Steady-State Analysis of Single Phase A.C. Circuit Steady-State Analysis of Single Phase A.C....

2.1 ALTERNATING QUANTITYAn alternating quantity is that which acts in alternate directions and whosemagnitude undergoes a definite cycle of changes in definite intervals of time.When a simple loop revolves in a magnetic field, an alternating emf is inducedin the loop. If the loop revolves with an uniform angular velocity the inducedalternating emf is sinusoidal in nature. The alternating quantity may havevarious other wave forms like triangular, semicircular, stepped, distorted, etc.as shown in Fig. 2.1(a), (b), (c) and (d), respectively. The graph repeats afterregular intervals. One complete set of positive and negative values of analternating quantity is called a cycle. The important alternating quantities, f (t)that will be discussed in the chapter are current and voltage.

2.2 ALTERNATING VOLTAGEAlternating voltage may be generated by

(A) By rotating a coil in a stationary magnetic field.(B) By rotating a magnetic field within a stationary coil.

The value of the voltage generated in each case depends on:(i) The number of turns in the coils.

(ii) The strength of the field.(iii) The speed at which the coil or magnetic field rotates.

(a) Maximum flux links with the coil when its plane is in verticalposition (perpendicular) to the direction of flux between the poles.

(b) When the plane of a coil is horizontal no flux links with the coil.

Steady-State Analysis of SinglePhase A.C. Circuit

2

2.4 ��

2.4 ADVANTAGES OF SINE WAVE

1. Any periodic non-sinusoidal wave can be expressed as the sum of anumber of sine wave of different frequencies.

2. Sine wave can be expressed in a simple mathematical form.

3. The resultant of two or more quantities varying sinusoidally at the samefrequency is another sinusoidal quantity of same frequency.

4. Rate of change of any sinusoidal quantity is also sinusoidal.

2.5 CYCLE

A cycle may be defined as one complete set of positive and negative values ofan alternating quantity repeating at equal intervals. Each complete cycle isspread over 360° electrical as shown in Fig. 2.5.

Fig. 2.5

2.6 PERIODIC TIME

The time taken by an alternating quantity in seconds to trace one completecycle is called periodic time or time-period. It is usually denoted by symbol T.

2.7 FREQUENCY

The number of cycles per second is called frequency and is denoted bysymbol f.

Thus, f = 1

T

or, T = 1

f

If the angular velocity w is expressed in radians per second, then

� = 2

T�

= 2� f

1- Cycle

0 180 360° 540° 720

� 2� 3� 4�

rad

degree

2.5��

2.8 PHASE DIFFERENCE

Let OP and OQ be the two vectors (morepreferred to be called phasors) representingtwo alternating quantities of the samefrequency at any instant. The angle � betweenthem is called the phase angle.

The direction of rotation in counter clock-wise direction is usually taken as positive. IfOQ and OP represent voltage and currentvectors, then

e = OQ sin �t

and, i = OP sin (�t – �)

where, � is called the phase difference. In above phasor OQ is said to lead thephasor OP.

The ‘phase’ of an AC wave may be defined as its position with respect to areference axis or reference wave.

Phase angle as the angle of lead or lag with respect to reference axis orwith respect to another wave.

Fig. 2.6

�t�

Q

P

O

Fig. 2.7

• A is � degree ahead of B.

• A attains its maxima � degrees before B or 2

�

�

T second or

� = �t tαω

� ��

� �

sec before B.

sec2

t

t T

� � ��

��

� ��

� ��

�

A B

�

�/2

� �

�

�

�

Phase shift = degrees

is ahead of because

attains its maxima or minima

before

leads

lags

A B

A

B

A B

B A

O�t

e

i

2.6 ��

This function has constant magnitude Vm and as �t moves through 0 ot 2�radians.

OA = Vm cos �, OB = Vm sin �

OC = (OA) + j(OB), � = tan–1 OBOA

by Euler theorem ej� = cos � + j sin �

(cos sin )� �� m mV V V j

In rectangular form

OC = OA + j OB

|OC| = x + j y where � = tan–1 y/x

|OC| = 2 2x y�

V1 = 1 1��mV , V2 =

2 2��mV

Then V1V2 = 1 2 1 2�� m mV V = Vm[cos (�1 + �2) + j sin (�1 + �2)]

V1/V2 = 1

2

1 2�� m

m

V

V = Vm [cos (�1 – �2) + j sin (�1 – �2)]

V1 + V2 = 1 1��mV +

2 2��mV

= 1mV (cos �1 + j sin �1) +

2mV (cos �2 + j sin �2)

� � � �1 2 1 21 2 1 2 1 2cos cos sin sinm m m mV V V V j V V� � � � � � � � �

Fig. 2.8

2.9 PHASOR NOTATION

Sinusoidal quantities can be represented by a function.f(t) = Vme j�t = Vm e j� = Vm ��

O�

XA

Y

B

Vm

��C

2.7��

Phasor diagram:

Let V1 = 1mV sin � =

1mV �O � = �t

V2 = 2mV sin (�t + �) =

2mV ��(means V2 leads, V1 by angle �°)

�

V2

V1

Fig. 2.9

2.10 MEASUREMENTS OF AC MAGNITUDE

So far we know that AC voltage alternates in polarity and AC current alternatesin direction. We also know that AC can alternate in a variety of different ways,and by tracing the alternation over time we can plot it as a “waveform”. We canmeasure the rate of alternation by measuring the time it takes for a wave toevolve before it repeats itself (the “period”), and express this as cycles per unittime, of “frequency”. In music, frequency is the same as pitch, which is theessential property distinguishing one note from another.

However, we encounter a measurement problem if we try to express howlarge or small an AC quantity is. With DC, where quantities of voltage andcurrent are generally stable, we have little trouble expressing how muchvoltage or current we have in any part of a circuit. But how do you grant asingle measurement of magnitude to something that is constantly changing?

One way to express the intensity, or magnitude (also called the amplitude),of an AC quantity is to measure its peak height on a waveform graph. This isknown as the peak or crest value of an AC waveform:

Fig. 2.10

Another way is to measure the total height between opposite peaks. This isknown as the peak-to-peak (P-P) value of an AC waveform.

2.8 ��

One way of expressing the amplitude of different wave-shapes in a moreequivalent fashion is to mathematically average the values of all the points onthe graph of a waveformto a single, aggregatenumber. This amplitudemeasure is known as theaverage value of thewaveform. If we averageall the points on thewaveform algebraically(that is, to consider theirsign, either positive ornegative), the average

Fig. 2.11

Fig. 2.12

Fig. 2.13

Unfortunately, either one of these expressions of waveform amplitude canbe misleading when comparing two different types of waves. For example, asquare wave peaking at 10 volts is obviously a greater amount of voltage for agreater amount of time than a triangle wave peaking at 10 volts. The effects ofthese two AC voltages powering a load would be quite different.

True average value of all points

(considering their signs) is zero.

+ +

+ +

+

+ +

+

+ ++

+ ++

–

–

––

–– – –

––

––

–

–

2.9��

value for most waveforms is technically zero, because all the positive pointscancel all the negative points over a full cycle.

This, of course, will be true for any waveform having equal-area portionsabove and below the “zero” line of a plot. However, as a practical measure of awaveform’s aggregate value, “average” is usually defined as the mathematicalmean of all the points’ absolute values over a cycle. In other words, wecalculate the practical average value of the waveform by considering all pointson the wave as positive quantities as if the waveform looked like this:

Fig. 2.14

Polarity-insensitive mechanical meter movements (meters designed torespond equally to the positive and negative half-cycles of an alternatingvoltage or current) register in proportion to the waveform’s (practical) averagevalue, because the inertia of the pointer against the tension of the springnaturally averages the force produced by the varying voltage/current valuesover time. Conversely polarity-sensitive meter movements vibrate uselessly ifexposed to AC voltage or current, their needles oscillating rapidly about thezero mark, indicating the true (algebraic) average value of zero for asymmetrical waveform. When the “average” value of a waveform is referencedin this text, it will be assumed that the “practical” definition of average isintended unless otherwise specified.

Another method of deriving an aggregate value for waveform amplitude isbased on the waveform’s ability to do useful work when applied to a loadresistance. Unfortunately, an AC measurement based on work performed by awaveform is not the same as that waveform’s average value, because the powerdissipated by a given load (work performed per unit time) is not directlyproportional to the magnitude of either the voltage or current impressed uponit. Rather, power is proportional to the square of the voltage or current appliedto a resistance (P = E2/R, and P = I2R). Although the mathematics of such anamplitude measurement might not be straightforward, the utility of it, is.

Current would produce the same amount of heat energy dissipationthrough an equal resistance:

2.10 ��

Fig. 2.15

In the two circuits above, we have the same amount of load resistance(2 �) dissipating the same amount of power in the form of heat (50 watts), onepowered by AC and the other by DC. Because the AC voltage source picturedabove is equivalent (in terms of power delivered to a load) to a 10 volt DCbattery, we would call this a “10 volt” AC source. More specifically, we woulddenote its voltage value as being 10 volts RMS. The qualifier “RMS” standsfor Root Mean Square, the algorithm used to obtain the DC equivalent valuefrom point on a graph (essentially, the procedure consists of squaring all thepositive and negative points on a waveform graph, averaging those squaredvalues, then taking the square root of the average to obtain the final answer).Sometimes the alternative terms equivalent or DC equivalent are used insteadof “RMS”, but the quantity and principle are both the same.

RMS amplitude measurement is the best way to relate AC quantities to DCquantities, or other AC quantities of differing waveform shapes, when dealingwith measurements of electric power. For other considerations, peak or peak-to-peak measurements may be the best to employ. For instance, whendetermining the proper size of wire (ampacity) to conduct electric power froma source to a load, RMS current measurement is the best to use, because theprincipal concern with current is overheating of the wire, which is a function ofpower dissipation caused by current through the resistance of the wire.However, when rating insulators for service in high-voltage AC applications,peak voltage measurements are the most appropriate, because the principalconcern here is insulator “flashover” caused by brief spikes of voltage,irrespective of time.

Peak and peak-to-peak measurements are best performed with anoscilloscope, which can capture the crests of the waveform with a high degreeof accuracy due to the fast action of the cathode-ray-tube in response to

2.11��

changes in voltage. For RMS measurements, analog meter movements (D’Arsonval, Weston, iron vane, electrodynamometer) will work so long as theyhave been calibrated in RMS figures. Because the mechanical inertia anddampening effects of an electromechanical meter movement makes thedeflection of the needle naturally proportional to the average value of the AC,not the true RMS value, analog meters must be specifically calibrated (or mis-calibrated depending on how you look at it) to indicate voltage or current inRMS units. The accuracy of this calibration depends on an assumedwaveshape, usually a sine wave.

Electronic meters specifically designed for RMS measurement are best forthe task. Some instrument manufacturers have designed ingenious methods fordetermining the RMS value of any waveform. One such manufacturerproduces “True-RMS” meters with a tiny resistive heating element powered bya voltage proportional to that being measured. The heating effect of thatresistance element is measured thermally to give a true RMS value with nomathematical calculations whatsoever, just the laws of physics in action infulfilment of the definition of RMS. The accuracy of this type of RMSmeasurement is independent of waveshape.

