CHAPTER 2 : AC CIRCUITS

INTRODUCTION

All alternating waveform changes its magnitude and direction periodically. Figure shows various
AC waveforms.

Many times alternating voltages and currents are represented by a sinusoidal waveforms. A
sinusoidal voltage can be represented as

v = Vmsinθ
= Vmsinωt
= Vmsin2πft (wkt f = 1/T)
𝟐𝝅
v=Vmsin 𝒕
𝑻

TERMS RELATED WITH ALTERNATING QUANTITIES

1. Waveform: A waveform is a graph in which the instantaneous value of any quantity is plotted against
time.
2. Cycle: One complete set of positive and negative values of an alternating quantity is termed as cycle
3. Frequency: The number of cycles per second of an alternating quantity is known as frequency. It is
denoted by ‘f’ and ins expressed in hertz (Hz) or cycles/sec (C/s)
4. Time Period: The time taken by an alternating quantity to complete one cycle is called time period. It
1
is denoted by ‘T’ and is expressed in seconds. ;T = 𝑓 sec
5. Amplitude: The max positive or negative values of an alternating quantity is called amplitude.
6. Phase : The phase of an alternating quantity is the time that has elapsed since the quantity has last
passed through zero point of reference.
7. Phase difference: This term is used to compare the phases of two alternating quantities. Two
alternating quantities are said to be in phase when they reach the max and zero values at the same time.
Their max value may be different in magnitude.
A leading alternating quantity is one which reaches its max or zero value earlier as compared to the
other quantity.
A lagging alternating quantity is one which attains its max or zero value later as compared to the
other quantity.
A plus (+) sign which used in connection with the phase difference denotes leading, whereas minus
(-) denotes lagging.
VA = Vmsinωt
VB = Vmsin (ωt + Φ)

1
 Here quantity B leads A by a phase angle Φ

Half wave rectified sine wave Full wave rectified sine wave
Average Value Vm 2Vm
Vav = Vav =
π π

RMS value 𝐼𝑚 𝐼𝑚
Irms = Ieff = Irms = Ieff =
2 √2

Form factor :It is defined as the ratio of rms value of a wave and the average value of the wave.
𝐫𝐦𝐬 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐰𝐚𝐯𝐞
Form factor =
𝐚𝐯𝐞𝐫𝐚𝐠𝐞 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐰𝐚𝐯𝐞

Peak factor or crest factor : It is the ratio of the peak value of the wave to its rms value.
𝐫𝐦𝐬 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐰𝐚𝐯𝐞
Peak factor =
𝐚𝐯𝐞𝐫𝐚𝐠𝐞 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐰𝐚𝐯𝐞

Form factor Peak factor

Sinusoidal wave 1.11 √2 = 1.414

Triangular wave 2 √3 = 1.732

√3

PHASOR DIAGRAMS:
• A geographical or pictorial representation of phases is known as phasor diagram.
• Phasor diagram is a pictorial representation of all the phase voltages and their respective
currents in a network.
• Phasor is a rotating vector, which rotates in the anticlockwise direction with angular
frequency ‘ω’ in the time domain.

2
 Fig: 2(a) VI sinusoidal wave 2(b) Phasor diagram of 2(a)

At each point on the time waveform, the angle of current lag is θ because the angle between the phasors
Vm and Im is at all time θ. Therefore, either the time waveform of the rotating phasors or the phasor
diagram, can be used to describe the system. Since both the diagrams, the time diagram and the phasor
diagram convey the same information, the phasor diagram being much more simpler, it is used for an
explanation in circuit theory analysis. Since electrical data is given in terms of rms value, we draw phasor
diagram with phasor values as rms rather than peak value used so far.

SINGLE PHASE SERIES CIRCUITS:

Single Phase series Circuits for pure passive elements
Circuit Voltage Current relation Phasor relation Phasor diagram
relation and power
factor(PF)
Resistor v = Vmsinωt i = Im sinωt • Current is in
phase with
voltage
• Power
factor(PF) is
unity
𝜋 •
Inductor v = Vmsinωt i = Imax (𝜔𝑡 − 2 ) Current lags
voltage by an
(XL= 2𝜋fL) angle 900

• Power
factor(PF) is
zero (lagging)
𝜋 •
Capacitor v = Vmsinωt I = Imsin(ωt + 2 ) Current leads
voltage by an
𝟏
(Xc = 𝟐𝜋fC) angle 900

• Power
factor(PF) is
zero (leading)

3
 1) AC through series R-L circuit:

Fig 3(a): series RL circuit

Consider a circuit consisting of a pure resistance(R) in ohms is connected in series with a pure
inductance(L) in henry as shown in fig 3(a).
The series combination is connected across AC supply given by v = Vmsinωt
Circuit draws current I, then the two voltage drops are given by
a) Drop across pure resistance VR = IR
b) Drop across pure inductance VL = IXL
Where, XL = 2πfL and I, VR , VL are the rms values.
KVL can be applied to get , V = V𝑅 + V𝐿 ……………………..(1)
V = 𝐼𝑅 + IX𝐿
Let us draw the phasor diagram, Current I is taken as reference as it is common in both the
elements.

Fig 3(b) : Phasor diagram Fig 3(c) : Voltage triangle

❖ Impedance: Impedance is defined as the opposition in the circuit for the flow of alternating current.
It is denoted by Z and its unit is ohms(Ω).
For the R-L series circuit, it can be observed from the phasor diagram that the current lags behind the
applied voltage by an angle Φ. From the voltage triangle, we can write,
𝐕𝐋 𝐗𝐋 𝐕𝐑 𝐑 𝐕𝐋 𝐗𝐋
𝐭𝐚𝐧 𝚽 = = ; 𝐜𝐨𝐬 𝚽 = = ; 𝐬𝐢𝐧 𝚽 = =
𝐕𝐑 𝐑 𝐕 𝐙 𝐕 𝐙

Fig 3(d) : Impedance triangle

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 From the impedance triangle, resistance R is given by R = Z 𝐜𝐨𝐬 𝜱
Inductive reactance XL is given by XL = Z 𝐬𝐢𝐧 𝜱
In rectangular form, impedance is given by Z = R + jXL Ω
In polar form, impedance is given by Z = |𝒁|∟𝜱
𝐗𝑳
Where, |𝐙| = √𝐑𝟐 + 𝐗 𝐋 𝟐 ; Φ = 𝐭𝐚𝐧−𝟏 ( )
𝑹

❖ Power and Power triangle:

Vm Im Vm Im
Average power Pav = cos Φ = ∗ *cos Φ
2 √2 √2
P = VI watts where V and I are rms values
If we multiply eqn (1) by I, we get
̅ = IVR + IVL
VI
̅ = ̅̅̅̅̅̅̅̅̅
VI ̅̅̅̅̅̅̅̅̅
VcosΦI + VsinΦI
From the above equation, power triangle can be obtained as shown in fig 3(e).

Fig 3(e) : Power triangle

So, the three sides of the triangle are (i) VI (ii) VIcosΦ (iii) VIsinΦ
❖ Apparent power (S): It is defined as the product of rms values of voltage(v) and current(I). It is
denoted by S. Its unit is volt-amp(VA) or kilo volt-amp(KVA) S = V*I VA
❖ Real or True power (P): It is defined as the product of applied voltage and active component of
current. It is the real component of apparent power. It is represented by ‘P’ and its unit is watts(W) or kilo
watts(KW). P = VI𝐜𝐨𝐬 𝚽 watts
❖ Reactive power (Q): It is defined as the product of applied voltage and reactive component of
current. It is also defined as the imaginary component of apparent power. It is represented by ‘Q’ and its
unit is volt-ampere reactive(VAR) or (KVAR). Q = VI𝐬𝐢𝐧 𝚽 VAR

❖ Power factor ( 𝐜𝐨𝐬 𝚽): It is defined as a factor by which apparent power must be multiplied in
order to obtain the true power.
True power VI cos Φ
Power factor = = = cos Φ
Apparent power VI
The numerical value of cosine of the phase angle between the applied voltage and the current drawn
from the supply voltage gives the power factor. It is always less than 1.
𝐑
It is also defined as the ratio of resistance to impedance. 𝐜𝐨𝐬 𝚽 =
𝐙

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 NOTE: If current lags voltage, power factor is said to be lagging, if current leads voltage, power factor
is said to be leading.
So ,for pure inductance, the power factor is 𝑐𝑜𝑠(90°) ie.,zero power factor(ZPF) lagging, while for
pure capacitance, the power factor is sin(90°) ie.,zero power factor(ZPF) leading. For a purely resistive
circuit volatge and current are in phase ie., Φ = 0°. Therefore, power factor is 𝑐𝑜𝑠(0°)= 1. Such circuit is
called unity power factor(UPF) circuit.
𝑷𝒐𝒘𝒆𝒓 𝒇𝒂𝒄𝒕𝒐𝒓 = 𝒄𝒐𝒔 𝜱 , where 𝛷 is the angle between voltage and current.

