Resonance - JNNCE ECE Manjunathcalled resonance. The frequency at which resonance takes place is...

Chapter 1

Resonance

1.0.1 Introduction:

Most of the transmission lines, electrical circuits and communication networks are made up of network elementslike resistor R,inductor, L and Capacitor C. The impedance of the inductor and capacitor depends on thefrequency of the applied sinusoidal voltage to the network. As we vary the frequency of the supply thenetwork impedance is purely resistive in which the impedance of the inductor is equal to the impedance ofthe capacitor. The phenomenon in which the the complex circuit behaves like a pure resistive iscalled resonance. The frequency at which resonance takes place is called the frequency of resonance ωr(radians/sec) or fr. (Hz)

Resonance may occur by varying frequency of the applied voltage to the complex network or by varyinginductance, L or Capacitance, C, by keeping the frequency constant. Under resonance the following conditionsoccur in the circuit.

• The impedance of the inductance is equal to the impedance Capacitance.

• The phase of the current in the circuit is in phase with the applied voltage.

• Maximum current will flow in the circuit.

• The voltage across the capacitor or inductor is I ×XC or I ×XL where I is the current at resonance andXC or XL is the impedance of the circuit.

• The total power is dissipated in the resistor and the absorbed average power is maximum.

———————————————————————————————————————-

1.1 Series Resonance

Consider a series circuit consists of resistor, inductorand capacitor as shown in Figure 1.1.

LC

R( )iv t ( )i t ( )ov t+

- -

+

Figure 1.1: Series resonance circuit

The impedance of the circuit is

Z = R+ j(XL −XC)

At resonance the imaginary part is zero

XL −XC = 0

ωrL−1

ωrC= 0

ωrL =1

ωrC

ω2r =

1

LC

ωr =1√LC

radians/sec

fr =1

2π√LC

Hz

1

1.1. SERIES RESONANCE Chapter 1. Resonance

The plot of the frequency response of series circuitis as shown in Figure 1.2. At resonant frequency ωrthe current is maximum.

I

mImI2

ω

1 r 2ω ω ω

Figure 1.2: Frequency response of series circuit

Z

ω

LX

rω

Impe

danc

e

R

C-X

Z

0

Figure 1.3: Frequency response of impedance of seriescircuit

At resonance frequency fr Z = R and current is ImAt half power frequencies f1 and f2 the current is Im√

2

Z =√

2R

Z = R+ jXL − JXC =√R2 + (XL −XC)2√

R2 + (XL −XC)2 =√

2R

R2 + (XL −XC)2 = 2R2

(XL −XC)2 = R2

XL −XC = R

At frequency ω1 the circuit impedance XC > XL

XC −XL = R1

ω1C− ω1L = R

1− ω21LC

ω1C= R

Rω1C − 1 + ω21LC = 0

ω21 +

R

Lω1 −

1

LC= 0

a = 1, b =R

L, c = − 1

LC

ω1 =−RL ±

√(RL

)2+ 4

LC

2= − R

2L±

√(R

2L

)2

+1

LC

Frequency is always positive

ω1 = − R

2L+

√(R

2L

)2

+1

LC

In terms of frequency f1

f1 =1

2π

− R

2L+

√(R

2L

)2

+1

LC

At frequency ω2 the circuit impedance XL > XC

XL −XC = R

ω2L−1

ω2C− = R

ω22LC − 1

ω2C= R

ω22LC −Rω2C − 1 = 0

ω22 −

R

Lω2 −

1

LC= 0

a = 1, b = −RL, c = − 1

LC

ω2 =

RL ±

√(RL

)2+ 4

LC

2=

R

2L±

√(R

2L

)2

+1

LC

Frequency is always positive

ω2 =R

2L+

√(R

2L

)2

+1

LC

In terms of frequency f2

f2 =1

2π

R2L

+

√(R

2L

)2

+1

LC

Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 2


Relation between ωr , ω1 and ω2

ω1 × ω2 =

=

R

2L+

√(R

2L

)2

+1

LC

×−R

2L+

√(R

2L

)2

+1

LC

=

1

LC

ωr =

√1

LC

ω2r =

1

LC= ω1.ω2

ωr =√ω1.ω2

fr =√f1f2

Relation between Bandwidth Quality factor

Bandwidth is B = ω2 − ω1

B = ω2 − ω1

=

R

2L+

√(R

2L

)2

+1

LC

−−R

2L+

√(R

2L

)2

+1

LC

=

R

Lradians

B =1

2π.R

L=

R

2πLHz

ωr =

√1

LC

ω1 = −B2

+

√(B

2

)2

+ ω2r

ω2 =B

2+

√(B

2

)2

+ ω2r

Quality factor Q

The quality factor Q is defined as the ratio of resonantfrequency to the bandwidth