For “pure” waveforms, simple conversion coefficients exist for equatingpeak, peak-to-peak, average (practical, not algebraic), and RMS measurementsto one another:

Fig. 2.16

RMS = 0.707 ( eak)

AVG = 0.637 ( eak)

P-P = 2 (Peak)

P

P

RMS = eak

= eak

P-P = 2 (Peak)

P

AVG P

RMS = 0.577 ( eak)

= 0.5 ( eak)

P-P = 2 (Peak)

P

AVG P

Sinusoidal wave

Square wave

Triangular wave

2.14 ��

Irms =

1/2

2

0

1T

i dtT

� �

� �

� ��

�

• For a sinusoidal wave

i = Im sin �t

Irms =

1/2

2

0

1( )

T

i dtT

� �

� �

� ��

�

=

1/2

2 2

0

1sin .

T

mI t d tT

� �

��

� ��

�

=

1/2

2

0

1 (1 cos2 )

2

T

m

tI d t

T

� ��

� ��

�

=

1/22

0

sin 2

2 4

TmI t t

T

� ��

� ��

=

1/22

02

mI T

T

� ��

� ��

2sin 2 sin 2.2 sin 2.t ft t

T��

� � � �

�

when t = T {sin 2�t = sin 4� = 0

= 2mI

• Ratio of maximum value to the RMS value is known as crest or peak

factor or amplitude factor. Peak factor = Maximum Value

RMS value

• Ratio of effective value to average value is known as form factor

form factor = RMS value

Average value

2.15��

2.11.2 Graphic Method

In Fig. 2.18(a) a positive half cycle of an unsymmetrical alternating current is

shown. Divide the period T into n equal intervals of time Tn

seconds. Let the

instantaneous middle values of current in the intervals be i1, i2, i3, ..., in. If R bethe resistance of the circuit through which varying current is passed, then:

Heat produced in:

1st interval = 21i R

Tn

watts

Fig. 2.18

Fig. 2.17

i2i1

i3

i4

i5

t TO(a)

M

N

T

t TO (b)

l

l2

D

Tn

i12 i2

2

i32

i42

i52

Tn

2.18 ��

(i) Iav = area of half cycle

interval

= 0

i d�

�

�

�

= 0

sinmI d�

� �

�

�

= � �0cosmI �

� �

�

= 2 mI�

= 0.637 Im

Similarly, for alternating sine voltage Eav = 2 mI�

.

Fig. 2.19

Consider small interval d� as shown in Fig. 2.19. If i is the average valueof current in the interval, then area of elementary strip = id�.

Total area of half cycle

= �

� id�

Hence, the average value of current is given by two ways.

i

d��

t

O

t

2.19��

(ii) Iav = /2

0

1T

i d tT �

= /2

0

1

/2

T

T � Im sin(wt).dt

= /2

0/2

T

mI

T � sinwtdt

= /2

0

cos

/2

TmI wt

T w

��

� ��

= .2

cos cos02

mI wT

Tw

� � ��

� ��

w = 2�f = 2

T�

= 2

2mI�

�

[cos � – cos 0]

av

20.637m

m

II I� �

�

2.12.2 Graphical Method

For an unsymmetrical wave as shown in Fig. 2.19(a), area of curve

= (i1 + i2 + i3 + ... + in).Tn

� Iav = 1 2 3( ... ).n

Ti i i i

nT

� � � �

= 1 2 3( ... )� � � � ni i i i

n

2.13 FORM FACTOR

The form factor is defined as the ratio of the effective value to the averagevalue of an alternating quantity.

2.21��

= 1

4V 2

m

Vrms = 1

2 Vm

From factor = VV

rms

av

= 0 5

0 318

.

.

VVm

m

= 1.572

Example 2: Find the average and effective values of the saw tooth waveform shown in Fig. 2.21 below.

Solution: From Fig. 2.21 below, the period is T.

v

Vm

0 T 2T 3T t

Fig. 2.21

Vav = 1

0T

VTm

T

� t dt

= 1

0T

VTm

T

� t dt

= 2

2 2 2m mV VT

T�

Effective values Vrms = 1

0

2

Tv dt

T

�

= 1

0

2

TVT

t dtT

m��

��

��

= Vm

3

2.22 ��

Example 3: Find the average and rms value of the full wave rectified sinewave shown in Fig. 2.22

v

5 V

0 � 2� 3� �t

Fig. 2.22

Solution: Average value Vav = 1

0�

�

� 5 sin �t d (�t)

= 3.185

Effective value or rms value = 1

50

2

�

� �

�

� ( sin ) ( )t d t

= 25

2 = 3.54

Example 4: The full wave rectified sine wave shown in Fig. 2.23 has a delayangle of 60°. Calculate Vav and Vrms.

v

10 V

0 � 2� 3�

�t60°

Fig. 2.23

Solution: Average value Vav = 1

0�

�

� 10 sin (�t) d (�t)

= 1

60�

�

�

� 10 sin �t d (�t)

2.23��

Vav = 10

�

(– cos �t)60� = 4.78

Effective value Vrms = 1

10 2

60�

� �

�

( sin ) ( )t d t�

�

= 60

100 1 cos2( )

2

td t

�

�

� ��

��

�

= 6.33

2.15 OPERATOR j

• An alternating voltage or current is a phasor quantity, but since theinstantaneous values are changing continuously, it must be representedby a rotating vector phasor.

• A phasor is a vector rotating at a constant angular velocity.

• j is defined as an operator which turns a phasor by 90° counter-clock-wise (CCW) without changing the magnitude of phasor

j = 1 �90°, jr = r �90°

2.16 CIRCUIT WITH PURE RESISTANCE ONLY

A pure resistance is that in which there is ohmic voltage drop only. Consider acircuit having a pure resistance R as shown is Fig. 2.24 below.

Let the instantaneous value of the alternating voltage applied be,

e = Em sin �t

The instantaneous value of current,

i = mEeR R� sin �t

Fig. 2.24

e

(a)

IE = IR

(c)

e.i

e

i

(b)

R( )�

i

2.25��

2.17 CIRCUIT WITH PURE INDUCTANCE ONLY

A pure inductive circuit possesses only inductance and no resistance orcapacitance as shown in Fig. 2.25. When an alternating voltage is applied to it,a back emf of self inductance is induced in it. As there is no ohmic resistancedrop, the applied voltage has to oppose the self induced emf only. So theapplied voltage is equal and opposite to the back emf at all instants.

Let the applied voltage

e = Em sin �t(1)

instantaneous value of self induced emf is e�

e� = – L didt

= –e

di = 1

L e dt

integrating both side, we get

�di =

1

Lsin .�

� mE t dt

i = �

mE

L(– cos �t)

i = �

mE

L sin

2

��

� ��

t �integration constant willcancel out from both side

i = Im sin2

��

� ��

t (2)�

��

��

mm

EI

L

observing (1) and (2) we find that the current lags the applied voltage by 90° or

2

�

radian.

� impedance Z = EI

= 0

2

22

� �

�

� �

m

m

E

E

Z = m

m

E

I 2

�

� = �L2

�

�

Li

e = E sin wtm

90°

I =

E

EwL

Fig. 2.25

2.26 ��

The quantity �L is called inductive reactance and is usually devoted bysymbol XL and units is ohm.

XL = �L ohms

where, L is in henry and � is in rad/sec.

Wave diagram and Phasor diagram for Pure inductance

e

i

o�/2 �

��/2 2�

i = I sin (wt – /2)m �

E sin wtm �/2E

I

Fig. 2.26

Average Power�

P = 2

0

1( )

2

�

�

�ei d wt

= 2

0

1sin . sin ( )

2 2

�

��

��

� m mE wt I wt d wt

= 2

0

1sin .cos . ( )

2

�

�

�

� m mE I wt wt d wt

= 2

0

sin 2. ( )

2 2

�

�

�

�

m mV I wtd wt

= 0

This shows power consumed in purely inductive circuit is zero.Hence, the average power consumption in an inductive circuit is zero and

is periodic with twice the supply frequency as expressed by equation (1). Thestored energy in the inductive circuit in one quarter of a cycle is released in thenext quarter.

2.28 ��

= C� Em sin2

t��

� ��

= .sin1 2

��

mEt

C

Comparing equations, we see that the current leads the voltage vector by90° as shown in Fig. 2.28.

Maximum value of current is given by,

Im = 1 �

mE

C

The quantity 1/C� is called inductive capacitance and is usually denotedby Xc. Its unit is ohm.

� Xc = 1

C�

ohms

where, C = Capacity in farads

� = angular velocity in rad/sec

Impedance Z = 0 2

90 2

��

�

m

m

EEI I

Z = m

m

E

I � – 90°

Since m

m

E

I= XC =

1

�C

� Z = XC �– 90°

= – j XC �

Average Power�instantaneous power P = vi

P = Vm sin �t.Im sin 2

��

t

= Vm Im sin �t.cos �t

= 2

m mV I sin 2�t (1)

Pav = 2

0

1( )

2

�

�

�

� P d t

2.29��

= 2

0

1sin 2 . ( )

2 2

�

� �

�

�m mV I

t d t

= 0This shows that the power consumed in purely capacitive circuit is zero.A capacitor receives energy during the first quarter cycle of voltage and

returns the same during the next quarter cycle.

2.19 CIRCUIT WITH RESISTANCE AND INDUCTANCE IN SERIES

Consider circuit of Fig. 2.29.Let R = Resistance in ohms in the circuit.

L = Inductance in henriesXL = Inductive reactance

= �LE = Effective value of applied emfI = Effective value of current in circuit.

Voltage drop across resistance,

ER = R.I in phase with current vector as shown in vector diagram ofFig. 2.30.

Voltage across reactance,

EL = I.�L = IXL, 90° ahead of vector I

Fig. 2.29 Fig. 2.30

Z = R + jXL

= 2 2 1tan�

� �L

L

XR X

R

here � = tan–1 LX

R

ER EL

R

I = E/RER

E

I =

90°

L

I

E

L�

e = Em sin wt

2.30 ��

�E

I

E=

IRR E = j XL L

and |Z | = 2 2� LR X

Z = | Z | ��

I = | |

��

�

��

E E

Z Z

� ��

EI

Z

instantaneous value of current is, i = Im sin (�t –��), where Im = EZ

hence in R-L circuit current lags the applied voltage by angle � = tan–1 LX

R

The applied voltage is therefore given by,

� E = 2 2r LE E�

= 2 2( ) ( )LIR IX�

� E = I 2 2LR X� = IZ

� = tan–1 LX

R = tan–1 �L

R

or, I = 2 2

L

E

R X�

The quantity 2 2LR X� is called impedance.

Since, the power is consumed by the resistance only, so the power in thecircuit is given by,

P = I2R = I.IR

= 2 2

L

E

R X�

. IR

or, P = E.I2 2

L

R

R X�

If � is the angle between E and I, then

cos � = 2 2

R

L

E IRE I R X

�

�

= RZ

2.31��

The applied voltage is, therefore, given by,

E = 2 2R CE E�

= I 2 2CR X� = IZ

Thus, ohm’s law is applicable to AC circuit also after replacing the termresistance by impedance.

Power = EI cos �

Cos � = 2 2

�

� C

R RZR X

...(2.53)

Z = R – j XC

Fig. 2.31 Fig. 2.32

� P = E.I cos �

Cos�� is called the power factor of the circuit. Obviously the power factoris lagging in an inductive circuit. So instantaneous current across R-L is

i = Im sin (�t – �).

2.20 CIRCUIT WITH RESISTANCE ANDCAPACITANCE IN SERIES

Consider circuit of Fig. 2.31.Voltage drop across resistance,ER = IR in phase with I as shown in vector diagram of a Fig. 2.32.

EC = I.1

�C = IXC, 90° lagging

with respect to the current vector.