2) AC through series R-C circuit:

Fig 4(a): series RC circuit

Consider a circuit consisting of a pure resistance(R) in ohms is connected in series with a pure
capacitance(C) in farad as shown in fig 4(a).
The series combination is connected across AC supply given by v = Vmsinωt
Circuit draws current I, then the two voltage drops are given by
a) Drop across pure resistance VR = IR
b) Drop across pure capacitance VC = IXC
1
Where, XC = and I, VR , VC are the rms values.
2πfC
KVL can be applied to get , V = V𝑅 + V𝐶 ……………………..(1)
V = 𝐼𝑅 + IX𝐶
Let us draw the phasor diagram, Current I is taken as reference as it is common in both the
elements.

Fig 4(b) : Phasor diagram Fig 4(c) : Voltage triangle

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 ❖ Impedance: Impedance is defined as the opposition in the circuit for the flow of alternating current.
It is denoted by Z and its unit is ohms(Ω).
For the R-C series circuit, it can be observed from the phasor diagram that the current leads the applied
voltage by an angle Φ. From the voltage triangle, we can write,
𝐕𝐂 𝐗𝐂 𝐕𝐑 𝐑 𝐕𝐂 𝐗𝐂
𝐭𝐚𝐧 𝚽 = = ; 𝐜𝐨𝐬 𝚽 = = ; 𝐬𝐢𝐧 𝚽 = =
𝐕𝐑 𝐑 𝐕 𝐙 𝐕 𝐙

Fig 4(d) : Impedance triangle

From the impedance triangle, resistance R is given by R = Z 𝐜𝐨𝐬 𝜱

Inductive reactance XL is given by XL = Z 𝐬𝐢𝐧 𝜱
In rectangular form, impedance is given by Z = R - jXC Ω
In polar form, impedance is given by Z = |𝒁|∟ − 𝜱 Ω
−𝐗 𝑪
Where, |𝐙| = √𝐑𝟐 + 𝐗 𝐂 𝟐 ; Φ = 𝐭𝐚𝐧−𝟏 ( )
𝑹

❖ Power and Power triangle:

Vm Im Vm Im
Average power Pav = cos Φ = ∗ *cos Φ
2 √2 √2
P = VI watts where V and I are rms values
n
If we multiply eq (1) by I, we get
̅ = IVR + IVL
VI
̅ = ̅̅̅̅̅̅̅̅̅
VI ̅̅̅̅̅̅̅̅̅
VcosΦI + VsinΦI
From the above equation, power triangle can be obtained as shown in fig 4(e).

Fig 4(e) : Power triangle

Thus, the various powers are,
a. Apparent power, S = VI VA
b. True or avg power, P = VI𝐜𝐨𝐬 𝚽 watts
c. Reactive power, Q = VI𝐬𝐢𝐧 𝚽 VAR

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 Note: Remember that, XL term appears to be positive in Z
ie., Z = R + jXL Ω = |𝑍|∟𝛷 , where Φ is positive for Z.
XC term appears to be negative in Z
ie., Z = R – jXC Ω = |𝑍|∟ − 𝛷, where Φ is negative for Z.

3) AC through series R-L-C circuit:

Fig 5(a) : Series RLC circuit

Consider a circuit consisting of a pure resistance(R) in ohms is connected in series with a pure
inductance(L) in henry and a pure capacitance(C) in farad as shown in fig 5(a).
The series combination is connected across AC supply given by v = Vmsinωt
Circuit draws current I, then the two voltage drops are given by
a) Drop across pure resistance VR = IR
b) Drop across pure inductance VL = IXL; Where, XL = 2πfL
1
c) Drop across pure capacitance VC = IXC Where, XC =
2πfC
The values of I, VR , VL, VC are the rms values.
The characteristics of three drops are
(a) VR is in phase with current I.
(b) VL leads current by 90°
(c) VC lags current by 90°
According to kirchoff’s law, we can write,
𝑉̅ = ̅̅̅
𝑉𝑅 + 𝑉̅𝐿 + 𝑉
̅̅̅𝐶
Let us draw the phasor diagram, Current I is taken as reference as it is common to all the elements.
The phasor diagram depends on the condition ofmagnitudes of VL and VC which ultimately depends on
the valurs of XL and XC.

Let us consider the different cases,

(i) XL > XC : When XL > XC, obviously IXL ie.,VL is greater than IXC ie., VC. So the resultant of VL
and VCwill be directed towards VL leading current I. Current I will lag the resultant of VL and VC
ie., (VL – VC).
The circuit is said to be inductive in nature. The phasor sum of VR and (VL – VC)
gives the resultant supply voltage V. This is shown in fig 5(b).

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 Fig 5(b) : Phasor diagram and voltage triangle for (XL > XC)
From the voltage triangle, V = √(VR )2 + (VL − VC )2 = √(IR)2 + (IXL − IXC )2
V = I √(R)2 + (XL − XC )2
V = IZ
Where, Z = √(R)2 + (XL − XC )2 ; (𝐗 𝐋 − 𝐗 𝐂 ) is positive.

(ii) XL < XC : When XL < XC, obviously IXC ie.,VC is greater than IXL ie., VL. So the resultant of VL
and VC will be directed towards VC lagging current I. Current I will leads the resultant of VL and
VC ie., (VC – VL).
circuit is said to be capacitive in nature. The phasor sum of VR and (VC – VL) gives the
resultant supply voltage V. This is shown in fig 5(c).

Fig 5(c) : Phasor diagram and voltage triangle for (XL < XC)
From the voltage triangle, V = √(VR )2 + (VC − VL )2 = √(IR)2 + (IXC − IXL )2
V = I √(R)2 + (XC − XL )2
V = IZ
Where, Z = √(R)2 + (XC − XL )2 ; (𝐗 𝐂 − 𝐗 𝐋 ) is positive

(iii) XL = XC : When XL = XC, obviously IXC ie.,VC is equal to IXL ie., VL ie., (VL – VC) = 0. Current
I be in phase with the voltage VR.
circuit is said to be resistive in nature. The phasor sum of VR and (VL – VC) gives the
resultant supply voltage V ie., VR itself. This is shown in fig 5(d).

V = IZ, where Z = R

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 1. AC through parallel R-L circuit

Fig :6(a) Fig: 6(b)

Consider a circuit consisting of a pure resistance(R) in ohms is connected in parallel across a pure
inductance(L) in henry as shown in fig 6(a).
We have taken here V as the reference quantity as it is the voltage which is common to both the
elements R and L and not the current. Now current through R will be in phase with the voltage across it
whereas current through inductor will lag by 90° the voltage across it. Hence, the phasor diagram in Fig.
6(b) follows and current I supplied by the source equals the phasor sum of the current I1 and I2.
V V V V
Now, the current is given by , IR = ; IL = ∟ − 90° ; I = IR + IL = +
R XL R jωL
1 1
Here, is the reciprocal of resistance and is termed as conductance(G), whereas ωL is the reciprocal of
R
the inductive reactance and is known as inductive susceptance(BL).
From the above equation, we have
I 1 1
= + ……………(1)
V R jωL
I
Now dimensionally is reciprocal of impedance and is known as admittance(Y). The above equation
V
can be rewritten as Y = G – jBL
1 1 1
Where, Impedance Y = ; Conducatnce G = ; Susceptance BL =
Z R ωL
It is to be noted that whereas for an inductive circuit Z = R + jXL, the admittance is Y= G – jBL

The units of admittance is mho or siemens and denoted by or S respectively.