Q =ωrB

=

√1LC

RL

=1

R

√L

C

I

mImI2

ω

1Q

2Q

1 2Q Q>

1 r 2ω ω ω

Figure 1.4: Plot of frequency verses Q

Resonance by varying circuit inductance

Consider a series RLC circuit as shown in Figure 1.5 isbecome resonant by varying inductance of the circuit.

LC

R( )iv t ( )i t+

-

Figure 1.5: Resonance by varying inductance

Let L1 is the inductance at ω

XC −XL = R1

ωC− ωL1 = R

L1 =1

ω2C− R

ω

Let L2 is the inductance at ω

XL −XC = R

ωL2 −1

ωC= R

L2 =1

ω2C+R

ω

Resonance by varying circuit capacitance

Consider a series RLC circuit as shown in Figure ?? isbecome resonant by varying capacitance of the circuit.

LC

R( )iv t ( )i t+

-

Figure 1.6: Resonance by varying capacitance

Let C1 is the capacitance at ω1

XC −XL = R ⇒ 1

ω1C1− ω1L = R

1

ω1C1= R+ ω1L

C1 =1

ω21L+ ω1R

Let C2 is the capacitance at ω2

XL −XC = R ⇒ ω2L−1

ω2C2= R

1

ω2C2= ω2L−R

C2 =1

ω22L− ω2R



Table 1.1: Important Formulae

Parameter Formula

At resonance Z = R, XL = XC Current Ir =ER

Resonance ωr =1√LC

fr =1

2π√LC

Half power frequency ω1 =−R2L +

√(R2L

)2+ 1

LC f1 =ω1

2π

ω2 =R2L +

√(R2L

)2+ 1

LC f2 =ω2

2π

Half power frequency ω1 =−B2 +

√(B2

)2+ ω2

r f1 =ω1

2π

ω2 =B2 +

√(B2

)2+ ω2

r f2 =ω2

2π

Bandwidth B = ω2 − ω1 =RL Radians

B = f2 − f1 =R

2πL Hz

Quality factor Q = ωrLR = 1

R

√LC

ωr ω1 ω2 ωr =√ω1ω2 OR fr =

√f1f2

Voltage acrosscapacitor/inductor

VLr = VCr = IXLr

Value of inductor at f1, f2L1 =

1ω2C −

Rω , L2 =

1ω2C + R

ω

Value of capacitor at f1, f2C1 =

1ω2L −

Rω , C2 =

1ω2L + R

ω

Selectivity: is property of circuit in which the circuit is allowed to select a band of frequenciesbetween f1 and f2.



1: For the circuit shown in Figure Find (a)The resonant and half power frequencies (b)Calculate the quality factor and bandwidth(c) Determine the amplitude of the current atωo, ω1, ω2

( )10sin

iv ttω

= ( )i t+

-

3.09C Fμ= 100L mH=

3R = Ω

Figure 1.7: Example 1

LC = 100× 10−3 × 3.09× 10−6 = 3.09× 10−7

The resonant frequency ωo is

ω0 =1√LC

=1√

3.09× 10−7= 1800 rad/s

The half power frequency ω1, ω2 is

R

2L=

3

2× 100× 10−3= 15

ω1 = − R

2L+

√(R

2L

)2

+1

LC

= −15 +

√225 +

1

3.09× 10−7

= −15 +√

225 + 3.236× 106

= −15 + 1798.96 = 1784 rad/s

ω2 =R

2L+

√(R

2L

)2

+1

LC

= 15 + 1798.96 = 1814 rad/s

Frequency in Hz is

f1 =ω1

2π=

1784

2π= 284 Hz

f2 =ω2

2π=

1814

2π= 289 Hz

Bandwidth B is

B = ω2 − ω1 = 1814− 1784 = 30 rad/s

Also B is

B =R

L=

3

100× 10−3= 30 rad/s

Quality factor Q is

Q =ωoB

=1800

30= 60

The amplitude of the current at ωo is

I =V

R=

10

3= 3.33A

The amplitude of the current at ω1, ω2 is

I =V√2R

=10√2× 3

= 2.36A

2: For the circuit shown in Figure findthe resonant frequency, quality factor andbandwidth for the circuit. Determine thechange in Q and the bandwidth if R is changedfrom R = 2 Ω to R = 0.4 Ω

