoidal illustrations tediousness, we rather use vectordiagrams to illustrate the relative magnitude and phase angles. A vector diagram for the

above solution is presented in figure 9.3.

Figure 9.3: Vector diagram of voltages and currents found in the RLC circuit.

Again note the magnitudes and angles of the voltages relative to each other and the

current. The vectors rotate anti-clockwise at an angular velocity of=2f radians per

second. Mentally rotating the vectors anti-clockwise clearly reveals which is leading r

lagging which.

From figures 9.3 and 9.2 you would have noticed something strange (if youre sharp).

The voltage across the inductor is greater than the voltage source itself!! Is this possible?

Yes, but before we look into the why, lets first have a look at the power.

9.2 Instantaneous and average power

The real time power is simply calculated as )()()( titvtp RR = , )()()( titvtp LL = and

)()()( titvtp CC = for the resistor, inductor and capacitor respectively and illustrated in

figure 9.4.

RV

LV

SV

CV

I

53.13

-36.87

-126.87

0

90

180

270

360


37/47


Figure 9.4: Real time power for Resistor, Inductor and Capacitor.

Observe the following in figure 9.4:

i. Power in resistor is always positive, has a frequency of 100Hz and an average of

PR=100W.


38/47


ii. Power in inductor also has a frequency of 100Hz but the average is 0.

iii. Power in capacitor also averages at 0 and is 180 out of phase with the inductors

power.

VERY IMPORTANT TO SEE: When the inductor is consuming/storing the energy, thecapacitor is releasing its energy. And when the capacitor is consuming/storing energy the

inductor releases its energy. This can also be seen as a continuous exchange of energy

between the capacitor and inductor through the flow of current.

Because the resistors power is always positive and has a positive average it means that the

resistor is always consuming energy. Both the inductor and capacitor has an average power

of 0 meaning that half the time it is consuming (storing) energy and the other half it acts as asource by releasing its energy to the other circuit elements. Because of this store/release

action found in inductors and capacitors resulting in a zero average power, it cannot be seen in

the same way as the power continuously consumed by the resistor. In order to distinguish

between the two types of power it is given specific names. It is stated that:

RESISTORS consume ACTIVE or REAL energy at a rate ofP measured in

WATT.

INDUCTORS and CAPACITORS consume REACTIVE or IMAGINARY

energy at a rate ofQ measured in VAR meaning Volt Ampere Reactive.

The name ACTIVE energy implies that that type of energy can be used for work, something

useful. It is this power that is converted into torque and heat inside a motor. The REACTIVE

energy does not physically contribute to the work. However the reactive energy is responsible

for creating and maintaining the magnetic fields in transformers and motors.

9.3 Complex power

The next step is finding the total instantaneous power of the RLC circuit. This can be

obtained by adding the three individual power components in figure 9.4. We can also first

add the two reactive powers before adding it to the active power. Figure 9.5 illustrates the

active and combined reactive powers (broken lines) alongside the total power, source voltage

and current of the RLC circuit. The total power is called:


39/47


APPARENT or COMPLEX power S measured in VA meaning Volt Ampere

Figure 9.5: Illustration of instantaneous source voltage, circuit current, active, reactive

and apparent power components within the RLC circuit

A couple of observations can be made from figure 9.5. First, because the inductors and

capacitors impedances are not equal, there is a non-zero resultant reactive power with an

amplitude of75 VAR. This resultant reactive power causes the total power to go negative for

a short period of time. This means that for a short time during each cycle the circuit as a

whole is actually acting as a source instead of an energy-consuming load. The second

observation is that the average total power equals the average active/real power, 100 W. This

however makes sense seeing that the average reactive power is still zero. The third

observation I want to highlight is the amplitude of the total power being 125 VA

Clearly now the APPARENT power consists of an ACTIVE component and a REACTIVE

component. But what is the mathematical relation between the three?


40/47


Up to now we have written the voltage and current waveforms as sine functions.