The power consumed by the circuit is again P = VI cosΦ where I is the total current and Φ is the angle
−BL G IR
between V and I. Φ = tan−1 ( ) ; Power factor , cosΦ = Y = (lag)
G I
1 1 1
From eqn (1), we have = +
Z R jωL
If Z1 = R and Z2 = jωL, we have
1 1 1 Z1 ∗ Z2
= + ; Z= ………(2)
Z 𝑍1 𝑍2 Z1 + Z2
i.e. if there are two impedances Z1, Z2 connected in parallel, their equivalent impedance equals the ratio
of the product of the two impedance to the sum of the two impedances. Also from eqn (2)
Y = Y1 + Y2
i.e. if there are two admittances Y1 and Y2 connected in parallel, the equivalent admittance is the sum of
the two admittances.

10
 2. AC through parallel R-C circuit

Fig 7(a) Fig 7(b)

Consider a circuit consisting of a pure resistance(R) in ohms is connected in parallel across a pure
capacitance(c) in farad as shown in fig 7(a).
We have taken here V as the reference quantity as it is the voltage which is common to both the
elements R and C and not the current. Now current through R will be in phase with the voltage across it
whereas current through capacitor will lead by 90° the voltage across it. Hence, the phasor diagram in Fig.
7(b) follows and current I supplied by the source equals the phasor sum of the current I1 and I2.
V V V
Now, the current is given by, IR = ; IC = ∟90° ; I = IR + IC = + jVωC
R XC R
1
Here, is the reciprocal of resistance and is termed as conductance(G), whereas ωC is the reciprocal of
R
the capacitive reactance and is known as capacitive susceptance(BC).
From the above equation, we have
I 1
= + jωC ………..(3)
V R
I
Now dimensionally is reciprocal of impedance and is known as admittance(Y). The above equation
V
can be rewritten as Y = G + jBC
1 1
Where, Impedance Y = ; Conductance G = ; Susceptance BC = ωC
Z R
It is to be noted that whereas for an inductive circuit Z = R – jXC , the admittance is
Y= G + jBC

The units of admittance is mho or simens and denoted by or S respectively.

The power consumed by the circuit is again P = VI cos Φ where I is the total current and Φ is the angle
B G IR
between V and I. Φ = tan−1 ( GC ) ; Power factor , cosΦ = = (lead)
Y I

3. AC through parallel R-L-C circuit

Fig 8(a) : parallel RLC circuit

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 Consider a circuit consisting of a pure resistance(R) in ohms is connected in parallel across a pure
inductance(L) in henry and a pure capacitance(C) in farad as shown in fig 8(a).
The parallel combination is connected across AC supply given by v = Vmsinωt
V V V
The current is given by , IR = ; IL = ∟ − 90° ; IC = ∟90°
R XL XC
I. BC > BL ( IC > IL )

Current, I = √(IR )2 + (IC − IL )2

IC − IL BC − BL
Admittance angle, Φ = tan−1 ( ) = tan−1 ( )
IR G
IR
Power factor, cosΦ = ( PF lead)
I
II. BC < BL ( IC < IL )

Current, I = √(IR )2 + (IL − IC )2

IL − IC BL − BC
Admittance angle, Φ = tan−1 ( ) = tan−1 ( )
IR G
IR
Power factor, cosΦ = ( ZPF lag)
I
III. BL = BC (IL = IC)

Current, I = IR ; Φ = 0° ; cosΦ = 1 (UPF)

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 Impedance(Z)
Circuit Voltage and current or Phase angle Phasor relation Power
relation Admittance(Y) Φ factor(PF)
= cosΦ

Series RL
circuit 𝑋𝐿 R VR
tan−1 ( ) =
Z = (R + jXL) 𝑅 I lags V by an angle Φ z V
V = √VR 2 + VL 2

Series RC
circuit −𝑋𝐶 R VR
tan−1 ( ) =
Z = (R - jXC) 𝑅 I leads V by an angle Φ z V
V = √VR 2 + Vc 2

Series
RLC 𝑋𝐿 − 𝑋𝐶 XL>XC; I lags V R VR
tan−1 ( ) =
circuit V =√(VR )2 + (VL − VC )2 Z=R + j(XL-XC) 𝑅 XL<XC; I leads V z V
XL=XC;I in phase with V

Parallel
RL circuit −𝐵𝐿 G IR
tan−1 ( ) =
Y = (G - jBL) 𝐺 I lags V by an angle Φ Y I
I = √IR 2 + IL 2

Parallel
RC circuit 𝐵𝐶 G IR
tan−1 ( ) =
I = √IR 2 + Ic 2
Y = (G + jBC) 𝐺 I leads V by an angle Φ Y I

Parallel
RLC 𝐵𝐶 − 𝐵𝐿 BL>BC; I lags V G IR
tan−1 ( ) =
circuit I = √(IR )2 + (IL − IC )2 Y=G + j(BC-BL) 𝐺 BL<BC; I leads V Y I
BL=BC;I in phase with V

R = Resistance ; X = Reactance; Z = Impedance

1 1 1
G= = Conductance; B= = susceptance; Y= = Admittance
R X Z

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 Concept of Time constant:
A. Time Constant of an RC Circuit
Let us take a simple RC circuit, as shown below.

Fig 9: Series RC circuit

Let us assume the capacitor is initially unchanged and the switch S is closed at time t = 0. After
closing the switch, electric current i(t) starts flowing through the circuit. Applying Kirchhoff Voltage Law
in that single mesh circuit, we get,
1
V = Ri(t) + 𝐶 ∫ 𝑖(𝑡)𝑑𝑡
Differentiating both sides with respect to time t, we get,
di(t) i(t)
R + =0
dt C
di(t) −i(t)
R =
dt C
di(t)
-RC = dt
i(t)
di(t) −1
= dt
i(t) RC
Integrating on both sides, we get,
di(t) −1
∫ i(t) = = dt
RC
−t
log e i(t) + log e i(k) =
RC
−t
log e (ki(t)) =
RC
−t
k*i(t) = e RC …………….(1)

Now, at t = 0, the capacitor behaves as a short circuit, so just after closing the switch. The current
V
through the circuit will be Io =
R
Now, substituting the above eq in eqn(1), we get
n

V R
k* = e0 ; k =
R V
Now, putting the value of k in eqn (1), we get
V −t
i(t) = eRC
R

14
 Now, if we put t = RC in the final expression of circuit current i(t), we get,
V −1 𝐕
I t= RC = = e = 0.367 = 0.364Io = 36.7% Io
R 𝐑
From the above mathematical expression, it is clear that RC is the time in second during which the
current in a charging capacitor diminishes to 36.7 percent from its initial value. Initial value means current
at the time of switching on the unchanged capacitor.
This term is quite significant in analyzing the behavior of capacitive as well as inductive circuits.
This term is known as the time constant.
So time constant is the duration in seconds during which the current through a capacities circuit
becomes 36.7 percent of its initial value. This is numerically equal to the product of resistance and
capacitance value of the circuit. The time constant is normally denoted by τ (tau).
So, τ = RC
In a complex RC circuit, the time constant will be the equivalent resistance and capacitance of the
circuit.
Let us discuss the significance of the time constant in more detail. To do this, let us first plot current
i(t).