ER

EL

R

I = E/R E

I =

90°

I

C

E

C�

90°�I

EE = I X

C C

E IRR=

2.32 ��

Z = 2 2� CR X 1tan�

��

� ��

CX

R

Z = |Z | � – � here 1tan CXR

��

� ��

I = EZ

= | |��

EZ

I = ��

E

Z = Im ��

instantaneuous value of current throw R-C is

i = Im sin (�t + �) where Im = EZ

hence current leads the voltages by aug � = tan–1 CX

R

��

� ��

.

2.21 SERIES R-L-C CIRCUIT

i

I

R L C

ECELER

EL

I EC

I

Fig. 2.33

Problems on alternating current circuits can be attempted easily by usingj-notation.

� Voltage across inductance = + jIXL = EL

Voltage across capacitance = – jIXC = EC

Net voltage across them = + j I (XL – XC) = j (EL – EC)

Resistance drop = IR = ER.

� Applied voltage in j-notation is represented by,

E = IR + j I(XL – XC)

2.33��

or, E = I 2 2( )L CR X X� �

Impedance in j-notation may be written as,

Z = R + j(XL – XC)

or, Z = 2 2( )L CR X X� �

I = EZ

E = 02� �

mE

Z = R + j (XL – XC)

Z = 2 2( ) | |� � �� L CR X X Z

where � = tan–1 �L CX X

R

� if XL > XC then � is +ve

if XL < XC then � is –ve

I = 0 | |2� � ��

mEZ

hence if XL > XC then current lags the applied volt.

I = I � � ��

EL

EC

E

ER

90°

90°

�

EL

– EC

I = IR

= I = IL C

Phasor diagram of series R–L–C Circuit

2.34 ��

2.22 POWER IN AC CIRCUITS

• When the current is out of phase with the voltage the power indicated bythe product of the applied voltage and the total current gives only whatis known as apparent power and measured in volt-ampere.

• Power that is returned to the source by the reactive components in thecircuit is called reactive power and is measured in VAR.

• Power that actually used in the circuit (dissipated in resistance) is trueor active power and is measured in watts or kW.

2.22.1 Active and Reactive Power and Apparent Power

Form Fig. given below

Impedance Z = R ± jX = |Z | �� = |Z | cos � + j |Z | sin �

R-L-C Ckt taking E as a refrence phasor when XL > XC

�

EL

E

I

EC

EL – EC

ER

Phasor diagram of a series R-L-C Ckt taking current I as a refrence phasor.

E

I

XR

90°

I Cos �

I Sin �

(a) (b)

�V

L C

I

2.35��

Magnitude or amplitude of impedance,

|Z| = 2 2R X�| | cos| | sin

� � �

� �

R ZX Z

Power factor of the circuit,

cos � = RZ

.

Current in the circuit I = EZ

.

This current has two components I cos � and I sin �. The componentI cos � is called in phase or wattfull component and I sin � is perpendicular toE and is called wattless component, as shown in Fig. 2.30(b). Then

Active (Real) Power = Voltage � Current � cos � watts

Since, the angle between the voltage and the wattless component of currentis 90°, hence the power absorbed by this component is zero. The power is onlyabsorbed by the wattful component.

The total power EI in volt amperes supplied to a circuit consists of twocomponents:

(a) Active power = EI cos � watts

(b) Reactive power = EI sin � volt amperes reactive or simply VAR.The above components can be shown in vector from in Fig. 2.34.

Fig. 2.34

From Fig. 2.31(b)

OA = Active power = EI cos � presented by watts

AB = Reactive power = EI sin � expressed by VAR

OB = Total power = EI expressed by VA

Obviously VA = 2 2Watts + VAR ...(2.4)

Z

R

X

VA

O A�

VAR

Watts

B

(b) (c)

kWA O�

KVA

KV

AR

B�

(a)

��

2.37��

cos �2 = 2

2 22 2

a

a b�

, sin �2 = 2

2 22 2

b

a b�

Active power:

= OA.OB cos AOB = 2 2 2 21 1 2 2a b a b� � � . cos (�2 – �1)

= 2 2 2 21 1 2 2a b a b� � � [cos �2.cos �1 + sin �2.sin �1]

= a12 + b1

2 2 2 2 1 2 12 2 2 2 2 2 2 2 2 2

2 2 1 1 2 2 1 1

a a b ba b

a b a b a b a b

� ��

� � � ��

� � � ��

= a1a2 + b1b2

Reactive power:

= OA.OB sin AOB = 2 2 2 21 1 2 2a b a b� � � .sin (�2 – �1)

= 2 2 2 21 1 2 2a b a b� � � [sin �2.cos �1 – cos �2.sin �1]

= 2 2 2 2 2 1 2 11 1 2 2 2 2 2 2 2 2 2 2

2 2 1 1 2 2 1 1

b a a ba b a b

a b a b a b a b

� ��

� � � � � ��

� � � ��

= a1b2 – a1b2

Note: V.I. = (a1 + jb1) (a2 + jb2)

= (a1a2 – b1b1) + j(a2b1 + a1b2)

If we write V � Conjugate of I

= (a1 + jb1) (a2 – jb2) = (a1a2 + b1b2) + j (a1b2 – b1a2)

= a1a2 + b1b2 + j (a1b2 – b1a2)(Active power) (Reactive power)

Note 1: Hence, the active and reactive powers would be given by the real andj ports of the vector product of voltage with the conjugate of the current vector.

Note 2: Active power can also be expressed by the sum of the algebraicproduct of the real parts of the current and the voltage and the algebraicproduct of the j parts of the current and voltage.

Alternate approach�Let E and I are the phasors given by

E = E ��1

2.38 ��

and I = I � ±�2+ sign for leading current

sign for lagging current

�

�

��

there complex power is given by S

S = E � I * 2

*2

if� � ��

�

� � ��

I I

I I= E � �1 . I� � �2

= EI 1 2��

S = EI cos (�1 � �2) + j EI sin (�1 � �2)

S P jQ� �

if V is the refrence phasor �1 = 0

S = EI cos �2 + j EI sin ( � �2)

S = EI cos �2 � j EI sin �2

S P jQ� �

ve for leading P.f load

+ve for lagging P.f load

S P jQ

S P jQ

��

��

�

�

� � ��

P = active powerQ = reactive power

2.24 POWER FACTOR

The power factor of an alternating-current device or circuit or electric powersystem is defined as the ratio of real or true power to the apparent power (VA)and is between 0 to 1.

Real power is the capacity of the circuit for performing work in a particulartime, and apparent power is the product of current and voltage of a system.Reactive power is the power that magnetic equipment (transformer, motor andrelays) needs to produce the magnetizing flux.

• In a single-phase circuit the power factor is also a measure of the phaseangle � between the phase voltage (Vph) and phase current (Iph)

Power factor = P

S

V I

V Iph ph

ph ph

� �

real power

apparent power

cos �

Power factor cos� �

2.39��

• Power factor is said to be lagging if the current lags behind voltage andleading if the current leads the voltage.

• The significance of power factor lies in the fact that utility companiessupply customers with volt-amperes, but bill them for watts.

2.24.1 Problems of Low Power Factor

(1) Power factor below 1.0 requires a utility to generate more than theminimum volt-amperes necessary to supply the real power (watts). Thisincreases generation and transmission cost.

(2) If the load power factor were as low as 0.7, the apparent power would be

real power

0.7��

��

1.4 times the real power used by the load. Line current in

the circuit would also be 1.4 times the current required at unity powerfactor, so the losses in the circuit would be doubled (proportional tosquare of current) result in all components of the system such asgenerator, conductors, transformers and switchgear would be increasedin size (cost) to carry the extra current.

(3) Higher current produces larger voltage drop in cables and otherapparatus. This results in poor voltage regulation. In practice, powerfactor is rarely corrected to unity because the cost of equipmentrequired to improve the power is usually greater than the saving ontariff.

2.24.2 Causes of Low Power Factor

Many alternating-current machines (transformer, induction motors) absorbreactive power to produce their magnetic fields, this decreases the powerfactor. Reactive power (kVAr) required by inductive loads increases theamount of apparent power (kVAr) in our, distribution system (Fig. 2.36). Thisincrease in reactive and apparent power results in a larger angle (measuredbetween kW and kVA). Recall that, as � increases, cosine ��(or power factor)decreases.

�

KVA

KW

KVAR

KVA

KW

KVAR

�

Fig. 2.36

So inductive loads (with large kVAr) results in low power factor.

2.41��

(3) The more increased voltage level in the electrical system and cooler, themore efficient motors will be.

As mentioned above, uncorrected power factor causes power systemlosses in the distribution system. As power losses increase, we may experiencevoltage drops. Excessive voltage drops can cause overheating and prematurefailure of motors and other inductive equipment.

So, by raising the power factor, these voltage drops can be minimizedalong feeder cables and avoid related problems. The motor will run cooler andmore efficiently, with a slight increase in capacity and starting torque.

2.24.4 Power Factor Correction

Power factor correction is the process of adjusting the characteristics ofelective loads in order to improve power factor closer to unity. A high powerfactor is generally desirable in a transmission system to reduce transmissionlosses and improve voltage regulation at the load.

• The presence of reactive power causes the real power to be less thanthe apparent power, and so the electrical load has a power factorless than unity.

• The reactive power increases the current flowing between thepower source and the load, which increase the power losses thoughtransmission and distribution lines.(1) Power factor correction can be done by supplying reactive power

of opposite sign adding capacitors or inductors which act to cancelthe inductive or capacitive effects of the load, respectively.For example, the inductive effect of motor loads may be offset bylocally connected capacitors, sometimes when the power factor isleading due to capacitive loading, inductors are used to correct thepower factor.

(2) Minimizing operation of idling or lightly loaded motors becauselow power factor is caused by running induction motor lightlyloaded.

(3) Avoiding operation of equipment above its rated voltage.(4) Replacing standard motors as they burn out with energy-efficient

motors.

(5) By using synchronous motor or synchronous condenser.Power Factor Correction by Static Capacitor: Consider an inducting loadconsisting of resistor R and an inductor L connected to an ac supply. Current I1lags the voltage V by angle �1 so PF is cos �1.

2.42 ��

R

L

V

I1

�1

V

I1

Let us now for improving the power factor connect a capacitor parallel to aload. This capacitor takes a leading current from the supply. The capacitorproduces a reactive power in the opposite direction hence net reactive powerdecreases.

R

L

V

I1

�2

V

I1

I2

ICV

�1

(kVAr)1

(kVAr)2

C

B

A

IC

(kVA)2

I2

(kVA)1

IC

kW

D

O

from the phasor diagram

OA = I1 cos �1 = I2 cos�2

I2 = I1 1

2

cos

cos

�

�

� �2 > �1 so cos �2 > cos �1

Since cos �2 > cos �1, I2 < I1

Hence, current drawn from the supply is less than the load current I1.Hence if power factor reduces then apparent power (VI) from the supply willalso reduce.� I2 cos �2 = I1 cos �1

� VI2 cos �2 = VI1 cos �1 = Real power

The above relation shows that active or true power taken from supply hasnot altered,

Example 4: A fluorescent lamp takes a current of 0.75 A when connectedacross a 240 V, 50 Hz ac supply. The power consumed by the lamp is 80 W.Calculate the values of the capacitance to be connected in parallel with thelamp to improve the power factor to (a) unity (b) 0.95 lagging.