Conventionally however they are written as cosine functions to make the power function more

user friendly. Writing the source voltage and circuit current as cosine functions we obtain:

AtIti

VtVtv

ip

vpS

)cos()()cos()(

=

=

Calculating the total instantaneous power can be done as follows:

)cos()cos()()()( ivppST ttIVtitvtp ==

Since [ ])cos()cos(2

1)cos()cos( vuvuvu ++= we can rewrite p(t) as:

)2cos()cos()2cos(

2

)cos(

2

)()()( +=+== tIVIVtIVIV

titvtp rmsrmsrmsrmspppp

S

Substituting the values of our RLC circuit into this equation results in:

)87.362cos(125)87.36cos(125)( += ttp

where:

PW=== 1008.0125)87.36cos(125

9.4 ALL IS REVEALED!!!

If you are interested in the complete mathematical derivation of what Im about to show,

you can find it in most textbooks discussing complex power. I will however now

continue by using the math we did up to now and figure 9.5 to reveal the essentials of

complex power calculations.

Note the following VERY important aspects:

The amplitude of the apparent power S is 125 VA, which equals rmsrmsIV .

The average power P in the RLC circuit is 100 W, which equals )cos(rmsrmsIV

where is the angle by which the current lags (or leads) the voltage.

Constructing a triangle based on the above trigonometry we get:

S = 125 VA

P = 100 W

= 36.87

S sin()


41/47


Where the remaining side of the triangle can be calculated as 75sin =S , which

equals the amplitude of the reactive power in figure 9.5.

The term )cos( is known as the POWER FACTOR, and as the POWER

FACTOR ANGLE being the angular difference between the voltage and current.

So, similarly to constructing an impedance triangle as illustrated in figure 7.1, we can

also construct what is called a Power triangle as illustrated in figure 9.6 applied to the

RLC circuit.

Figure 9.6: Power triangles for (a) an inductive load and (b) a capacitive load.

If the total impedance of an RLC circuit has a positive angle and reactance, it is an inductive

load. If the total impedance has a negative angle and reactance, it is a capacitive load.

On the final page you will find a summary of all the formulas discussed and required.

Exercise:

Complete the analysis of the RLC circuit by doing the following:

i. Calculate the power of each element

ii. Calculate the total power of the circuit

iii. Show that the sum of the individual elements power equals the total power

iv. Draw a power triangle for the circuit

S

pf

Q>0

P

S

pf

Q


42/47


10. Balanced three phase AC circuit analysis and power

10.1 What does a balanced 3-phase AC circuit look like?

If you understand single phase, you can do three phase circuits. Three phase circuits is

just three single phase circuits connected together in on of two ways, either Wye (Y) or

Delta () as shown in figure 10.1.

Figure 10.1: Wye and Delta three phase connections.

The three phases are named A, B and C. In a Wye connection there is a neutral point

and hence each phase can be seen and analysed as a single-phase circuit. In the Delta

connection there is no neutral point and hence needs a transformation to a Wye

equivalent before it can be analysed as using single-phase analysis. We will first

discuss all the aspects with regard to the Wye connected circuit before later discussing

the Delta connection.

10.2 Analysis

Looking at the voltages

Figure 10.2 shows the instantaneous voltages ofvAN, vBNand vCN. Notice that the

voltage waveforms are 120s out of phase. Writing the voltage as phasors we obtain:

VV

VV

VV

CN

BN

AN

=

=

=

120231

120231

0231

These voltage values are called line-to-neutral or phase voltages since they are

measured between neutral and the line connecting the source to the load.

A

BC

N

a

b c

n

vBN

vAN

vCN

Wye connection

iAa

iBb

iCc

A

BC

a

b

c

VCA vAB

VBC

Delta connection

iAa

iBb

iCc


43/47


If we measure between two lines it is called the line-to-line or line voltage. Line

voltages are noted as follows and displayed in figure 10.3 alongside the phase voltages:

VVVV

VVVV

VVVV

ANCNCA

CNBNBC

BNANAB

==

==

==

150400

90400

30400

Figure 10.2: Instantaneous phase voltages for phases A, B and C

Figure 10.3: Instantaneous line-to-line voltages


44/47


Notice the 30 degree shift between the phase and line voltages as well as the increase

in amplitude from phase to line voltage.