𝐕
At t = 0, the current through the capacitor circuit is Io =
𝐑
𝐕
At t = RC, the current through the capacitor is 0.364Io = 0.367
𝐑
B. Time Constant of an RC Circuit
Let us take a simple RC circuit, as shown below

Fig 10 : Series RL circuit

15
Applying Kirchhoff Voltage Law in the above circuit. We get,
di(t)
V = Ri(t) + L
dt
The equation can also be solved Laplace Transformation technique. For that, we have to take Laplace
Transformation of the equation at both sides,
V
= L[sI(s) – i(0)] + RI(s)
s
V
= LsI(s) + RI(s)
s
V
I(s) =
s(sL+R)
𝑉𝑜 1
Now, I(s) =
𝑅 s(sL+R)
Taking inverse laplace transform of the above equation, we get
−Rt
Vo
i(t) = (1- e L )
R
L 𝐿 Vo Vo
Now, if we put t = , we get, i(𝑅) = (1- e−1 ) = 0.633
R R R

At the RL circuit, at time t = L/R sec, the current becomes 63.3% of its final steady-state value. The
𝑳
L/R is known as the time constant of an LR circuit ie., τ =
𝑹
Let us plot the current of the inductor circuit.

Resonance: Resonance in electric circuit is because of the presence of both energy storing elements
ie., inductor and capacitor.
Under resonance, for an electric circuit, if the applied voltage and resultant current, both are in
phase (Φ = 0°) , then the system is under resonace.
Z = Resistive (ie., imaginary part = 0)
(OR) The total impedance of any electrical circuit is only resistive (imaginary part = 0) then
circuit is under resonance.
(OR) If any electrical circuit is operating at UPF( ie., cos Φ = 0), then the circuit is under
resonance.

16
 Resonance

Series Resonance Parallel Resonance

➢ R-L-C Parallel circuit
➢ R-L & C parallel circuit
➢ R-L & R-C parallel circuit
➢ Series Resonance
❖ R-L-C series circuit:

Fig 11: series RLC circuit for resonance

The impedance of the circuit at any frequency ω is given as
Z = ZR + ZL + ZC
1
Z = R + j(ωL - )
𝜔𝐶
Since resistance is independent of frequency the circuit will have minimum impedance at some
frequency.
1 𝟏 𝟏
When Z=R ; Or when, ωo L - =0; ωo = ; fo =
ωo C √𝐋 𝐂 𝟐𝛑√𝐋 𝐂
Here fo is the frequency of resonance i.e. if a circuit has fixed values of R, L and C, resonance will
1
take place if the supply frequency fo = , when the impedance of the circuit is purely resistive i.e. the
2π√L C
power factor of the circuit is unity, the supply voltage and current are in phase. However, it is to be noted
that the phase relations between the voltage and current in the individual elements R, L and C are not same.
The current in the inductor lags its voltage by 90° and in the capacitor it leads its voltage by 90° •

17
 Since R is independent of frequency it is shown by a horizontal line Z = R. Also XL = ωL the
inductive reactance is a straight line passing through the origin and inductive reactance is taken as +ve,
1
where as XC = the capacitive reactance as a function of ω is a rectangular hyperbola and the reactance
ωC
is taken as -ve. The net impedance is shown a positive quantity. The resonance frequency is f0 where (ωL -
1
) is zero and at this frequency the impedance curve has minimum value equal to R.
𝜔𝐶

Fig 12: Z Vs ω for series RLC circuit

The variation of current is also shown in Fig 12 as a function of frequency and is maximum at f 0
whereas on either side the current decreases. It is to be noted that at ω = 0, the current in the RLC series
circuit is zero as the capacitor reactance is infinite and, therefore, the graph starts from origin whereas it is
again zero at ω = ∞ and hence the graph should not be passing through zero rather it should have some
finite value as indicated in the diagram.
Again, it can be seen that the series circuit is capacitive for all frequency ω < ω0, inductive for all
frequency ω> ω0 and at ω = ω0 the circuit is resistive.
There are various applications of a series resonant circuit where the frequency is fixed and either L
or C is varied to obtain the condition of resonance. A typical example is that of tuning a radio receiver to a
particular desired station that is operating at a fixed frequency. Here a circuit or L and C is adjusted to
resonance at the operating frequency of the desired station. The capacitor C (parallel plate) is variable in
most portable radio receivers and the inductance of the coil is usually varied in tuning of an automobile
radio receiver.
V
Under resonance, Io =
R
V
Therefore, VR = IoR = *R = V is the supply voltage
R
V ωo L
VL = IoωL = ωL = V
R R
where Q is known as the quality factor of the series network. Usually Q >> 1, hence it is also
known as voltage gain, as the voltage across the inductor is much greater than the supply voltage.
IO 1
Also, VC = =V = VQ; VL = Vc >> V
ωO C ωO CR
Therefore, extreme care must be taken when working on series circuits that may become resonant
when connected to power line sources .

18
 Quality factor (Q):
In a practical circuit R is essentially resistance of the coil since practical capacitors have very low loss
in comparison to practical inductor. Hence Q is a measure of the energy storage property (LI2) in relation to
the energy dissipation property (I2R) of a coil or a circuit.
The Q is, therefore, defined as
𝐦𝐚𝐱𝐢𝐦𝐮𝐦 𝐞𝐧𝐞𝐫𝐠𝐲 𝐬𝐭𝐨𝐫𝐞𝐝
Q = 2π
𝐞𝐧𝐞𝐫𝐠𝐲 𝐝𝐢𝐬𝐬𝐢𝐩𝐚𝐭𝐞𝐝 𝐩𝐞𝐫 𝐜𝐲𝐜𝐥𝐞

In electric circuit energy is stored in the form of electromagnetic field in the inductance where as in
electrostatic form of energy across a capacitance. It can be proved that at any instant at a certain frequency
the sum of energy stored by the inductor and the capacitor is constant. At the extreme situation when the
current through the inductor is maximum, the voltage across the capacitor is zero hence the total energy is
1
L(√2 I)2 = LI2
2
where √2I is the instantaneous maximum value of the current. At this since VC is zero, therefore
maximum energy stored is LI2 . The power consumed per cycle is the energy per sec divided
I2 R
by f0 under resonance condition. Therefore PR =
fo
2πLI2 𝛚𝐨 𝐋
Hence Q= ; Q= (Q for series RL circuit)
I2 R 𝐑
NOTE:
Q can also be looked as a ratio of
𝑇𝑖𝑚𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑 𝑅𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑝𝑜𝑤𝑒𝑟 𝑎𝑏𝑠𝑜𝑟𝑒𝑑 𝑏𝑦 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟
Q= =
𝑇𝑖𝑚𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 𝐴𝑐𝑡𝑖𝑣𝑒 𝑝𝑜𝑤𝑒𝑟 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑏𝑦 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟

The great advantage of this definition of Q is that it is also applicable to more complicated lumped
circuits, to distributed circuits such as transmission lines and to non-electrical circuits.

Q is also a measure of the frequency selectivity of the circuit. A circuit with high Q will have a very
sharp current response curve as compared to one which has a low value of Q. To understand this let us
consider Fig. 13 . Here we find that the current response is maximum at f0 and on either side of f0, the
current decreases sharply.

Fig 13 : Frequency selectivity

Frequencies f1 and f2 are known as half-power frequencies as at these frequencies the power
dissipated by the circuit is half of that dissipated at fo.
The band width of a resonant circuit is defined as the frequency range between the 70.7% current
points. Bandwidth, BW =f2 - f1

19
 Resonant frequency , ωo = √𝛚𝟏 𝛚𝟐
This means the resonance frequency is the geometric mean of the half-power frequencies.
𝛚𝟐 − 𝛚𝟏 𝟏 1
= 𝛚oCR = ( where, Q = ωoCR is the Q factor value for RC series circuit)
𝛚𝒐 𝐐
𝐟𝐨
f2 – f1 =
𝐐
Bandwidth is thus given by the ratio of the frequency of resonance to the quality factor and
selectivity is defined as the ratio of resonant frequency to the bandwidth f 0/(f2 – f1). This, therefore, shows
that the larger the value of Q the smaller is (f2 – f1) and hence sharper is the current response.
𝟏 𝒇𝒐 𝝎𝒐
Selectivity ∝ Q-factor ∝ ;Q= ; Q=
𝐛𝐚𝐧𝐝𝐰𝐢𝐝𝐭𝐡 𝒇𝟐 −𝒇𝟏 𝝎𝟐 −𝝎𝟏
NOTE
Q factor for
𝝎𝒐 𝑳
❖ RL series circuit: Q =
𝑹
1
❖ RC series circuit: Q =
𝜔𝑜𝐶𝑅
1 𝐿
❖ RLC series circuit: Q = √
𝑅 𝐶