( )iv t ( )i t+

-

5C Fμ= 2L mH=

2R = Ω

Figure 1.8: Example

LC = 2× 10−3 × 5× 10−6 = 10× 10−9


ω0 =1√LC

=1√

10× 10−9= 10× 103 rad/s

B is

B =R

L=

2

2× 10−3= 1000 rad/s

Quality factor Q is

Q =ωoB

=10× 103

1000= 10

When R = 0.4Ω B is

B =R

L=

0.4

2× 10−3= 200 rad/s

Quality factor Q is

Q =ωoB

=10× 103

200= 50

3: For the circuit shown in Figure find thefollowing (a) The resonant frequency fo (b)Quality factor Q (c) fc1, fc2(d) Bandwidth B

( )iv t ( )i t+

-

1.25pF 312mH

50kΩ

12.5kΩ

( )ov t



Figure 1.9: Example

LC = 312× 10−3 × 1.25× 10−12 = 0.39× 10−12


ω0 =1√LC

=1√

0.39× 10−12= 1.6× 106 rad/s

fo =ω0

2π=

1.6× 106

2π= 254× 103 Hz

B is

B =R

L=

62.5× 103

312× 10−3= 200× 103

Quality factor Q is

Q =ωoB

=1.6× 106

200× 103= 8

The half power frequency ω1, ω1 is

R

2L=

62.5× 103

2× 312× 10−3= 1× 105

ω1 = − R

2L+

√(R

2L

)2

+1

LC

ω1 = −1× 105 +

√(1× 105)2 +

1

0.39× 10−12

= −1× 105 +√

1× 1010 + 2.5641× 1012

= −1× 105 + 1.6× 106 = 1.5× 106 rad/s

ω2 =R

2L+

√(R

2L)2 +

1

LC

ω2 = 1× 105 + 1.6× 106 = 1.7× 106 rad/s

Frequency in Hz is

fc1 =ω1

2π=

1.5× 106

2π= 238.73× 103 Hz

fc2 =ω2

2π=

260× 103

2π= 271.56× 103 Hz

Bandwidth B is

B = ω2−ω1 = 1.7× 106− 1.5× 106 = 200× 103 rad/s

Bandwidth B in Hz is

B =200× 103

2× π= 31.83× 103 Hz

4: A variable frequency voltage source drivesthe network shown in Figure. Find the resonantfrequency fo, Quality factor Q, Bandwidth

B and the average power dissipated by thenetwork at resonance

12 cos ωt ( )i t+

-4Ω

50μF 50mH

Figure 1.10: Example

LC = 50× 10−3 × 5× 10−6 = 2.5× 10−6


ω0 =1√LC

=1√

2.5× 10−6= 632.45 rad/s

fo =ω0

2π=

632.45

2π= 100 Hz

B is

B =R

L=

4

50× 10−3= 80

Quality factor Q is

Q =ωoB

=632.45

80= 7.9

The average power dissipated at resonance

P =1

2× V 2

R

P =1

2× 122

4= 18 Watts

5: For the circuit shown in Figure themaximum amplitude of current is 10A, circuitquality factor Q is 100 and L=0.1H. If theapplied voltage is 100 V find the value ofcapacitance

( )v t ( )i t+

-RΩ

C 0.1H


Solution:The maximum value of current flows in a circuit

when the circuit is in resonance and the impedance ofthe circuit is pure resistor and its value is

R =V

Im=

100

10= 10 Ω

The relation between Q and Bandwidth B is

Q =1

R

√L

C⇒

√L

C= Q×R



L

C= (Q×R)2 = (100× 10)2 = 1× 106

C =L

1× 106=

0.1

1× 106= 0.1µF

6: The Q of a series circuit network is 10. Themaximum amplitude of current at resonance is1 A when applied voltage is 10 V. If L=0.1Hfind the value of capacitance