Solution: I1 = 0.75 A, V = 240 V, P = 80 W; VI1 cos �1 = P

I1 cos �1 = PV

� �

80

240

1

3; cos �1 =

1

3

1

3 0 751I�

� . = 0.444

2.43��

� �1 = 63.61°, tan �1 = 2.0155(a) cos �2 = 1, �2 = 0, tan �2 = 0

IC = I1 cos �1(tan �1 – tan �2) = 1

3(2.0155 – 0) = 0.6718 A

C = I

VC

� �

�

� �

0 6718

240 2 50

. = 8.91 � 10–6 F = 8.91 �F

(b) cos �2 = 0.95, �2 = 18.19°, tan �2 = 0.3287

IC = I1 cos �1 (tan �1 – tan �2) = 1

3(2.0155 – 0.3287) = 0.5623 A

C = I

VC

� �

�

� �

0 5623

240 2 50

. = 7.457 � 10–6 F = 7.457 �F

Example 5: A single-phase 50 Hz motor takes 20A at 0.75 power factor froma 230 V sinusoidal supply. Calculate the kVAr and capacitance to be connectedin parallel to raise the power factor to 0.9 lagging. What is the new supplycurrent?Solution: I1 = 20A, f = 50Hz

cos �1 = 0.75, �1 = 41.4°, tan �1 = 0.8819cos �2 = 0.90, �2 = 25.84°, tan �2 = 0.4843

IC = I1 cos �1 (tan �1 – tan �2) = 20 � 0.75 (0.8819 – 0.4843)= 5.9637 A

C = 5.9637

2 50 230CI

V�

� � � �

= 82. 53 � 10–6 F = 82.53 �F

QC = VIC = 230 � 5.9637 = 1371.65 VAr = 1.3716 kVAr

Let I2 be the new supply current. Since the active component of supplycurrent remains changed.

I2 cos �2 = I1 cos �1

I2 = I1cos

cos

.

.

�

�

1

2

200 75

0 9� � = 16.67 A

Example 6: A factory draws an apparent power of 300 kVA at a power factorof 65% (lagging). Calculate the kVAr capacity of the capacitor bank that mustbe installed at the service entrance to bring the overall power factor to (a) unity(b) 85 percent lagging.

Solution: (a) Apparent power absorbed by the plant isS = 30 kVA

Active power absorbed by the plant isP = S cos �

= 300 � 0.65 = 195 kW

2.45��

or, Y = Y1 + Y2 + Y2 + ... ...(2.63)

The impedance Z has two components resistance R and reactance X.Admittance has also two components, the conductance ‘g’ and suceptance ‘b’.The impedance and admittance triangles are similar as shown in Fig. 2.37.

Fig. 2.37

From Fig. 2.37(b), conductance is given by,

g = Y cos �

Since, Y = 1

Z

and, cos � = RZ

from Fig. 2.33(a)

� g = 1

Z�

2

R RZ Z�

= 2 2

R

R X�

...(2.64)

Similarly, susceptance is given by,

b = Y sin �

= 2

1.X X

Z Z Z�

= 2 2

X

R X�

...(2.65)

If y1, y2, y3, ... are equal to g1 + jb1, g2 + jb2, g3 + jb3 ... then,

g = g1 + g2 + g3 + ... mho ...(2.66)

�

Z

R

X

Impedance tr angle

(a)

i

b

y

g

Admittance triangle

(b)

�

2.46 ��

b = b1 + b2 + b3 + ... mho ...(2.67)

y = g + jb ...(2.68)

= 2g + b2 ...(2.69)

Total current

I =EZ

= E.Y ...(2.70)

Power factor angle,

� = tan–1 b

g...(2.71)

Power factor will be lagging if b is + vePower factor will be leading if b is – veNote: Inductive suceptance b is assigned + ve sign and capacitivesusceptance –ve sign.

2.25.2 Vector-method

Consider a parallel circuit shown in Fig. 2.38(a)

Fig. 2.38

Branch I.

Impedance Z1 = 2 21 LR X� ...(2.72)

� I1 = 1

EZ

...(2.73)

�1 = tan–1 1

LXR

lagging ...(2.74)

– jXC

R2I

2

R1I

1jX

L

E

(a)

I

(b)

�

�1

�2

I1

E

I2

I

2.47��

Take E as reference vector. Draw I1 lagging at an angle �1 with E as shownin vector diagram of Fig. 2.34(b).

Branch II.

Impedance Z2 = 2 22 ( )cR X� � = 2 2

2 cR X� ...(2.75)

� I2 = 2

EZ

...(2.72)

�2 = tan–1 2

cXR leading ...(2.77)

Draw I2 leading E by �2. The resultant of I1 and I2 will give total current Iand the angle between E and I will give the p.f. angle.

Thus, a parallel circuit can be solved easily in this way.

2.25.3 j-Method

Consider a parallel CKT of Fig. 2.34, we can express the various impedancesin j form as under.

Z1 = R1 + j XL

Z2 = R2 – j XC

1

Z=

1 2

1 1

Z Z� =

1 2

1 1

L cR jx R jx�

� �

Z = 1 2

1 2

( ) ( )( ) ( )

L c

L c

R jx R jxR R j X X

� �

� � �

Z = � �2 1 1 21 2

2 2 2 21 2 1 2

( ) ( ) ( )( )

( ) ( ) ( ) ( )L c L cL c

L c L c

j X R X R R R X XR R X X

R R X X R R X X

� � � ��

� � � � � �

Z = R + jX

� Total current drawn = EZ

Power factor cos � = RZ

Power factor be lagging if X is +veleading if X is –ve

2.49��

voltage across capacitance

VC = I

j c�

= ( )

V

Z j c� =

1( )

V

j c R j Lc

� ��

� � � ��

| VC | = 12 2

2 1

V

c R Lc

� ��

� � � ��

Frequency fc at which VC is maximum can be obtained CdV

d� = 0

12 2

2

1 1

2 2c

Rf

LC L

� �

� ��

��

VL = I ( j�L) = 12 2

2 1

LV

R Lc

�

� � � �

� � ��

12 2 2

1

22

Lf

R CLC

�

� �

� ��

frequency at which VL is max

� it has been found that at resonance the values of VL and VC may be higher

even then the supply voltage at resonance OLV =

OCV .

Phasor diagram at resonance �

IO

V = VRO

VLO

VCO

Fig. 2.40

2.50 ��

2.28 BAND WIDTH

Band width of a series CKT is defined as the range of frequency for which thepower delivered to the resistance is greater than or equal to half the powerdelivered at resonance.

Fig. 2.41

Curve between current and frequency is known as resonance curve.

Band width = f2 – f1= �2 – �1

�1 and �2 are the angular frequencies at which the power delivered is half thepower delivered at resonance. These are also known as half power frequencies.

� At resonance Z = o

VI

= R

At half power point Z =

2o

VI

= 2o

VI

= R 2

Z = 2 2R X�

R 2 = 2 2R X� so at this point X = R

� Lower half power frequency �1 XC > XL

XC – XL = R

1

1

c�

– �1L = R

Less R

High RX >> X

C L

f1

f f2

I

I0

I0

Curve

(Resonance curve)

2

I = Current at resonance0

f

frequency

Capacitive

regioninductive

region

2.51��

�21 +

RL�1 –

1

LC= 0

�1 = 2

RL

� ±

12 21

2

RL LC

� ��

��

12 2 2

1 0[( )]� � ��

11 ,

2 oR

L LC

��

�

–ve frequency is meaningless so we take only +ve frequency.

At upper half frequency �1 XL – XC = R

�2L – 2

1

c�

= R

�22 –

RL�2 –

1

LC= 0

�2 =

12 21

2 2

R RL L LC

� ��

� ��

12 2 2

2 0( )� � � � � � � take +ve value

Band width = �2 – �1 = 2� = RL

and, �1� ��2 = �20

2.29 QUALITY FACTOR AND SELECTIVITY

Ratio of resonant frequency to band width is an indication of the degree ofselectivity of the CKT and this is known as Quality factor, Q.

1

selectivity= Q = 0 0

2 1

L

R

� � �

� � �

�

�

or, Q = 20

0 2 1( )

�

� � � �

=

0

1LC

RL

� ��

� �

= 0

1

cR�

2.52 ��

Higher values of Q o L

R

��

� �

� �

Resonance curve is very narrow and sharp (�)

� Sharpness of the curve depends on the parameters R and L. By changing C,the resonance can be made to occur at different values of frequencies.

� Q = oL

R

�

=

20 0

20

1(2 )

212

f LI

I R

� �

� � ��

L L

o

L

VV X

RV

LR

V QV

ω

�

��

�

��

�

�

��

��Hence at series resonance voltage across inductance and capacitancebecomes Q times of applied voltage. So it is also called voltage resonance.

Q =

20

20

0

1( )

2( )

2

LI

I Rf

� 2� = 2� � total stored energy

energy dissipated/cycle

Q = oL

R

�

= 2

2

L

LC R

� �

� �

= 1 LR C

Selectivity = o�

��

= oL

R

�

� A CKT with a flat frequency response curve (high R) will be moreresponsive and therefore less selective at frequencies in the neighbourhood ofthe resonant frequency.

2.30 PARALLEL RESONANCE OR CURRENT RESONANCE

A parallel combination of R, L and C or (R, L) and C branches connected to asource will produce a parallel resonance (anti-resonance) when the resultantcurrent through the combination is in phase with applied voltage at resonancepower factor is unity for this.

2.54 ��

�W

o=

1 LC

rad

/sec

�f o

= �1 2

1 LC H

z

adm

ittan

ce a

t re

sona

nce

is y

o =

1 R.

Thu

s,

the

is m

inim

um a

t re

sona

nce.

� A

CK

T c

onsi

stin

g of

par

alle

l R, L

, and

C is

calle

d a

seco

nd

orde

r pa

ralle

l re

sona

ntci

rcui

t�

Par

alle

l LC

com

bina

tion

is k

now

n as

tank

circ

uit.

Wo

= �

2 2

1L

RLC

L

f o=

�

2 2

11

2L

RH

LCL

Y=

��

��

��

��

��

22

22

L

LL

XR

jW

CR

XR

X

Y=

G +

jBat

res

onan

ce B

= 0

, W =

Wo

WoC

–

�2

22

o

o

WL

RW

L =

0

Wo

=

�

2 2

1R

LCL

f o =

�1 2

�

2 2

1R

LCL

if R

is s

mal

l—

f o =

�1 2

1 LC

2.55��

2.32 IMPEDANCE AT RESONANCE

Y = 2 2 2

R

R w L�

Z = 2 2 2R w L

R

�

At resonance w = wo = 2

2

1 RLC L

�

At resonance Zd = Resistive part

Zd = Rd is dynamic resistance

Zd = R + 2 2

2

1L RR LC L

� �

��

Zd = R + L

RC – R � Zd =

LRC

dL

ZRC

�

Zd is called dynamic impedance, this is pure resistive. It is seen lower the Rhigher the Zd. Hence the value of impedance at resonance is maximum and theresultant current is minimum. A parallel resonant circuit is also called arejector circuit since the current at resonance is minimum or tank circuit almostrejects the current at resonance.

Io = Current at resonance = d

VZ

= VCR

L, it R = 0 ckt will draw no current

at resonance. The supply current is zero and large current circulates in parallelckt at resonance.

2.33 CURRENT MAGNIFICATION

Current drawn from supply at resonance is I = d

VR

or, I = VCR

L

2.56 ��

So circulating current is VwoC.