➢ Parallel Resonance
❖ Parallel R-L-C circuit

Fig 14 : Parallel RLC circuit

Consider a circuit consisting of a pure resistance(R) in ohms is connected in parallel across
a pure inductance(L) in henry and a pure capacitance(C) in farad as shown in fig 14.
Wkt, Y = G + j(BC – BL)
Under resonance, (BC – BL) = 0 ; BC = BL
1 1 1
BC = = ωo C ; BL = =
XC XL 𝜔𝑜 𝐿
1 1 𝟏 𝟏
ωo C = = ; ωo 2 = ; ωo = rad/sec ; fo = Hz
𝜔𝑜 𝐿 LC √𝐋𝐂 𝟐𝛑√𝐋𝐂
𝟏
Therefore, the expression for resonant frequency is fo = Hz
𝟐𝛑√𝐋𝐂

20
❖ R-L & C parallel circuit

Fig 15: R-L & C parallel circuit

Consider the circuit in which one branch consists of resistor R in series with the inductor L.
So it is series R-L circuit with impedance ZRL. The other branch is pure capacitive with capacitor C.
Both the branches are connected in parallel across a variable frequency constant voltage source.
The current drawn by inductive branch is IL, while drawn by capacitive branch is IC.
V
IL = , where ZL = R + jXL
ZRL
V 1
IC = , where ZC =
ZC 2πfC

1 1 1 1
Y= + = +
ZRL ZC R+jXL −jXC

1 1 R−jXL j
Y= + = +
R+jXL −jXC R2 + XL 2 XC

𝑅 1 XL
Y= 2 2 +j(X − )
R + XL C R + XL 2
2

At resonance, imaginary part is equal to zero.

1 XL 1 XL
− = 0; =
XC R + XL 2
2 XC R + XL 2
2

1 L
R2 + XL 2 = XLXC (XLXC = ωoL* = )
ωo C C

L L
R2 + X L 2 = ; X L 2 = - R2
C C

L 2
XL = √ C − R

1 L 𝟏 𝐑𝟐
ωo = √ C − R2 ; 𝛚o = √ 𝐋𝐂 − rad/sec
L 𝐋𝟐

𝟏 𝟏 𝐑𝟐
f0 =
𝟐𝝅
√
𝐋𝐂
− 𝐋𝟐
Hz

21
❖ R-L and R-C parallel circuit:

Fig 16: R-L and R-C parallel circuit

Consider the circuit in which one branch consists of resistor R in series with the inductor L,
So it is series R-L circuit with impedance ZRL. The other branch is a resistor R in series with a pure
capacitive with capacitor C, So it is series R-C circuit with impedance ZRC. Both the branches are
connected in parallel across a variable frequency constant voltage source v.
1 1 1 1
Y= + = +
ZRL ZRC
R+jXL R−jXC
1 1RL −jXL RC +jXC
Y= + = 2 2 +
RL +jXL RC −jXC RL + XL RC 2 + XC 2
𝑅𝐿 𝑅𝐶 X X
Y= 2 2 + 2 2 +j( 2 C 2 − 2 L 2 )
R L + XL R C + XC R C + XC R L + XL
At resonance, imaginary part is equal to zero.
XC XL
2 2 − =0
RC + XC RL + XL 2
2

XC XL
2 2 =
R C + XC R L + XL 2
2

X C (R L 2 + X L 2 ) = X L (R C 2 + X C 2 )
XC R L 2 + XC XL 2 = XL R C 2 + XL XC 2
RL 2 L 2 L
+ ωo L = R C ωoL + XC
ωo C C C

Multiplying by ωo C, we get,
L L
R L2 + ωo2 LC = RC2 ωo2 LC +
C C
L L
ωo LC (RC - ) = RL -
2 2 2
C C
2 L
1 RL − C 𝐑𝐋 𝟐 − 𝐋⁄𝐂
𝟏
ωo =
2
; ωo = √ √ 𝟐 𝐋 rad/sec
LC RC 2 − L 𝐋𝐂 𝐑𝐂 − ⁄𝐂
C

𝟏 𝐑𝐋 𝟐 − 𝐋⁄𝐂
fo = √ Hz
𝟐𝝅√𝐋𝐂 𝐑𝐂 𝟐 − 𝐋⁄𝐂

22
NOTE:
Dynamic impedance at resonance (ZD): The impedance offered by the parallel circuit at resistance is
denoted as dynamic resistance (ZD). This is maximum at resistance.
𝑳
ZD = = dynamic resistance
𝑹𝑪
Quality factor of parallel circuit: Parallel circuit is used to magnify the current and hence known as
current resonance circuit.
Q factor for
𝐑
➢ RL parallel circuit: Q =
𝛚𝐨 𝐋
➢ RC parallel circuit: Q = ωo CR
c
➢ RLC parallel circuit: Q = R√
L

Q factor
Series Parallel
R-L ωo L X R R
= L =
R R ωo L XL
R-C 1 ωo CR =
R
XC
ωo CR
R-L-C 1 L c
QS = √ QP = R√
R C L

Resonant frequency:

Series RLC 1 1
fo = Hz ωo = rad/sec
2π√L C √L C

Parallel RLC 1 1
fo = Hz ωo = rad/sec
2π√LC √LC

R-L & C in parallel 1 R2

2 ω0 = √ − rad/sec
f0 =
1
√ 1 − R2 Hz LC L2
2𝜋 LC L

R-L & R-C in parallel 1 RL 2 − L⁄C 1 RL 2 − L⁄C

fo = √ Hz ω0 = √ rad/sec
2𝜋√LC RC 2 − L⁄C √LC RC 2 − L⁄C

NOTE:
Series RLC resonant circuit – acceptor circuit
Parallel RLC resonant circuit – rejector circuit
23
Concept of power factor improvement:

Significance of power factor: Apparent power drawn by the circuit has two components
(i) Active power or true power(P)
(ii) Reactive power(Q)
True power should be as large as possible because it does useful work in the circuit. This is possible
if the reactive power component is small.
The greater the power factor(PF) of the circuit, greater will be the abilty to utilize apparent power. ∴
PF of of the circuit should be nearly equal to 1.
P
P = √3VLIL cosΦ → IL =
√3VL cosΦ
n
From the above eq for fixed power and voltage, the load current is inversly proportional to the PF,
ie.,smaller the PF, higher is the load current and vice versa.

Power factor improvement: there are three main ways to improve power factor
1. Capacitor bank: Improving PF means reducing the phase difference between voltage and current.
Since the majority of loads are of inductive nature, they require some amount of reactive power for
them to function.
A capacitor or bank of capacitors installed parallel to the load provides this reactive power.
They act as a source of local reactive power, thus less reactive power flows through the line.
Capacitor banks reduce the phase difference between the voltage and current.

2. Synchronous condenser: Synchronous condensers are 3-Φ synchronous motor with no load
attached to its shaft.
The synchronous motor has the characteristics of operating under any power factor ie.,
leading or lagging or UPF depending upon the excitation. For inductive loads, a synchronous
condenser is connected towards load side and is overexcited.
Synchronous condensers make it behave like a capacitor. It draws the lagging current from
the supply or supplies the reactive power.

3. Phase advancers: This is an AC exciter used to improve the PF of an induction motor. They are
mounted on the shaft of the motor and are connected to the rotor circuit of the motor. It improves te
PF by providing excitating ampere turns to produce the required flux at the given slip frequency.
Further, if ampere turns increases, it can be made to operate at leading power factor.

Measurement of power in three phase (3-Φ) circuits:

Following are the methods used for measurement of active power in 3-Φ circuits
1. Three wattmeter method – both balanced and unbalanced loads.
2. Two wattmeter method – both balanced and unbalanced loads.
3. One wattmeter method – balanced loads only.

24
 Two wattmeter method:
Figure (a) shows the two wattmeter method for the measurement of power. This is the most
commonly used method for the measurement of power as this can be used for both balanced
and unbalanced loads and requires two watt meters.