Solution:

The maximum value of current flows in a circuitwhen the circuit is in resonance and the impedance ofthe circuit is pure resistor and its value is

R =V

Im=

10

1= 10 Ω

The relation between Q and Bandwidth B is

Q =1

R

√L

C⇒

√L

C= Q×R

L

C= (Q×R)2 = (10× 10)2 = 10× 103

C =L

10× 103=

0.1

10× 103= 10µF

7: A coil of inductance 9H and resistance 50 Ω in serieswith a capacitor is supplied at constant voltage froma variable frequency source. If the maximum currentis 1A at 75 Hz, find the frequency when the current is0.5A

Solution:

8: Design a series resonant circuit to have ωr =2500 rad/sec Z(ωr) = 100 Ω and B = 500 rad/secSolution:

B = ωr − ω1 =R

L

L =R

B=

100

500= 0.2H

The resonant frequency ωr is

ωr =1√LC

√LC =

1

ωr=

1

2500= 4× 10−4

LC = 1.6× 10−7

C =1.6× 10−7

0.2× 10−3= 0.8× 10−6 = 0.8µF

R = 100Ω L = 0.2H C = 0.8µF

9: A series resonant RLC circuit has resonantfrequency of 80 K rad/sec and a quality factorof 8. Find the bandwidth, the upper cutofffrequency and lower cutoff frequency

Solution:

Q =ωrB

B =ωrQ

=80× 103

8= 10× 103 rad/sec

Bandwidth in Hertz

B =B

2π=

10× 103

2× π= 1.59× 103 Hz

B = ω2 − ω1 =R

L

ω1 = −B2

+

√(B

2

)2

+ ω2r

=−10× 103

2+

√(10× 103

2

)2

+ (80× 103)2

= −5× 103 +√

25× 106 + 6.4× 109

= −5× 103 + 80.156× 103 = 75156 rad/sec

ω2 =B

2+

√(B

2

)2

+ ω2r

= 5× 103 + 80.156× 103 = 85156 rad/sec

Frequency in Hertz

f1 =ω1

2π=

75156

2× π= 11.96× 103 Hz

f2 =ω1

2π=

85156

2× π= 13.55× 103 Hz

10: A series RLC circuit has R=10 Ω,L=0.1H, and C=100 µFand is connected across200 V, variable frequency source. Find (a)the resonant frequency (b)impedance at thisfrequency (c) the voltage across drop across



inductance and capacitance at this frequency(d) quality factor and (e) bandwidth.

Solution: (a) Resonant frequency

fr =1

2π√LC

=1

2π√

0.1× 100× 10−6= 50.33Hz

(b)Impedance at resonance Z = R = 10Ω

I =V

R=

200

10= 20A

(c) Voltage across inductance/capacitance

XL = XC = 2πfr × L = 2π × 50.33× 0.1 = 31.62Ω

VL = IXL = 20× 31.62 = 632.46V

(d) Quality factor

Q =XL

R=

31.62

0.1= 3.162

(e) Bandwidth

B =R

2πL=

10

2× π × 0.1= 15.9 Hz

11: A coil of resistance R=20 Ω and inductanceL=0.2 H is connected in series with acapacitance across 230 V supply Find (a) thevalue of the capacitance for which resonanceoccurs at 100 Hz frequency (b) the currentthrough and voltage across the capacitor (c) Qfactor of the coil


fr =1

2π√LC

√LC =

1

2πfr=

1

2π × 100= 1.6× 10−3

LC = 2.53× 10−6

C =2.53× 10−6

0.2= 12.66µF

Impedance at resonance is Z = R = 10Ω(b) Current through and voltage across the capacitor

I =V

R=

230

20= 11.5A

XC =1

2πfr × C=

1

2π × 100× 12.66× 10−6= 125.7Ω

Voltage drop across capacitance

VC = IXC = 11.5× 125.7 = 1445.7V

(c) Quality factor

Q =XL

R=

2πfr × LR

=2π × 100× 0.2

20= 6.28

12: An RLC series circuit has a resistance ofR=10 Ω and a variable inductance. is connectedin series with a capacitance across 230 V supplyFind (a) the value of the inductance for whichthe voltage across the resistance is maximum.resonance occurs at 100 Hz frequency (b) Qfactor of the coil (c) Voltage across R, L, C.The applied voltage is 230 V 50 Hz.