Q = Circulating current

Current drawn from supply =

.

oVw CR

VCL

= oLwR

� Parallel tuned circuit exhibits a current amplification of Q, whereas seriesckt exhibits voltage amplification of Q.

2.34 SELECTIVITY AND BAND WIDTH

At half power frequency w1 and w2. ckt impedance is 2dR

{At reasonance V = I. Rd

{At half power V = 2

I. Rd so

2dR

Z� �

��

� �

Band width = w2 – w1 = ow

Q

Q = ow LR

= 1

ow CR =

1 LR C

Comparison between Parallel and Series Resonance

Parallel Resonance Series Resonance

(i) Net susceptance is zero Net reactance is zero(ii) Admittance is equal to conductance Impedance is equal to resistance

(iii) Impedance is L

CRImpedance is R

Fig. 2.42

R

IL

Ic

L

–j c�V

I

2.58 ��

If R is small enough, whose squares may be neglected, then

g = 2L

RX

...(2.86)

SOLVED EXAMPLES

Example 7: A two element series circuit is connected across an AC sourceV = 300 cos (314t + 20°) volts. The current is drawn 15 cos (314 t – 10°) Amp.Determine circuit impedance magnitude and phase angle. What is the averagepower drawn? (U.P. Tech 2003-04)

Solution:Given, V = 300 cos (314 t + 20°) [cos � = sin (� + 90°)

V = 300 sin (314 t + 110°)

In polar form V = 300

2��110°

i = 15 cos (314t – 10°)= 15 sin (314 t + 80°)

I = 15

2��+80°

Z (circuit impedance) =

300110

2115

802

� �

��

Z = 20 30� �

Hence, the angle between voltage and current is 30° and current lags

VI

Z� �

��

the voltage by 30°. Phase angle = 30°

Pav = 2

m mV I cos � (in R-L ckt)

= 1

2 � 15 � 300 � cos 30°

= 1949.85 watt.

2.59��

Example 8: A 120 V, 100 W lamp is to be connected to a 220 V, 50 Hz ACsupply. What value of pure inductance should be connected in series in order torun the lamp on rated voltage? (2003-04)

Solution:

� Z = R + jXL

� Vsupply = V + jVL = 2 2LV V�

� 220 = 2 2120 LV�

� VL = 184.39

Fig. 2.43

� Current through the lamp and inductance is same. Current through lamp

I = PV

= 100

120

�

100

120= L

L

VX

� XL = 120

100LV�

= 120

100 � 184.39 = 221.269 ohm.

� XL = 2�fL

L = 2

2

X

f� = 221.269

2 3.14 50� �

0.7046 henry�L

V

VL

L henry

120 Volt 100 W

V = 220, 50 Hzsupply

2.60 ��

Example 9: For the circuit shown in figure, find the current and powerdrawn from the source. (2004-05)

Fig. 2.44

Solution:

Let Z1 = 3 + j4 = 5 �53.13 �

Z2 = 6 + j8 = 10 �53.13 �

Z1 + Z2 = 9 + j12 = 15 �53.13 �

Both Z1 and Z2 are parallel hence net impedance of the circuit is Z

Z = 1 2

1 2

Z ZZ Z�

= 5 10 106.26

15 53.13

� �

�

3.33 53.13� �Z

� Current drawn from the ckt is I = VZ

I = 0�

��

V

Z =

230

3.33 53.13�

69 53.13 Amp� � �I

Hence, net current lags the net voltage by 53.13° and circuit is inductive innature.

� Power drawn from source = VI cos �

= 230 � 69 � cos (53.13)

= 9.522 kw Ans.

230 V 50 HZ1

3 � j4

6 � j8

2.61��

Example 10: A coil connected to 100 V DC supply draws 10 Amp and thesame coil connected 100V, AC voltage of frequency 50 Hz draws 5 Amp.Calculate the parameters of the coil and power factor. [2004-05]

Solution:� Coil means a resistance and inductance both.

Let impedance of a coil Z = R + jXL

� When DC supply is connected to coil inductance behave like a short circuit(XL = 2�fL = 2� � 0 � L = 0 �)

So resistance of coil R = dcV

I =

100

10 = 10 ohm.

� When AC is applied across the same coil.Given V = 100 volt of 50 Hz frequency.

I = 5 amp.

� V = IZ

� Z = VI

= 100

5 = 20 �

� Z2 = R2 + X2L

� X2L = Z2 – R2 = 202 – 102

XL = 300 = 17.32 �

L = 2

LX

f� = 17.32

2 3.14 50� �

= 0.05 henry

� Power factor of coil = RZ

= 10

20 = 0.5 lagging Ans.

Example 11: Discuss the effects of varying the frequency upon the currentdrawn and the power factor in a RLC series circuit, a series RLC circuit withR = 10 �, L = 0.02 Hz, and C = 2�f is connected to 100 V variable frequencysource. Find the frequency for which the current is maximum. (2004-05)

Solution:

V,f

R L C

I

Fig. 2.45

2.62 ��

Impedance Z = R + jXL – jXC

Z = R + j(XL – XC)

Z = 2 2( )L CR X X� � 1tan��

�L CX X

R = |Z | ��

I = | | ��

VZ

� |Z | = 2 2( )L CR X X� � and P.F. cos � = cos tan–1 ( )L CX X

R

�

(1) when XL = XC source frequency f = resonant freq (fr)� |Z | = R so current is maximum and power factor is unity.

(2) now if we increase the frequency from resonance frequency f > fr. Then

XC = 1

2 fC�

will decrease and XL increases. Impedance increases hence

current will decrease and power factor decreases and becomes lagging.

(3) If frequency decreases below resonance frequency ( f < fr), then XLdecrease and XC increases but net impedance will increase, so currentwill decrease and power factor will also decrease.

cos � = RZ

but it becomes capacitive.

� Current is maximum at resonance so at resonance frequency

fr = 1

2 LC�

Fig. 2.46

Capacitive Inductive

fr

Io

I

Z = R

2.63��

fr = 6 2

1

2 2 10 2 10� �

� � � �

= 795.5 Hz

Example 12: A load having impedance of (1 + j1) � is connected to an AC

voltage represented as V = 20 2 cos (�t + 10°) volt. Find the current in loadexpressed in the form of i = Im sin (�t + �) A. Find the real and apparent power.

(2004-05)

Solution:

Load impedance Z = 1 + j1 = 2 45� �

Voltage across the load V = 20 2 cos (�t + 10°)

= 20 2 sin (�t + 100°)

V = 20 2

1002

� � = 20 �100°

Current through load I = VZ

= 20 100

2 45� �

14.144 55� � �I is rms value of current

i = Im sin (�t + �), Im = 14.144

25

� �

� ��

i = 2 . I sin (wt + 55°)

i = 20 sin (�t + 55)

(ii) Real power = Vrms.Irms cos �

= 1

2 Vm Im cos �

= 1

220 2 .20.cos 45

200 wattP �

2.64 ��

(iii) Apparent power = 1

2 Vm Im

= 1

2 � 20 2 � 20

= 282.84 VAR

Example 13: An emf given by 100 sin 3144

t��

��

volts in applied to a

circuit and the current is 20 sin (314t – 1.5808) ampere. Find (i) frequency(ii) circuit elements. [2005-06]

Solution:(i) Let instantaneous emf be e

e = 100 sin 3144

t��

��

� �t = 314t

� 2�f = 314 � f = 50 Hz

(ii) E = 100

42

�

� �

i = 20 sin (314t – 1.5808)

= 20 sin 1.5808 180

3143.14

t� ��

��

i = 20 sin 3142

t��

��

I = 20

22

�

� �

Circuit impedance Z = 4

2

�

��

�

��

V

I

Fig. 2.47

�/2

�/4

V

I

reference axis

2.66 ��

Z = 100 �– 60° = R + jXL (ckt is inductive)

100 cos 60 + j100 sin 60 = R + jXL

� R = 100 cos 60 = 100 � 1

2 = 50 �

XL = 100 sin 60 = 100 3

2 = 86.6 �

� Frequency of supply = 50 Hz

L = 2

LX

f� =

86.6

314 = 2.758 � 10–1 H

(ii) Now the choke coil is connected to 100 V, 25 Hz supply.R and L will be same as above.

Now, XL = 2�fL = 2 � 3.14 � 25 � 2.758 � 10–1

= 43.3 �

Now, Z = 2 2LR X� = 2 250 43.3� = 66.1 1 43.3

tan50

�

�

= 66.1 �40.89

and current from the coil = VZ

I = 100

66.1 40.89�

= 1.5 40.89� � Amp

Power consumed = VI cos � = 100 � 1.5 � 0.75 = 112.5 W

or, I2R = (1.5)2 � 50 = 112.5 W

Example 15: Two coils of 5 � and 10 � and inductances 0.04 H and 0.05 Hrespectively are connecting in parallel across a 200 V, 50 Hz supply. Calculate:

(i) Conductance, susceptance and admittance of each coil.

(ii) Total current drawn by the circuit and its power factor.(iii) Power absorbed by the circuit.

2.67��

(iv) The value of resistance and inductance of single coil which will take thesame current and power as taken by the original circuit.

[2005-06]

Solution: Given

Fig. 2.49

(i) Z1 = R1 + j1LX

Z1 = 5 + j12.56 = 13.52 �68.29

Z2 = R2 + j2LX

= 10 + j15.7 = 18.62 �57.51

Admittance of coil (1) is y1 = G1 + jB1 = 1

1

Z

�

1

1

Z=

1

13.52 68.29�

= 0.074 � – 68.29

� Y1 = 0.074 68.29� �

Y1 = 0.0274 – j0.069

� G1 = 0.0274 and susceptance B1 = 0.069

Admittance of coil (2) is Y2 = G2 + jB2

Y2 = 2

1

Z =

1

18.62 57.51�

= 0.0537 �– 57.51

Y2 = 0.029 – j0.0453

XL1= 2�fL1

= 2 � 3.14 � 50 � 0.04

= 12.56 �XL2

= 2�fL2

= 2 � 3.14 � 50 � 0.05

= 15.7 �

5 � 0.04 HI1

10 � 0.05 H

I2

200 V 50 HZ1

I

2.68 ��

So conductance G2 = 0.029, susceptance B2 = 0.0453

(ii) Total admittance of ckt is Y = Y1 + Y2

Y = 0.0274 – j0.069 + 0.029 – j0.0453

= 0.0564 – j0.1143

Y = 0.1275 � – 63.74

Current drawn by the circuit is I = VY

I = 200 � 0.1275 � – 63.74

25.5 63.74 Amp� ��I

Hence, current lags the supply voltage by 63.74°.

So power factor = cos (63.74) = 0.443

(iii) Power absorbed by the circuit P = VI cos �

P = 200 � 25.5 cos (63.74)

= 2.256 kW Ans.