Fig (a) : Two wattmeter method for measuring 3-Φ power

The voltage across the pressure coil of W1 is (E1 – E2) and the current I1 whereas voltage across the
pressure coil of W2, the voltage is (E3 – E2) and the current I3. Hence, instantaneous power recorded by the
two watt meters.
= (E1 – E2)I1 + (E3 – E2)I3
= E1I1 + E3I3 – E2 (I1 + I3)
= E1I1 + E3I3 + E2I2
As I1 + I3 = -I2
Hence the two wattmeters connected in this fashion i.e. current coil in phase 1 and 3 and potential
coils between phase 1 and 2 & 3 and 2, measures instantaneous 3-phase power.
However, if the load is balanced, the following analysis can be used to calculate the power factor of
the balanced load.
Fig (b) shows the phasor diagram for the connections of Fig(a) for balanced loads. Phasors have
been drawn taking rms values into considerations.

Fig (b) : Phasor diagram for fig (a)

25
 Power W = W1 + W2 = E12 I1 cos(30° + Φ) + E23 I3 cos(30° - Φ)
Since it is a balanced circuit,
|E12 | = |E23 | = √3 ∗ |Eph |

|I1 | = |I3 | = |I|

Therefore, W = W1 + W2 = √3 Eph I cos(30° + Φ) + √3 Eph I cos(30° - Φ)
= √3 Eph I [(cos(30° + Φ) + cos(30° - Φ)] = 2* √3 Eph I * cos30° * cosΦ
W = W1 + W2 = 3 Eph I cosΦ
Which is nothing but the three phase power in a balanced system.

It can be seen that if the p.f. of the load i s 0.5 ie., . Φ = 60° the reading of W1 becomes zero and if
the p.f. is less than 0.5, the p.f. angle is greater than 60° and with the usual connection of the wattmeter W1
will read negative and hence the connection of either the current coil or the pressure coil should be reversed
to obtain its reading and in such a situation this reading is to be reckoned negative and the total power
consumed by the load is the algebraic sum of the two wattmeter readings.
Now, to find out W1 – W2
= √3 Eph I [(cos(30° + Φ) - cos(30° - Φ)]
= √3 Eph I [-2sinΦ sin30°]
1
(or) W2 – W1 = √3 Eph I [ 2sinΦ sin30°] = √3 Eph I * 2 * * sinΦ = √3 Eph I sinΦ
2

W2 – W1 = √𝟑 Eph I sinΦ
W2 − W1 √3 Eph I sinΦ tanΦ
= =
W2 + W1 3 Eph I cosΦ √3
𝐖 −𝐖
tanΦ = √𝟑 ∗ 𝐖𝟐+ 𝐖𝟏
𝟐 𝟏
Hence, Power factor cosΦ of the circuit can be calculated.
For reactive power , Q = 3 Eph I sinΦ
Advantages of 3Φ system:
In the three phase system, alternator armature has three windings and it produces three independent
alternating voltages. The magnitude and frequency of all of them is equal but they have a phase differnce of
120° between each other. Such a three phase system has following advantages over single phase system:
1. The output of three phase machine is always greater than single phase machine of same size,
approximately 1.5 times. So for a given size and voltage a three phase alternator occupies less space and
has less cost too than single phase having same rating.
2. For transmission and distribution, three phase system needs less copper or less conducting material than
single phase system for given volt amperes and voltage rating, so transmission becomes very much
economical.
3. It is possible to produce rotating magnetic field with stationary coils by using three phase system. Hence
three phase motors are self starting.

26
4. In a single phase system, the instantaneous power is a function of time and hence fluctuates w.r.t time.
This fluctuating power causesconsiderable vibrations in single phase motor. Hence the performance of
single phase motor is poor. While instantaneous power in symmetrical three phase system is constant.
5. Three phase systems give steady output.
6. Single phase supply can be obtained from three phase but three phase canont be obtained from single
phase.
7. Power factor of a single phase motors is poor than three phase motors of same ratings.
8. For converting machines like rectifiers, the DC output voltage becomes smoother if number of phases
are increased.
But it is found that optimum number of phases required to get all the above said advantages is three.
Any further increase in number of phases cause a lot of complications. Hence three phase system is
accepted as standard polyphase system throught the world.

NOTE:
❖ W1 = √3 Eph I (cos(30° + Φ) = EL IL (cos(30° + Φ)
W2 = √3 Eph I (cos(30° - Φ) = EL IL (cos(30° - Φ)
1. W1 = W2 , cosΦ = 1
2. W1 - +ve, W2 - +ve (W1 ≠ W2) , 0.5 < cosΦ < 1
3. W1 - -ve, W2 - +ve (W1 ≠ W2) , 0 < cosΦ < 0.5
4. W1 = 0 , cosΦ = 0.5
5. W2 = -W1 , cosΦ = 0

❖ For star connection, EL = √3 Eph , IL = Iph

❖ For delta connection, EL = Eph , IL = √3 Iph

27
 Problems on RL, RC and RLC circuits:

1. A 50Hz, alternating voltage of 150V (rms) is applied independently to (i) Resistance of 10Ω
(ii) Inducatance of 0.2H (iii) Capacitance of 50μF.Find the expression for instantaneous current in each
case. Draw the phasor diagram in each case.
Solution: (i) Given : R = 10 Ω
V = Vmsinωt
Vm = √2 Vrms = √2 * 150 = 212.13V
Vm 212.13
Im = = = 21.213A
R 10
In a pure resistive circuit, current is in phase with voltage,ie., Φ = 0°
i = Imsinωt = Imsin(2πft)
i = 21.213sin(100πt) A

(ii) Given ; L = 0.2H

Inductive reactance XL = 2πfL = 2π*50*0.2 = 62.83 Ω
Vm 212.13
Im = = = 3.37A
XL 62.83
π
In a pure inductive circuit, current lags voltage by an angle 90° ie., Φ = 90° =
2
i = Imsin(ωt – Φ)
𝛑
i = 3.37sin(2πft - ) A
𝟐

(iii) Given C = 50𝝁F

1 1
Capacitive reactance XC = = = 63.66 Ω
2πfC 2π∗50∗50∗10−6
Vm 212.13
Im = = = 3.33A
XC 63.66
π
In a pure capacitive circuit, current leads voltage by an angle 90° ie., Φ = 90° =
2
𝛑
i = Imsin(ωt + Φ) i = 3.33sin(2πft + ) A
𝟐
Phasor diagrams

(i) Pure resistive circuit (ii) Pure inductive circuit (iii) Pure capacitive circuit

28
 2. An alternating current, i = 1.414sin(2π*50*t)A is passed through a series circuit consisting of a
resistance of 100Ω and inductance of 0.31831H. Find the expression for instantaneous values of volatge
across (i) resistance (ii)inducatabce (iii) combination
n
Sol :

Given : i = 1.414sin(2π*50*t)A , f = 50Hz , ω = 2πf = 2*π*50 , R = 100Ω , L = 0.31831H ,

XL = 2πfL = 2*π*50*0.31831 = 100Ω
(i) The voltage across resistance is,
VR = iR = 1.414sin(2π*50*t)*100
VR = 141.4sin(2π*50*t) V
(ii) The voltage across inductor leads current by an angle 90°,
VL = iXL = 1.414sin(2π*50t + Φ) * 100
VL = 141.4sin(2π*50t + 90°)V
(iii) From the expression of VR, we can write,
VR 141.4
VR(rms) = = = 100V , Φ = 0°
√2 √2
VR(rms) = 100∟0° = 100 + j0 V
VL 141.4
VL(rms) = = = 100V , Φ = 90°
√2 √2
VL(rms) = 100∟90° = 0 + j100 V
V = 100 + j0 + 0 + j100
= 100 + j100 = 141.42∟45°
Vm = √2 * 141.42 = 200V
Hence the expression for instantaneous value of resultant voltage is,
V = 200 sin(2π*50t + 45°)V