fr =1

2π√LC

√LC =

1

2πfr=

1

2π × 50= 3.18× 10−3

LC = 1× 10−5

L =1× 10−5

100× 10−6= 101 mH

(b) Quality factor

Q =XL

R=

2πfr × LR

=2π × 100× 100× 10−3

10= 3.14

Impedance at resonance is Z = R = 10Ω(c) Current through resistance is

I =V

R=

230

10= 23A

Voltage drop across resistor is

V = IR = 23× 10 = 230V

Voltage drop across Inductor/capacitance is

XL = 2πfr × L = 2π × 50× 101× 10−3 = 31.714Ω

XL = XC

VL = VC = IXL = 23× 31.714 = 729.2V


1.2. PARALLEL RESONANCE Chapter 1. Resonance

1.2 Parallel Resonance

Consider a parallel circuit consists of resistor andinductor in one branch and capacitor in another branchwhich is as shown in Figure 1.12.

RC

LCILII RI

Figure 1.12: General parallel resonant circuit

The total admittance of the circuit is

Y = G+ j

(ωC − 1

ωL

)When the circuit is at resonance the imaginary part

is zero

ωrC −1

ωrL= 0

ω2r =

1

LC

ωr =

√1

LC

fr =1

2π

√1

LC

1.3 Practical Parallel Resonant cir-cuit


R

CL

CI LII


The impedance of the inductor branch is

ZL = R+ jωL

The admittance of the inductor branch is

YL =1

ZL=

1

R+ jωL

YL =1

ZL=

1

R+ jωL

=1

R+ jωL× R− jωLR− jωL

=R− jωLR2 + ω2L2

Similarly the impedance of the capacitor branch is

ZC =1

jωC

The admittance of the capacitor branch is

YC =1

ZC= jωC

Total admittance of the circuit is

Y = YL + YC =R− jωLR2 + ω2L2

+ jωC

Separating real and imaginary parts

Y =R

R2 + ω2L2+ j

[ωC − ωL

R2 + ω2L2

]

ωrC −ωrL

R2 + ω2rL

2= 0

ωrC =ωrL

R2 + ω2rL

2

R2 + ω2rL

2 =ωrL

ωrC=L

C

ω2rL

2 =L

C−R2

ω2r =

LC −R

2

L2=

1

LC− R2

L2

ωr =

√1

LC− R2

L2

fr =1

2π

√1

LC− R2

L2


1.4. PARALLEL RESONANT CIRCUIT BY CONSIDERING CAPACITANCE RESISTANCEChapter 1. Resonance

1.4 Parallel Resonant circuit byconsidering capacitance resis-tance


CR

C L

CI LII

LR


The impedance of the inductor branch is

ZL = RL + jωL


YL =1

ZL=

1

RL + jωL

YL =1

ZL=

1

RL + jωL

=1

RL + jωL× RL − jωLRL − jωL

=RL − jωLR2L + ω2L2

Similarly the impedance of the capacitor branch is

ZC = RC − j1

ωC


YC =1

ZC=

1

RC − j 1ωC

=1

RC − j 1ωC

×RC + j 1

ωC

RC + j 1ωC

=RC + j 1

ωC

R2c + 1

ω2C2

Total admittance of the circuit is

Y = YL + YC =RL − jωLR2L + ω2L2

+RC + j 1

ωC

R2c + 1

ω2C2

Separating real and imaginary partsReal part is

RLR2L + ω2L2

+RC

R2c + 1

ω2C2

Imaginary part is

1ωC

R2C + 1

ω2C2

− ωL

R2L + ω2L2

1ωC

R2C + 1

ω2C2

− ωL

R2L + ω2L2

= 0

1ωC

R2C + 1

ω2C2

=ωL

R2L + ω2L2

1

ωrC

(R2L + ω2

rL2)

= ωrL

(R2C +

1

ω2rC

2

)1

LC

(R2L + ω2

rL2)

= ω2r

(R2C +

1

ω2rC

2

)R2L

LC+ ω2

r

L

C= ω2

rR2C +

1

C2

ω2r

(R2C −

L

C

)=R2L

LC− 1

C2

ω2r

(R2C −

L

C

)=

1

LC

(R2L −

L

C

)

ω2r =

1

LC×(R2L −

LC

)(R2C −

LC

)ωr =

√1

LC

√R2L −

LC

R2C −

LC

fr =1

2π√LC

√R2L −

LC

R2C −

LC

The frequency response of parallel graph is as shownin Figure 1.15. From the figure it is observed thatthe current is minimum at resonance. The parallelcircuit is called as a rejector circuit. The circuitimpedance is maximum at the resonance. The halfpower frequencies are at

√2Ir.