(iv) Current taken by original circuit is I = 25.5 � – 63.74 Amp

V = 200 V

Impedance of coil Z = R + jXL = VI

= 200

25.5 63.74��

Z = 7.843 � 63.74

= 3.47 + j7.034

So R = 3.47 � and X2 = 7.034 �,

Power = I2R

= (25.5)2 � 3.47

= 2.256 kW

Example 16: An AC voltage e(t) = 141.4 sin 120 t is applied to a series RCcircuit. The current through the circuit is obtained as

i(t) = 14.14 sin 120 t + 7.07 cos (120t + 30°).(2004-05)

2.71��

(ii) Power factor = RZ

= 12

20 = 0.6

(iii) Voltage across the coil is VL, then

VL = I (2 + jXL)

VL = 10(2 + j 16) = 161.245 �82.87 volt

Example 18: For the given figure shown

Fig. 2.51

(i) Admittance of each parallel branch(ii) Total circuit impedance

(iii) Supply current and power factor(iv) Total power supplied by the source. (2005-06)

Solution:

Z3 = 1.6 + j7.2 = 7.375 �77.47

Z1 = 4 + j3 = 5 �36.86

Z2 = 6 – j8 = 10 �–53.13

(i) Admittance of each parallel branch is Y1 and Y2, then

Y1 = 1

1

Z =

1

7.375 77.47�

= 0.1356 77.47��

1 0.029 7.18Y j� �

Y2 = 2

1

Z =

1

5 36.86�

= 0.2 36.86� � �

4 � j 3 �

6 � – j �8

Z2

1.6 � j 7.2 �

Z3100 V

50 Hz

Z1

2.72 ��

Y2 = 0.2 [cos (–36.86)2 + j sin (–36.86)]

= 0.16 – j0.119

(ii) Total circuit impedance is Z = (Z1 || Z2) + Z3

Z = Z3 + 2 1

2 1

Z ZZ Z�

= 1.6 + j7.2 + 5 36.86 10 53.13

(4 3) (6 8)

� � ��

� � �j j

= 1.6 + j7.2 + 50 16.27

10 5

��

� j

= 1.6 + j7.2 + 50 16.27

11.18 26.56

��

��

= 1.6 + j7.2 + 4.47 10.29�

= 1.6 + j 7.2 + 4.398 + j0.798

5.998 11.598 13.06 62.65� � � �Z j Ans.

(iii) Supply current I = VZ

= 100

13.06 62.65�

7.65 62.65 Amp� ��I

Power factor = cos 62.65 = 0.459

(iv) Power supplied by source = VI cos �

P = 100 � 7.65 cos (62.65)

= 351.13 watt

or, P = I2R = (7.65)2 � 5.998

= 351.02 watt

Example 19: For the circuit shown below, determine:(i) Resonant frequency

(ii) Total impedance at resonance

2.74 ��

Example 20: Draw the phasor diagram showing the following voltage andfind the RMS value of resultant voltage.

V1 = 100 sin 500 t, V2 = 200 sin 5003

t��

��

V3 = –50 cos (500t), V4 = 150 sin 5004

t��

��

Solution:

� If V = Vm sin (wt + �) can be represented in a polar for V = 2mV

� and

shown in X-Y plane

Similarly � V1 = 100 sin 500t

� V1 = 100

02� �

� V2 = 200 sin 5003

t��

��

� V2 = 200

32

�

�

� V3 = –50 cos (500 t)

= –50 sin 5002

t��

��

� V3 = 50

22

� �

�

� V4 = 150 sin 5004

t��

��

� V4 = 150

42

��

�

Phasor diagram

Fig. 2.53

Vm

2

�Ref axis

Fig. 2.54

�/2

�/4

�/3

V1

V4

V3

V2

ref-axis

2.75��

Resultant voltage

V = V1 + V2 + V3 + V4

V = 100 200 50 150

03 2 42 2 2 2

� � � ��

� ��

= 1

100 200 cos 200 sin 50 150 cos3 3 42

j j� � � ��

� � � � � ��

�

150 sin4

j��

� � � ��

�

= 1

2[100 + 100 + j173.20 – j50 + 106.06 – j106.06]

= 1

2[306.06 + j17.14]

V = 306.54

32.052

� �

RMS value of resultant voltage = 306.54

2 = 216.756 volt

and resultant voltage leads from reference axis by 32.05°.

Instantaneous voltage V = Vm sin (wt + �)

306.54 sin (500 32.05)V t� �

Example 21: A series R-L-C circuit has R = 10 �, L = 0.1 H and C = 8 F.Determine,

(i) Resonant frequency

(ii) Q-factor of circuit at resonance(iii) The half power frequencies

Solution: Given: R = 10 �, L = 0.1 H, C = 8 � 10–6 F

(i) For a series R-L-C circuit resonant frequency fr is given by

fr = 1

2 LC�

2.76 ��

fr = 6

1

2 3.14 0.1 8 10�

� � �

= 177.94 Hz.

(ii) Q–factor at resonant = rw LR

= 2 2 3.14 177.94 0.1

10rf L

R

� � � �

�

= 11.17

(iii) B.W = f2 – f1 { f1 and f2 are half power frequencies.

� f1 = fr – .

2

BW

= fr – 4

R

L� = 177.94 – 10

4 3.14 0.1� �

= 169.99 Hz

� f2 = fr + .2

BW = 177.94 + 7.95 = 185.89 Hz.

Example 22: An alternating current of frequency 50 Hz, has a maximum of

100 A. Calculate (a) its value 1

600 second after the instant the current is zero

and its value decreasing thereafter (b) How many seconds after the instant thecurrent is zero (increasing therefore words)? Will the current attain the value of86.6 A? (Elect. Tech. Allah. Univ. 1991).

Solution: The equation of the alternating current (assumed sinusoidal) withrespect to the origin of Fig. 2.51.

i = 100 sin 2� � 50t = 100 sin 100�t.

(a) It should be noted that, in this case, time is being measured from point Aand not from O.

If the above equation is to be utilized, then, this time must be referred to

point O. For this purpose, half period i.e., 1

100 sec. has to be added to

2.77��

1

600 sec. The given time as referred to point O becomes =

1

100 +

1

600

= 7

600 sec.

� i = 100 sin 100 � 180 � 7

600 = 100 sin 210°.

= 100 � 1

2� ��

� � = –50 A ...Point B.

(b) In this case the reference point is O

� 86.6 = 100 sin 100 � 180 t or sin 18,000 t = 0.866

or 18,000 t = sin–1 (0.866) = 60°

� t = 60

18000 =

1

300 second.

Example 23: An alternating voltage e = 200 sin 314t is applied to a devicewhich offers an ohmic resistance of 20 � to the flow current in one direction,while preventing the flow of current in opposite direction. Calculate RMSvalue, average value and form factor for the current over one cycle.

(Elect. Engg. Nagpur Univ. 1992).

Solution: Comparing the given voltage equation with the standard form ofalternating voltage equation, we find that

Vm = 200 V, R = 20 �, Im = 200

20 = 10 A.

Fig. 2.55

1

50Sec

210°

60°

100 A

86

.6A

A

B

– 50A

O t

i

2.79��

= 350 sin (2 � 180 � 50 � 0.018)

= –205.72 Volt Ans.

Example 25: A sinusoidal alternating current of frequency 25 Hz has amaximum value of 100 A. How long will it take for the current to attain valuesof 20, and 100 A?

Solution:

For AC current i = Im sin (wt)

Given Im = 100 A, f = 25 Hz

i = 100 sin 50�t

(a) When current attain value of 20 amp, means instantaneous valuei = 20 amp.

20 = 100 sin 50�t

sin 50�t = 0.2

50�t = sin–1 0.2 = 11.5°

t = 11.5 11.550 50 180

� �

�

� � �

= 0.00128 sec

(b) When instantaneous current i = 100 amp

i = 100 sin 50�t

100 = 100 sin 50�t

� 50�t = sin–1 = 90°

� t = 90

50 180

�

� �

= 0.01 sec.

Example 26: The voltage across and current through a circuit are given byv = 250 sin (314t – 10°) volt and i = 10 sin (314t + 50°) A. Calculate, theimpedance, resistance, reactance and power factor of the circuit.

Solution: given v = 250 sin (314t – 10°) volt

i = 10 sin (314t + 50°) amp

2.80 ��

above voltage are in time domain we can write in polar form

V = 250

102

� � �

I = 10

502� �

� Impedance of ckt = Z = VI

Z =

25010

210

502

��

��

25 60� �� Z

From this it is clear that current leads the voltage by 60°.

So power factor = cos 60° = 0.5

� Z = R – jXC = 25[cos 60 + j sin (–60)]

R – jXC = 12.5 – j21.65

Comparing real and imaginary part

R = 12.5 �, XC = 21.65 � Ans.

Example 27:

Fig. 2.56

2 � j3 �

4 � j2I3

3 V3

V = 10 0

I2

B

1 � – j5

V2

C

I

1

2

Find (a) Total impedance

(b) Current drawn from supply

2.82 ��

(d) 2 = 2

BCVZ

= 6.65 36.8

1 5

��

� j = 1.30 41.9� �

current I2 leads the VBC by (41.9° + 36.8°)

Im = 3

BCVZ

= 6.65 36.8

4 2

��

� j = 1.49 63.4� � �

(e) Power factor cos � = cos (15.7) = 0.963 lagging

or, cos � = RZ

= 5.65

5.87 = 0.963

(f) Apparent power s = VI

= 10 � 1.70 = 17.0 VA

True power = I2R = 1.7 � 1.7 � 5.65 = 16.32 W.

= VI cos � = 10 � 1.7 � 0.963 = 16.34 W.

Reactive power = I2X = 1.7 � 1.7 � 1.59 = 4.59 vars

= VI sin � = 10 � 1.7 � sin (15.7°) = 4.6 vars.

(g) Phasor diagram

Let V = 10 0� is a reference.

Fig. 2.57

41.9°

15.7°

36. °�

63.4°

V3

I3

V

VBC

I2

V1

I

2.83��

Example 28: An alternating current of frequency 60 Hz, has a maximumvalue of 120 A. Write down the equation for its instantaneous value.Reknocking time from the instant the current is zero and is becoming positiveFind

(a) The instantaneous value after 1/360 second and(b) The time taken to reach 96 A for the first time.

Solution:The instantaneous current equation is

� i = 120 sin 2��f t = 120 sin 120 �t.

Now, when t = 1/360 second, then

(a) i = 120 sin (120 � � � 1/360) ... angle in radians

= 120 sin (120 � 180 � 1/360) ... angle in degree

= 120 sin 60° = 103.9 A.

(b) 96 = 120 � sin 2 � 180 � 60 � t angle in degree

or, sin (360 � 60 � t) = 96/120 = 0.8

� 360 � 60 � t = sin–1 0.8 = 53° (approx.)

� t = 0/2� f = 53/360 � 60 = 0.00245 second.

Example 29: An alternating current varying sinusoidally with a frequency of50 Hz, has an RMS value of 20A. Write down the equation for theinstantaneous value and find this value.

(a) 0.0025 second (b) 0.0125 second after passing through a positivemaximum value. At what time, measured from a positive maximum value, willthe instantaneous current be 14.14 A? [Elect. Sc. I Allah. Univ. 1992]

Solution:

Im = 20 2 = 28.2 A, W = 2� � 50 = 100� red/sec.

Fig. 2.58

225°

60°

A BO

–20 A

D

45°

+20A

14.14 A

C Q

28.2 A

i

2.85��

This gives us the equation for the function for one cycle.

Yav = 1

T 0

Tydt

� =

1

T 0

1010

Tt

T� �

��

�dt

= 1

T 0

1010 . .

Tdt t dt

T� �

��

� =

1

T

2

0

510

Tt

tT

� = 15

Mean square value= 1

T2

0

Ty dt

� =

2

0

1010

Tt

T� �

��

�dt

= 1T

220

100 200100

Tt t

TT� �

� ��

�dt

= 1

T

3 2

2

0

100 100100

3

Tt t

tTT

� � = 700

3

or, RMS value = 10 7/3 = 15.2

Example 31: Determine average value, effective value and form factor of asinusoidally varying alternating current whose half wave is rectified in eachcycle.