3. A resistance of 120Ω and a capacitive reactance of 250 Ω are connected in series across a AC voltage
source. If the current of 0.9A is flowing in a circuit, then find (i) Power factor (ii) Supply voltage
(iii) Voltage across resistance and capacitance (iv) Active and reactive power
n
Sol :

Given : R = 120 Ω , XC = 250 Ω , I = 0.9∟0°A

Z = R – jXC = 120 – j250 Ω = 277.308∟-64.358°
(i) Power factor = cosΦ = cos(-64.358°) ; PF = 0.4327 leading
(ii) Supply voltage, V = I*Z = (0.9∟0°)*(277.308∟-64.358°)

29
 V = 249.5772∟-64.358° V
(iii) VR = IR = 0.9*120 = 108V
VC = IXC = 0.9*250 = 225V
(iv) P = VIcosΦ = 249.5772*0.9*0.4327
P = 97.1928W
Q = VIsinΦ = 249.5772*0.9*sin(-64.358°)
Q = -202.49VAR

4. A series circuit having pure resistance of 40Ω, pure inductance of 50.07mH and a capacitor is connected
across a 400V, 50Hz, AC supply. This R, L , C combination draws acurrent of 10A. Calculate (i) Power
factor of the circuit (ii) Capacitor value
Soln:

Given: R = 40 Ω , L = 50.07mH , I = 10A

XL = 2πfL = 2π ∗ 50*50.07*10-3 = 15.73 Ω
Z = R + j(XL – XC)
|𝑍| = √(R)2 + (XL − XC )2 = √(40)2 + (15.73 − X C )2
|V| 400
|𝑍| =
|I|
= = 40 Ω
10
40 = √(40)2 + (15.73 − XC )2 = 1600 = 402 + (15.73 – XC)2
(15.73 – XC)2 = 0 ; XC = 15.73 Ω
1 1 1
XC = ;C= = = 2.02358*10-4
2πfC 2πfXC 2π∗50∗15.73
C = 202.358µF

Z = 40 + j(15.73 – 15.73) = 40 + j0 Ω = 40∟0°

PF = cosΦ = cos(0°); PF = 1 (unity power factor)
5. A voltage v = 200sin100πt is applied to a load having R = 200Ω in series with L = 638mH. Estimate
(i) Expression for current in i = Imsin(ωt± Φ) form (Ans: i = 0.7071sin(100πt – 45.06°)
(ii) Power consumed by the load (Ans: P = 50W)
(iii) Reactive power of the load (Ans: Q = 50VAR)
(iv) Voltage across R and L (Ans: VR = 100V ; VL = 100.21V)

6. The impedance of a circuit placed across a 120V, 50 Hz source is (10 + j20). Find current and the power.
(Ans: i = 2.4 – j4.8A , P = 288W)

7. Calculate the resistance, inductance or capacitance in series for each of the following impedances.
Assume the frequency to be 60Hz. (i) ( 12 + j30)Ω (ii) ( -j60) Ω (iii) 20∟60°
(Ans: (i) R = 10 Ω , L = 79.58mH; (ii) R = 0 Ω, C = 44.209µF; (iii) R = 10 Ω, L = 45.94mH)

30
 π
8. The waveform of volatge and current of a circuit are given by e = 120sin(314t)V and i = 10sin(314t + ).
6
Calculate the values of resistance, capacitance which are connected in series to form the circuit and also
the power consumed by the circuit. (Ans: R = 10.393Ω, C = 530.45µF, P = 519.52W)

9. A 15V source is applied to a capacitive circuit that has an impedance of (10 - j25)Ω . Determine the value
of current and active and reactive power in the circuit. (Ans: i = 0.55∟68.2°, P = 3W , Q = -7.7VAR)

10. Two impedances Z1 and Z2 are connected in parallel across a applied voltage of (100+j200)volts. The
total power supplied to the circuit is 5KW. The first branch takes aleading current of 16A and has a
resistance of 5Ω while the second branch takes a lagging current at 0.8 power factor. Calculate (i) current
in second branch (ii) Total current (iii) Circuit constants
(Ans: I2 = 20.79∟26.55° , I = 22.48∟69.70° , R1 = 5 Ω, R2 = 8.60 Ω, X1 = 13.04 Ω, X2 = 6.4531 Ω)

11. A parallel circuit having resistance of 25 Ω, inducatance of 64mH and a capacitance of 80µF are
connected across 110V, 50Hz single phase AC supply. Calculate the current in individual elements, the
total current from the supply and the overall power factor of the circuit. Draw the phasor diagram showing
̅, I̅R , I̅L , I̅C and I.̅
V
(Ans: IR = 4.4∟0°, IL = 5.47∟-90°, IC = 2.76∟90°, I = 5.1676∟-31.62°, PF = 0.851 lagging)

12. The impedance of a parallel RC circuit is (10 - j30)Ω at 1 MHz. Determine the values of the components.
(Ans: R = 100 Ω, C = 4.7*10-9F)

13. Determine the line current and the total impedance and admittance of the circuit having R and L in series
and that in parallel with C where R = 25 Ω , XL = 50 Ω , XC = 40 Ω across volatge source of 100V.
(Ans: Z = 83.12∟-48.4° , Y = 1.2*10-2∟48.4°, I = 1.2∟48.4° A)

14. A 4700 Ω resistor and a 2μF capacitor are connected in parallel across a 240 V 60 Hz source. Determine
the circuit impedance and the line current. (Ans: Z = 1276∟-74.3° Ω, I = 0.188∟74.3°)
15. A room heater of 2KW, 125V rating is to be operated at 230V , 50Hz AC supply. Calculate the value of
inductance, that must be connected in series with the heater, so that the heater will not get damaged due too
over heating. (Ans: L = 0.0384H)

16. A choke coil and a pure resistance are connected in series across 230V, 50Hz AC supply. If the voltage
drop across the coil is 190V and across resistance is 80V while the current drawn by the circuit is 5A.
Calculate (i) internal resistance of coil (ii) inductance of coil (iii) Resistance R (iv) Power factor of circuit
(v) Power consumed by the circuit (Ans: r = 13Ω, L = 0.1136H, R = 16Ω, PF = 0.6304lag, P = 724.96W)

17. Two coils A and B are connected in series across 200V, 50Hz AC supply. The power input to the circuit is
2.2KW and 1.5KVAR. If the resistance of coil A is 4Ω and the reactance id 8Ω. Calculate the resiatance
and the reactance of the coil B. Also calculate the active power consumed by coil A and coil B, total
impedance of the circuit. (Ans: RB = 8.4 Ω, XB = 0.5Ω, PA = 707.56W, PB = 1.4858KW, ZT = 15.037Ω)

31
18. A sereis R-C is connected across 200V, 50Hz AC supply draws a current of 20A. When the frequency of
the supply is increased to 100Hz, the current increases to 23.40821a. Calculate the value of resistance and
capaciatnce of the circuit. (Ans: R = 8Ω, L = 0.5305mH)

19. Two Impedance Z1 and Z2 havind same numerical value are connected in series. If Z1 is having power
factor of 0.866 lagging and Z2 is having power factor of 0.6 leading. Calculate the PF of the series
combination. (PF = cosΦ = 0.9796 leading)

20. A circuit is shown in the figure Draw its equivalent admittance circuit. Also calculate admittance,
conductance and susceptance of the circuit.