I

Ir

2Ir

f1 r 2f f f

Figure 1.15: plot of parallel resonant circuit



Calculate the resonant frequency of the circuitshown in Figure 1.16

AC

AL

AR


ZLC = jXL + jXC = jωL+1

jωC=

1− ω2LC

jωC

ZLC =j(ω2LC − 1)

ωC

ω2LC = 1

ω2 =1√LC

1: For the circuit as shown in Figure 1.17 find theresonant frequency and the corresponding current ineach branch.

CI LII

6 Ω

1 mH

4 Ω

20 F μ200 V


Solution:

L

C=

1× 10−3

20× 10−6= 50

fr =1

2π√LC

√R2L −

LC

R2C −

LC

=1

2π√

2× 10−8

√62 − 50

42 − 50

= 1125.4× 0.641 = 721 Hz

XL = 2π × f × L = 2π × 721× 1× 10−3 = 4.53Ω

XC =1

2π × f × C=

1

2π × 721× 20× 10−6= 11.03Ω

IL =V

ZL=

200

R+ jXL=

200

6 + j4.53

=200

7.51, 6 37= 26.636 − 37

IC =V

ZC=

200

R+ jXC=

200

4− j11.03

=200

11.73, 6 − 70= 17.056 70

2: Find the value of L for which the circuit as shownin Figure 1.18 is resonance at ω = 5000 rad/sec.

CI LI

8 Ω

L

4 Ω

-j12 Ω


Solution: The admittance of circuit is given by

Y =1

4 + jXL+

1

8− j12

=4 + jXL

42 +X2L

+8− 12

82 + 122

At resonance

XL

42 +X2L

=12

82 + 122

12(42 +X2L) = XL(82 + 122)

192 + 12X2L = 208XL

12X2L − 208XL + 192 = 0

3X2L − 52XL + 48 = 0

XL1 =52±

√522 − (4× 3× 38)

2× 3

XL1 = 16.36 Ω

XL2 = 0.978 Ω

L1 =XL1

ω=

16.36

5000= 3.27mH

L2 =XL2

ω=

0.978

5000= 0.196mH

3: Find the value of L for which the circuit as shownin Figure 1.19 is resonance at 1000 Hz.

8 Ω

L

10 Ω

50 F μ





XC =1

2πfrC=

1

2π1000× 50× 10−6= 3.18 Ω

Y =1

10 + jXL+

1

8− j3.18

=10− jXL

102 +X2L

+8 + j3.18

82 + 3.182

At resonance

XL

102 +X2L

=3.18

82 + 3.182

3.18(100 +X2L) = XL(64 + 10.11)

318 + 3.18X2L = 74XL

3.18X2L − 74XL + 318 = 0

XL =74±

√742 − (4× 3.18× 318)

2× 3.18

=74±

√5476− (4045)

6.36

=74± 37.8

6.36

XL1 = 17.57 Ω

XL2 = 5.69 Ω

L1 =XL

ω=

17.57

2π × 1000= 2.797mH

L2 =XL

ω=

5.69

2π × 1000= 0.9mH

4: Find the value of C for which the circuit as shownin Figure 1.26 is resonance at 750 Hz.

CI LI

10 Ω6 Ω

C j8 Ω



Y =1

10 + j8+

1

6− jXC

=10 + j8

102 + 82+

6− jXC

62 +X2C

=

(10

164+

6

36 +X2C

)+ j

(XC

36 +X2C

− 8

164

)

At resonance

XC

36 +X2C

− 8

164= 0

164XC − 8X2C − 288 = 0

2X2C − 41XC + 78 = 0

XC1 =41±

√412 − (4× 2× 72)

2× 2

XC1 = 18.56 Ω

XC2 = 1.94 Ω

C1 =1

ωXC1=

1

2× π × 750× 18.56= 11.44 µF

C2 =1

ωXC2=

1

2× π × 750× 1.94= 109.45 µF

5: Find the value of C for which the circuit as shownin Figure 1.21 is resonance at 3000 rad/sec.