Solution: Average value of current is given by,

Iav = area of rectified wave

interval

Fig. 2.60

�O

i

2� �

2.86 ��

= 0

2

id�

�

�

�

= 0sin

2

mI d�

� �

�

� = 0[ cos ]

2mI

�

� �

�

= mI�

...(2.24)

Effective value of current,

I =

2

0

2

i d�

�

�

� = 2

2

0sin

2m

I d�

� �

�

�

= 2mI

�0

(1 cos 2 )

2

� � �

�d�

= 2mI

� 0

1 sin 22 2

�

��

= 2mI

�

= 2

�

= 2mI

� Form factor = /2

/m

m

II �

= 2

�

= 1.57 Ans.

Example 32: Three coils of resistances 20, 30 and 40 � and inductance 0.5,0.3 and 0.2H, respectively are connected in series across a 230 V, 50 c/s supply.Calculate the total current, power factor and the power consumed in the circuit.

Solution:

Total resistance R = 20 + 30 + 40 = 90 �

Total resistance L = 0.5 + 0.3 + 0.2 = 1.0 �

� XL = 2� fL = 2� � 50 � 1.0 = 314 �

Impedance Z = 2 2LR X�

= 2 290 314� = 327 �

2.87��

� Current I = EZ

= 230

327 = 0.704 A.

Power factor cos � = RZ

= 90

327.

= 0.275 lagging.

Power consumed = EI cos �.

= 230 � 0.704 � 0.275.

= 44.5 watts.

Example 33: A resistance of 100 � and a capacitance of 40 �F areconnected in series across a 400 V supply of 50 c/s. Find the current, powerfactor and the power consumed in the circuit. Draw the vector diagram.

Solution:

R = 100 �

XC = 1

2 . .f c�

= 6

1

2 50 40 10�

� � � �

= 79.5 �

Impedance Z = 2 2100 79.5� = 127.8 �

� Current = 400

127.8 = 3.13 A.

Fig. 2.61

90°I = 3.13 A� = 38.5°

F = 400V I X c = 248.5

IR = 313 V

2.90 ��

= 240 � 1.725 � 0.72

= 298 watts. Ans.

(f) Resonance will occur, when

XL = XC

or, 2� fL = 1

2 . .f c�

� fo = 1

2 LC�

= 6

1

2 .2 20 10�

� � �

= 39.8 c/s. Ans.

Example 35: A circuit consisting of resistance of 10 � in series with anXL = 15 � is connected in parallel with another circuit consisting of resistanceof 12 � and capacitive reactance of 20 � combination is connected across a230 V, 50 Hz supply. Find (a) Total current taken from supply (b) Power factorof circuit.

Solution:(a) The given circuit is shown in Fig. 2.63.

Fig. 2.64

230V

I

II

10� 15�

12� –20�

Branch I:

Conductance g1 = 121

RZ

= 2 2

10

10 15�

= 0.0307 �

2.91��

Susceptance b1 = 2 2

15

10 15

�

�

= – 0.0461 �

Branch II:

Conductance g2 = 222

R

Z = 2 2

12

12 20�

= 0.022 �

Since, branch II has capacitive susceptance, so it will be assigned –ve sign.

� Susceptance b2 = 2 2

20

12 20

�

�

= + 0.0368 �

Combined circuit:

Total conductance g = 0.0307 + 0.022 = 0.0527 �

Total susceptance b = b2 – b1 = 0.0368 – 0.0461 = – 0.0093 �

� Total admittance Y = 2 2g b�

= 2 20.0527 0.0093�

= 0.0535 �

� Current taken from supply,

I = E.Y = 230 � 0.0535 = 12.3 A Ans.

(b) Power factor

cos � = gY

= 0.0527

0.0535 = 0.985 lagging Ans.

Example 36: In a parallel circuit, branch I consists of a resistance of 20 � inseries with an inductive reactance of 15 � and branch II has a perfectcondenser of 50 � reactance. The combination is connected across 200 V,60 c/s supply. Calculate:

(a) Current taken by each branch.(b) Total current taken.(c) P.F. of the combination.

Draw vector diagram.

2.92 ��

(a) Branch I:

Z1 = 2 220 15� = 25 �

� I1 = 200

25 = 8 A

�1 = tan–1 15

20 = 36.9° lagging.

Branch II:

Za = 20 ( 50)� � = 50 �.

� I2 = 200

50 = 4 A.

�2 = tan–1 50

0 = tan–1

�

= 90° leading.(b) Combined circuit:

Total current I is the vector sum of the two branch currents I1 and I2.Resolving the currents along E (i.e., in their active components).

I cos � = I1 cos �1 + I2 cos �2

= 8 cos 36.9° + 4 cos 90°

= 8 � 0.8 = 6.4 A.

Fig. 2.65

Solution:

Fig. 2.66

I = 4A2

�2

= 90°

�

�1

= 36.9°

I = 8A1

I

E = 200V

200V

I1

20 � j5 �

– j5 �I2

2.95��

equivalent impedance of the parallel circuit,

23

1

Z=

2 3

1 1

Z Z�

= 1 1

4 3 3 4j j�

� �

= (3 4) (4 3)

(4 3) (3 4)

j j

j j

� � �

� �

= 7 124 7

jj

�

�

= 24 7

7 1

j

j

�

�

= (24 7) 7 1

7 1 7 1

j j

j j

� �

�

� �

= 2 2

175 25 175 25507 1

j j� �

�

�

= 3.5 – j 0.5.

Symbolic expression of the total impedance,

Z13 = Z1 + Z23

= (2.5 + j 1.5) + (3.5 – j 0.5)

= 6 + j 1. Ans.

Taking the voltage as reference vector,

= E = 200 + j 0.

Total current I� = 13

E

Z – 200 0

6 1j

j�

�

= 200 (6 – j 1) = 32.4 – 5.4 = 62 + 1

� I = 2 232.4 5.4� = 328 A Ans.

and, � = tan–1 5.4

32.4� ��

� ��

= tan–1 (–0.1665) = –9.5°

2.96 ��

Voltage across the series branch,

E = I� .Z1 = (32.4 – j 5.4) (2.5 + j 1.5)

= 89.1 + j 35.1

�12 = tan–1 35.1

89.1� �

� ��

= tan–1 0.394 = 21.5°

� E23 = E13 – E12

= 200 + j 0 – (89.1 + j 35.1)

= 110.9 – j 35.1

�23 = tan–1 35.1

110.9� ��

� ��

= tan–1(–0.317) = – 17.6°

� E23 = 2 2110.9 35.1�

= 116 V.

Current in upper parallel branch,

I�1 = 23

2

E

Z = 110.9 35.1

4 3

j

j

�

�

= 2 2

(110.9 35.1) (4 3) 338.3 473.1

254 3

j j j� � �

�

�

= 13.55 – j 18.9

� I1 = 2 213.5 18.9� = 23.2 A Ans.

and, �1 = tan–1 18.9

13.55� ��

� ��

= tan–1(–1.395) = –54.4°.

Current in lower parallel branch,

I2 = 23

3

E

Z = 110.9 35.1 3 4

3 4 3 4

j j

j j

� �

�

� �

= 473.1 338.3

25j�

= 18.9 + j 13.55.

2.99��

= 100 5

2 2� cos 45° = 176.78 W

Example 40: A 100 V, 60 W lamp is to be operated on a 250 V 50 Hz supply.Calculate the value of (a) non-inductive resistor, (b) pure inductance, lamp inorder that it may be used at its rated voltage. What would be required to placein series with the lamp in order that it may be used as its rated voltage.

Solution:Current taken by the lamp

I = 1

PV

= 60

100 = 0.6 A

If R1 is the resistance of the lamp,P = I2R1

R1 = 2

PI

= 2

60

(0.6) = 166.66 �

(a) Non-inductive resistor RWhen a non-inductive resistance R is placed in series with the lamp, the

total resistance of the circuit becomes RT (say), where

RT = R1 + R = 166.66 + R

Since, the circuit is purely resistive

V = RTI

250 = (166.66 + R) � 0.6

R = 250

0.6 – 166.66 = 250 �

(b) Pure inductance LWhen a pure inductance L placed in series with the lamp the total

impedance of the circuit is given byZ2 = R1 + jXL = 166.66 + j 2� � 50L

By Ohm’s lawV = Z2I

250 = (166.66 + j 2� � 50L) � 0.6

250

0.6= 2 2(166.66) (2 50 )L� � �

2.100 ��

22500.6

� �

� ��

– (166.66)2 = (2� � 50L)2

(173611 27775)� = 2� � 50L

L = 1.2155 H

Example 41: Three sinusoidally alternating currents of RMS values 5, 7, 5and 10 A are having same frequency of 50 Hz. with phase angles of 30°, – 60°and 45°.

(i) Find their average values.(ii) Write equation for their instantaneous values.

(iii) Draw wave forms and phasor diagrams taking first current as thereference.

(iv) Find the instantaneous values at 100 m sec from the original reference.[Nagpur Univ. Nov. 1996]

Solution:(i) Average value of alternating quantity in case of sinusoidal nature of

variation = (RMS values)/1.11

Average value of 1st current = 5/1.11 = 4.50 A

Average value of 2nd current = 7.5/1.11 = 6.76 A

Average value of 3rd current = 10/1.11 = 9.00 A

(ii) Instantaneous values

i1(t) = 5 2 sin (314 t + 30°)

i2(t) = 7.5 2 sin (314 t – 60°)

i3(t) = 10 2 sin (314 t + 45°)

(iii) First current is to be taken as a reference, none form the expressionsecond current lags, behind the first current by 90°. Third current leadsthe first current by 15° wave form with this description are drawn in Fig.2.71 (a) and the phasor diagrams in Fig. 2.71 (b).

(iv) A 50 Hz AC quantity completes a cycle in 20 m sec. In 100 m sec, itcompletes five cycles original reference is the starting point required forthis purpose. Hence, at 100 m sec from the reference.

2.101��

(v) Instantaneous value of i1(t) = 5 2 sin 30° = 3.53 A

Instantaneous value of i2(t) = 7.5 2 sin (–60°) = –9.816 A

Instantaneous value of i3(t) = 10 2 sin (45°) = 10 A

Example 42: A resultant current wave is made up of two components.5 A DC component and a 50 Hz AC component. Which is of sinusoidal waveform and which has a maximum value of 5 A.

(i) Draw a sketch of the resultant wave.

(ii) Write an analytical expression for the current wave, reckoning t = 0 at apoint where the AC component is at zero value and when di/dt is +ve.

(iii) What is the average value of the resultant current over a cycle?(iv) What is the effective or RMS value after resultant current?

wt

360°

i2

130°

i3

i1

O90°

i

270°

Fig. 2.70(a)

Fig. 2.70(b)

5°

I = 5A1

I = 10A3

I =7.5A2

2.103��

� RMS value I = 37.5 = 6.12 A.

Alternate�Let the effective value of resultant current is I.

Instantaneous current i = 5 + 5 sin �t

� I2R = 52R + 2

5

2

� �

� ��

R

I = 2 255 37.5

2� �

I = 6.12 amp

Example 43: If the current in a 20� resistor is given by i = 4 + 5 sin�t – 3 cos 3 �t. Determine the power consumed by the resistor.