(Ans: G = 0.1mho, B = 0.05mho, Y = 0.1118mho)

21. A heater operates at 100V, 50Hz AC supply and takes a current of 8A and consumes 1.2KW power. A
choke coil is having ratio of reactance to resistance as 10, is connected in series with the heater. The series
combination is connected across 230V, 50Hz AC supply. Calculate (i) Resistance of choke coil (ii)
Reactanc eof choke coil (iii) Power consumed by the choke coil (iv) Total power consumed.
(Ans: r = 2.576Ω, XL = 25.576 Ω, Pcoil = 164.87W, Ptotal = 964.864W)

22. Two impedances (8 + j6)Ω and (3 – j4)Ω are connected in parallel. If the total current drawn by the
combination is 25A, find the current and power taken by each impedance.
(Ans: I1 = 11.1803∟-63.4349° A, I2 = 22.3607∟26.5642° A, P1 = 1000W, P2 = 1500W)

23. Two impedances Z1 = (12 + j15)Ω and Z2 = (8 – j4)Ω are connected in parallel across a potential
difference of (230 + j0)V. Calculate (i) total current drawn (ii) total power and branch power consumed
(iii) overall power factor of the circuit.
(Ans: IT= 30.55∟4.03°A, PT = 7.0091KW, P1 = 7.719KW, P2 = 5.289KW, cosΦ = 0.9975leading)

24. A parallel circuit consists of two branches. Branch (i) consists of R of 100Ω connected in series with
inducatnce of 1H and branch (ii) consists of R of 50Ω in series with capacitance of 79.5µF. This parallel
circuit is connected across single phase 200V, 50Hz AC supply. Calculate (i) Branch currents (ii) total
current of the circuit (iii) total power factor of the circuit (iv) total power drawn by the circuit.(Ans: I1=
0.6066∟-72.343°A,I2 = 3.1223∟38.687°A, IT = 2.9592∟27.6563°A, PF = 08857leading ,P = 524W)

25. Two impedances Z1 = 40∟30°Ω and Z2 = 30∟60°Ω are connected in series across a 230V, 50Hz AC
supply. Calculate (i) current (ii) Power factor (iii) Power consumed by the circuit
(Ans: I = 3.399∟-42.807°A, cosϕ = .7336 lagging, P = 573.543watts)

32
 Problems on time constant:
1) Obtain the value of time contant for the given circuit.
a.

𝟑
(Ans: τ = 𝟒 𝐬𝐞𝐜)
b.

𝟒
(Ans: τ = 2sec) (Ans: τ = sec)
𝟏𝟏
Problems on Resonance:
1) A series RLC circuit is connected to 230V AC supply. The current drawn by the circuit ar resonance is
25A. The voltage drop across the capacitor is 4000V. At series resonace, calculate the value of resistance,
inductance if the value of capacitance is 5µF, also calculate the resonant frequency.
(Ans: R = 9.2Ω, L = 0.128H, fo = 198.943Hz)

2) An inductive coil of resistance 10Ω and an inductance of 0.1H is connected in parallel with 150µF
capacitor to a variable frequency, 200V supply. Find the resonant frequency at which the total current
taken from the supply is in phase with the supply voltage. Also find the value of current in the circuit and
draw the phasor diagram for the same. (fo = 37.8865Hz, I = 3A)

3) For the Fig(3) shown determine the maximum current, the frequency at which it occurs and the resulting
voltage across the inductance and capacitance.

Fig(3)
(Ans:I = 2A, fo= 251.64KHz, VL = VC = 12.63V)

4) In a series RLC circuit shown in fig(4), determine (i) The necessary value of capacitor, (ii) The supply
1
voltage to produce a voltage of 5V across the capacitor if fo = 5KHz (iii) If the capacitance is made 2 at (i),
determine fo , the Q value of new circuit.

33
 (Ans: C = 0.0507 µF, V = 0.05V, fo = 7071Hz, Q = 141.49 )

5) For the circuit shown, determine the value of inductance for resonance if Q = 50 and fo = 175KHz. Also
find the circuit current, the voltage across the capacitor and the bandwidth of the circuit.

(Ans: L=2.58mH, I = 14.96mA, VC = 42.5V, BW = 3.5KHz)

6) For a series RLC circuit having R = 10Ω, L = 100mH, C = 0.01µF, determine the impedance magnitude at
resonance, 1KHz below resonance and at 1KHz above resosnace.
(Ans: Z1 = 1.42KΩ, Z2 = 1.15KΩ)

7) For the circuit shown in Fig R1 = 0.5 Ω, R2 = 0.5Ω, R3 = 1.5 Ω, C1 = 6 μF and C2 = 12 μF, L1 = 25 mH and
L2 = 15mH. Determine (i) the frequency of resonance (ii) Q of the circuit (iii) Q of coil 1 and coil 2
individually.

(Ans: ωo = 2.5*103 rad/sec, Q = 40, Q1 = 62.5, Q2 = 25)

8) A coil having a 5Ω resistor is connected in series with a 50 μF capacitor. The circuit resonates at 100 Hz
(a) Determine the inductance of the coil. (b) If the circuit is connected across a 200 V 100 Hz a.c. source,
determine the power delivered to the coil (c) the voltage across the capacitor and the coil (d) the bandwidth
of the circuit.
(Ans: L = 50.66mH, P = 8KW, VC = 1273.24V,VL = 1256V, Q = 6.3, BW = 16Hz)

9) A coil having a resistance of 50Ω and inductances 10mH is connected in series with a capacitor and is
supplied at constant voltage and variable frequency source. The maximum current is 1A at 750Hz.
Determine the bandwidth and half power frequencies. (Ans: BW = 795.8Hz, f2 = 1246.9Hz, f1 = 451Hz)

10) Two branches of a parallel circuit have element RL = 6Ω, L = 1mH and RC = 4 Ω and C = 20 μF.
Determine the frequency at resonance. (Ans: ωo = 4537 rad/sec)

34
 Problems on two wattmter method:

1. A 500 volt 3-phase motor has an output of 37.3 KW and operates at a p.f. of 0.85 with an efficiency of
90%. Calculate the reading on each of the two wattmeters connected to measure the input.
(Ans: W1 = 13.31KW , W2 = 28.13KW)
2. Two wattmetrs reading 4KW and 2KW are connected for measuring the active power of 3−Φ circuit.
Calculate the power factor of the circuit. (Ans: cosΦ = 0.866lag)

3. EL = 400V , IL = 10A, cosΦ = 0.5lag. Calculate the two wattmter readings. (Ans: W1 =0 , W2 = 3.464KW)

4. A balanced star connected lod of (8+j6)Ω is connected to a 3 phase 230V supply. Find the line current,
power factor, reactive voltamperes and total voltamperes.
(Ans: IL = 13.279∟-36.86°, P = 4.231KW, Q = 3.173KVAR, S = 5289.95)

5. Three 100Ω resistors are connected (i) star (ii) delta across a 415V 50Hz 3-phase supply.Calculate the line
current and power consumed in each case.
(Ans: (i) star: IL = 2.396A, P = 1.72KW; (ii) Delta: IL = 4.15A, P = 2.98KW)

6. The current of average value 18.019A is flowing in a circuit in which a voltage of peak value of 141.42V is
π
applied, V lags I by radians. Determine (i) Z = R + jX (ii) Power (Ans: Z = 7.85∟30° Ω, P = 2.206KW)
6
7. Three volatges are connected as shown in the figure below, If Va = 17.32 + j10 V, Vb = 30∟80° V,
Vc = 15∟-100° V. Find (i) V12 , (ii) V23 , (iii) V34

(Ans: V12 = 7.804+j44.312V, V23 = 9.52-j34.31V, V34 = -12.12+j19.54V)

8. How is current of 10A shared by three impedance Z1 = (2-j5)Ω, Z2 = 6.708∟26.56° Ω , Z3 = (3+j4)Ω all
connected in parallel? (Ans: I1 = 5.8∟77.76°A, I2 = 4.65∟-17° A, I3 = 6.25∟-43.57° A )

9.A 3-phase star connected supply with phase voltage of 230V is supplying a balanced delta (∆) load. The
load draws a power of 15KW at 0.8 PF lagging. Find the line current and the current in each phase of the
load. What is the load impedance per phase? (Ans: IL = 27.17A, Iph = 15.68A, Zph = 14.66∟0.8° Ω)

10. Power is measured in a 3 phase balanced load using two wattmetrs. The line voltage is 400V. The load
and its power factor is so adjusted that the line current is always 10A. Find the readings of the wattmeter
when the PF is (i) unity (Ans: W1 = 3.46KW, W2 = 3.46KW
(ii) 0.866 W1 = 2 KW, W2 = 4KW
(iii) 0.5 W1 = 0 watts, W2 = 3.464KW
(iv) Zero W1 = -2000 watts, W2 = 2000 watts)

35