From the above results we can find a VERY IMPORTANT relation between the phase

and line voltage values. If we calculate

37316.1231

400=

V

VL

So to convert between phase and line voltage value we use:

3

3

L

L

VV

VV

=

=

Adding a load and looking at the currents

Since we are only looking at balanced three phase systems, the load connected to each

phase must be identical. Let us connect an inductive load with a power factor of 0.9

lagging such that:

=+= 263.13345.588.119 &jZ

To calculate any of the line currents, AaI , BbI or CcI one simply takes the

corresponding phase voltage and the impedance using Ohms law. For example

finding AaI :

AZ

V

I

AN

Aa=

2673.1

Similarly you will find the other two to be:

AI

AI

Cc

Bb

9473.1

14673.1

Note that each of the currents is lagging their respective voltages by 26 because of the

0.9 lagging power factor. Also note that the line currents are flowing through the loads

of each phase. Hence in Y connected circuits, the Line currents are also the Phase

currents (IL = I). This is not the case in Delta connected circuits.

DO NOT CONFUSE THIS WITH CALCULATING THE RMS VALUE OF A

SINUSOIDAL WAVEFORM, WHICH USES A FACTOR OF 2 . THE VALUES

USED IN AC CIRCUITS ARE ALREADY RMS VALUES


45/47


10.3 Three phase power

Knowing the voltage and current of each phase, it is possible to calculate the complex

power in each phase. This is done preferably using the I2R relation, i.e. use the current

flowing through the load rather that the voltage across the load. You will make fewer

mistakes this way!

VAjZIS L 35.1755.35926400263.1337325.122

+==== &

To find the total 3-phase power we simply multiply the power of a single phase by

three to find:

VAjVASS 526107926120033 +===

VARQ

WP

526

1079

3

3

=

=

Other formulas that can be used to calculate the magnitude of the total power are:

ZIZ

V

Z

VIVIVS LL

222

3 33

33

=====

pfpf SSS == 333

In closing figure 10.4 shows the instantaneous single-phase power for each phase and

the sum of the three single-phase powers. Notice that the sum equals a constant value.

And just guess what that value is The total average 3-phase power: WP 10793 =

Figure 10.4: Instantaneous single phase powers and sum there of equalling P3


46/47


Three-phase circuit analysis and calculation are usually done per-phase. This means

that if a circuit is connected Y-Y, you only need to analyse one of the phases since all

three phase are identical except for the 120 phase shift. When talking about per phase

values, such as the per phase impedance or phase current or phase voltage, it refers to a

Y-Y connected circuit.

If a circuit is not connected Y-Y, for instance Y-, or -Y or -, then that circuit is

first transformed to a Y-Y circuit so that per-phase analysis can be done.

10.4 Transforming from a Delta to Wye circuit

When given a delta source and/or delta load, one must first transform it to its

equivalent Wye before one can analyse it using the techniques discussed up to thispoint.

Transforming a Delta source to a Wye source has already been discussed earlier and is

very easily done using:

3

LVV =

Regarding the phase angle, dont worry, you simply select VANas your reference

voltage giving it an angle of 0s. Transforming a Delta load to a Wye load is done

using:

3

=

ZZ

(If you want to, you are welcome to prove it.)

It was mentioned earlier that the phase currents in Delta circuits are not equal to the

line currents. Figure 10.5 gives the relation using Kirchoffs current law:

caabAaIII =

WhereIab andIca are phase currents.

Going through the math you will find that

=

303

LII

A

B

C

a

b

c

iAa

iBb

iCc

iab

ibc

ica


47/47

AC CIRCUIT ANALYSIS AND POWER FORMULAS:

ZVector= ZScalar=

Reactance:

fLLXL 2== fCCXC

2

11==

Voltage, Current and Impedance (Ohms Law):

jXRX

RZZ +=

=

1tan jXRX

RZZ =

=

1tan*

ZIVV Vrms == Z

VII Irms ==

**

Z

VII

Irms ==

Power factor and angle:

R

X

P

QIVpf

11 tantan ===

Z

R

S

Ppf pf === )cos(

Apparent Power:

22

2

QPpf

PIVIVS

pp

rmsrms +====

ZIZ

VIVjQPSS rms

rms

pf

2

2

** ===+==

Active / Real Power:

RIQSSpfSIVP rmspfpfrmsrms222)cos()cos( =====

Reactive / Imaginary Power:

XIPSSIVQ rmspfpfrmsrms222)sin()sin( ====

Balanced three phase circuits:

VVL 3= 3

=

ZZ

3

LVV =

ZIZ

V

Z

VIVIVS LL

222

3 33

33

=====

alternating current circuit fundamentals

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