Solution: P = P0 + P1 + P2

= 42 � 20 + 20 � 2

5

2

� �

� ��

+ 20 � 2

3

2

� �

� ��

= (16 + 12.5 + 4.5) � 20 = 660 watt.

effective value of current = 660

3320

� = 5.7 Amp

Example 44: A large coil of inductance 1.405 H and resistance of 40 � isconnected in series with a capacitor of capacitance 20 µF. Calculate thefrequency at which the circuit resonates. If a voltage of 100 V is applied to thecircuit at resonant condition, calculate the current drawn from the supply andthe voltage across the coil and the capacitor, quality factor, band width.

Solution:R = 40 �, L = 1.405 H, C = 20 � 10–6 F

resonant frequency f0 = 1

2 LC�

= 6

1

2 1.405 20 10�

� � �

= 30 Hz.

At resonance current I0 = 100

40

VR� = 2.5 A

At resonance impedance Z0 = R + j0LX

2.104 ��

Z0 = 0

2 2LR X� = 2 240 264.8� = 267.8 �

� Voltage across the coil at resonance is 0LX

0LV = I0 Z0 = 2.5 � 267.8 � = 669.5 volt

� Capacitive reactance at resonance is 0CX

0CX = 60

1 12 2 30 20 10f C �

�

� � � � �

= 265.2 �

� Voltage across the capacitor

0CV = 0CX I0 = 265.2 � 2.5 = 663 V

� Quality factor Q0 = 0W LR

= 02 2 30 1.405

40

f LR

� � � �� = 6.6175 Ans.

� Band width = 40

1.405

RL� = 28.469

Example 45: A current of 120 – j 50 flows through a circuit when theapplied voltage is 8 + j 2, determine (i) impedance (ii) power factor (iii) powerconsumed and reactive power.

Solution:V = (8 + j 2)V = 8.25 � 14° V

I = (120 – j 50)A = 130 � –22.62° A

(i) Z = 8.25 14

130 22.62

� ��

� � �

VI

= 0.0635 � 36.62° �

� Z = 0.0635 �

(ii) � = 36.62° lag.

� p.f. = cos � = cos 36.62° = 0.803 lag

(iii) Complex VA, S = Phasor voltage � conjugate of phasor currentor p + jQ = 8.25 ��14° � 130 ��22.62° = 1072.5 � 36.62° VA

= 1072.5 (cos 36.62° + j sin 36.62°)

= (860.8 + j 639.75)VA

� Power consumed, P = 860.8 W

Reactive power, Q = 639.75 VAr.

2.105��

Example 46: In an R – L series circuit R = 10 � and XL = 8.66 � if current inthe circuit is (5 – j 10)A, find (i) the applied voltage (ii) power factor and(iii) active power and reactive power.

Solution:Z = R + jXL = (10 + j 8.66) � = 13.23 � 40.9° �

I = (5 – j 10)A = 11.18 � –63.43° A

(i) V = IZ = 11.18 � –63.43° � 13.23 � 40.9° = 148 � –22.53° V

� V = 148 Volts.

(ii) � = 63.43° – 22.53° = 40.9°

p.f. = cos � = cos 40.9° = 0.756 lag.

(iii) S = phasor voltage � conjugate of phasor current.

or P + jQ = 148 � –22.53° � 11.18 � 63.43° = 1654.64 � 40.9° VA.

= (1250.66 + j 1083.36)VA

� Active power, P = 1250.66 W

Reactive power Q = 1083.36 VAr.

Example 47: Two circuits having the same numerical ohmic impedance arejoined in parallel. The power factor of one circuit is 0.8 and the other 0.6. Whatis the power factor of the combination?

Solution:Let Z be the impedance of each circuit

Z1 = Z � cos–1 0.8 = Z � 36.87° = Z (0.8 + j 0.6)

Z2 = Z � cos–1 0.6 = Z � 53.13° = Z (0.6 + j 0.8)

Since, the two impedances are connected in parallel, the equivalentimpedance of the combination is given by

Zp = Z1 || Z2 = 1 2

1 2

Z ZZ Z�

= 2( 36.87 ) ( 53.13 ) 90

(0.8 0.6 0.6 0.8) (1.4 1.4)Z Z Z

Z j j Z j� � � � � �

�

� � � �

= 2 90

(1.98 45 )Z

Z� �

� �

= 45

1.98

� �Z

� The power factor of the combination is

cos � = cos 45° = 0.707

2.107��

Y1 = 1

1 1 6 5

6 5 (6 5) (6 5)

jZ j j j

��

� � �

= 2 2

6 5

6 5

j�

�

= 0.09836 – j 0.08196 S

Z2 = 8 – j 6 = 10 � – 36.87° �

Y2 = 2 22

1 1 8 6 8 68 6 (8 6) (8 6) 8 6

j jZ j j j

� ��

� � � �

= 0.08 + j 0.06 S

Z3 = 8 + j10 = 12.8 � 51.34° �

Y3 = 3

1

Z = 2 2

1 8 10 8 10

8 10 (8 10) (8 10) 8 10

j jj j j

� ��

� � � �

= 0.04878 – j 0.06097 S

Total admittance of the circuit

Y = Y1 + Y2 + Y3

= 0.09836 – j 0.08196 + 0.08 – j 0.06 + 0.04878 – j 0.06097

= 0.22714 – j 0.08293 = 0.2418 20.06� � � S

Total circuit voltage

V = IZ = IY

= 20 0

0.2418 20.06

� �

� � �

= 82.71 � 20.06° V

I1 = 1

VZ

= 82.71 20.06

7.81 39.8

� �

� �

= 10.59 � – 19.74° A = 9.967 – j 3.576 A

I2 = 2

VZ

= 82.71 20.06

10 36.87

� �

� � �

= 8.271 � 56.93° A = 4.513 + j 6.930 A

I3 = 3

VZ

= 82.71 20.06

12.8 51.34

� �

� � �

2.108 ��

= 6.46 � – 31.28° A = 5.52 – j 3.35 A

I1 + I2 + I3 = 20 + j 0 = I

Example 50: A single phase circuit consists of three parallel branches, theadmittance of the branches are

Y1 = 0.4 + j 0.6

Y2 = 0.1 + j 0.4

Y3 = 0.06 + j 0.23

Determine the total admittance and impedance of the circuit.

Solution: Since, the admittances are in parallel

Y = Y1 + Y2 + Y3

= (0.4 + j 0.6) + (0.1 + j 0.42) + (0.06 + j 0.23)

= (0.4 + 0.1 + 0.06) + j (0.6 + 0.42 + 0.23)

= 0.56 + j 1.25 = 1.369 65.86�

Impedance = Z = 1Y

= 1

1.369 65.86�

= 0.73 65.86� �

Z = 0.298 – j 0.666

Example 51: In the network shown in Fig. 2.74, determine (a) the totalimpedance, (b) the total current (c) the current in each branch, (d) the overallpower factor, (e) volt-amperes, (f) active power, and (g) reactive volt-amperes.

1

212 �

7 �

I1

I2 180 F�

0.015 H

C

230 V,

50 Hz

3

IA 5 � 0.01 H

Fig. 2.72

Solution:(a) Branch 1:

R1 = 7 � L1 = 0.015 H

1LX = 2��fL1 = 2� � 50 � 0.015 = 4.71 �

Z1 = R1 + j1LX = 7 + j 4.71 = 8.437 33.93� ��

2.109��

Branch 2:

R2 = 12 � C2 = 180 �F = 180 � 10–6 F

2CX = 2

12 f C�

= 6

1

2 50 180 10�

� � � �

= 17.68 �

Z2 = R2 – j2CX = 12 – j 17.68 = 21.37 � – 55.83° �

Since, Z1 and Z2 are connected in parallel, their equivalent impedance Zp isgiven by

Zp = Z1 || Z2 = 1 2

1 2

Z ZZ Z�

= (8.437 33.93 ) (21.37 55.83 )

(7 4.71) (12 17.68)j j

� � � � �

� � �

= 180.3 21.9 180.3 21.9

19 12.97 23 34.3j

� � � � � �

�

� � � �

= 7.839 � + 12.4° = 7.656 + j 1.68 �

Branch 3:

R3 = 5 �, L3 = 0.01 H

3LX = 2��fL3 = 2� � 50 � 0.01 = 3.14 �

Z3 = R3 + j3LX = 5 + j 3.14 = 5.9 � 32.13° �

Since, Z3 and Zp are connected in series, the total impedance of the circuitis

Z = Z3 + Zp = 5 + j 3.14 + 7.656 + j 1.68 = 12.656 + j 4.82 = 13.54 20.85� ��

(b) Let the supply voltage V be taken as reference phasor.

V = 230 � 0° V = 230 + j 0 V.

By Ohm’s law total circuit current is

I = VZ

= 230 0

13.54 20.85

� �

� �

= 16.99 � –20.85° A

= 15.87 – j 6.046 A

2.111��

Fig. 2.73

(i) � Z = 230 30

44.7 3.46

V

I

� ��

��

= 5.14 �33.46° �

(ii) P = V.I cos � = 230 � 44.7 � cos 33.46° = 8577 W.

Example 53: A parallel circuit consists of a 2.5 �F capacitor and a coilwhose resistance and inductance are 15 � and 260 mH, respectively.Determine (i) the resonant frequency (ii) Q-factor of the circuit at resonance(iii) dynamic impedance of the circuit.

Solution:(i) Resonant frequency,

fr = 2

2

1 1

2

R

LC L�

�

= 6 2

2

1 10 (15)

2 0.260 2.5 (0.260)

�

� �

= 197 Hz

(ii) Q-factor = 2 2 197 0.260

15rf L

R

� � � �� = 21.45

(iii) Zr = 6

0.260

2.5 10 15

LCR �

�

� �

= 6933 �

EXERCISE

1. What is meant by an alternating quantity? Explain how a sine wave isproduced.

2. Define: cycle, periodic time and frequency.

230 30°V�

I 20 60°A1

� I 40 –30°AL

�

I

2.112 ��

3. What is understood by “phase difference” between two alternatingquantities? Explain the term lagging current and leading current withthe aid of suitable curves.

4. Define RMS value of an alternating current.Derive RMS value in case of a:(a) Sinusoidal wave

(b) Rectangular wave(c) Triangular wave(d) Semicircular wave

(e) Trapezoidal wave(g) Stepped wave.

5. Define average value of an alternating current.

Derive average value in case of a:(a) Sinusoidal wave(b) Rectangular wave

(c) Triangular wave(d) Semicircular wave(e) Trapezoidal wave

(g) Stepped wave.Also define form factor and find its value in case of all the above waves.

6. Define peak or crest factor and state its practical utility.7. Determine average value, effective value and form factor of a triangular

wave whose half wave is suppressed in each cycle.

8. Two waves represented by e1 = 3 sin �t. and e2 = 4 sin 3

t��

� ��

are

acting in a circuit. Find an expression of their resultant and check theresult by a graphical construction. Also find the peak and RMS valuesof the resultant.

[ 37 sin (�t. – 0.605); 37 , 4.3]

9. An alternating current is given in amperes by the expression,i = 50 sin 44 �t.

Find (a) frequency.

(b) w in radians per second(c) maximum value of the current(d) effective value of the current

[(a) 70 c/s, (b) 440, (c) 50 A and (d) 35.35 A]

2.115��

Calculate also the total current supplied in each case if the appliedvoltage is 240 V. [6.74 �F, 16 A & 0.0648 A]

26. Define effective, equivalent or dynamic impedance of a rejecter circuit.Find the current in a parallel circuit at resonance after making practicalassumptions.

27. Define Q-factor and determine its value in:

(a) Series resonant circuit(b) Parallel resonant circuit

Steady-State Analysis of Single Phase A.C. Circuit Steady-State Analysis of Single Phase A.C....

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