10 Ω8 Ω

C 0.005H


Solution:

XL = ω × L = 3000× 0.005 = 15 Ω

The admittance of circuit is given by

Y =1

10 + j15+

1

8− jXC

=10− j8

102 + 152+

8 + jXC

82 +X2C

=

(10

325+

8

64 +X2C

)+ j

(XC

64 +X2C

− 8

325

)At resonance

XC

64 +X2C

− 8

325= 0

325XC = 512 + 8X2C

8X2C − 325XC + 512 = 0



325XC =325±

√3252 − (4× 8× 512)

2× 8

=325±

√105625− 16384

16

=325± 298.7

16

XC1 = 38.98 Ω

XC2 = 1.64 Ω

C1 =1

ωC=

1

3000× 38.98= 8.55 µF

C2 =1

ωC=

1

3000× 1.64= 203 µF

6: Find the value of RL for which the circuit as shownin Figure 1.22 is resonant .

CI LI

LR10 Ω

-j15 Ω j10 Ω



Y =1

RL + j10+

1

10− j15

=RL − j10

R2L + 100

+10 + j15

100 + 225

=

(RL

R2L + 100

+10

325

)+ j

(15

325− 10

R2L + 100

)At resonance

15

325− 10

R2L + 100

= 0

15(R2L + 100) = 3250

R2L = 116.6

RL = 10.8

7: Two impedances (10+j12) Ω and (20-j15)Ω areconnected in parallel and this combination is

connected in series with an impedance (5− jXC) Ω.

Find the value of XC , for which resonance occurs.

10 Ω20 Ω

-j15 Ω j12 Ω

5 Ω CjX


Solution: The impedance of circuit is given by

Z = (5− jXC) + [(10 + j12)||(20− j15)]

= (5− jXC) +(10 + j12)(20− j15)

(10 + j12) + (20− j15)

= (5− jXC) +(10 + j12)(20− j15)

(30− j3)

= (5− jXC) +380 + j90

30− j3

= (5− jXC) +390.516 13.325

30.146 − 5.71= (5− jXC) + 13 6 19.03

= (5− jXC) + 12.78 + j2.33

= (5− jXC) + 17.78 + j(2.33− jXC)

At resonance imaginary part is zero

XC = 2.33

Z = (5− jXC) + [(10 + j12)||(20− j15)]

= (5− jXC) +(10 + j12)(20− j15)

(10 + j12) + (20− j15)

= (5− jXC) +(10 + j12)(20− j15)

(30− j3)

= (5− jXC) +380 + j90

30− j3

= (5− jXC) +390.516 13.325

30.146 − 5.71= (5− jXC) + 13 6 19.03

= (5− jXC) + 12.78 + j2.33

= (5− jXC) + 17.78 + j(2.33− jXC)



8: (2015-JAN) Find the value of C for which the circuitas shown in Figure 1.24 is resonance at 50 Hz.

20 Ω10 Ω

CX j37.7 Ω


Solution:The admittance of circuit is given by

Y =1

20 + j37.7+

1

10− jXC

=20− j37.7

202 + 37.72+

10 + jXC

102 +X2C

=20− j37.7

1821.29+

10 + jXC

102 +X2C

At resonance imaginary part is

XC

100 +X2C

− 37.7

1821.29= 0

XC

100 +X2C

=37.7

1821.29

1821.29XC = 37.7X2C + 3770

37.7X2C − 1821.29XC + 3770 = 0

XC =1821.29±

√1821.292 − (4× 37.7× 3770)

2× 37.7

=1821.29±

√3317097− 568516

75.4

=1821.29± 1657.8

75.4XC1 = 46.14 Ω

XC2 = 2.16 Ω

ω = 2× π × f = 2× π × 50 = 314.16

C1 =1

ωXC1=

1

314.16× 46.14= 68.98 µF

C2 =1

ωXC2=

1

314.16× 2.16= 1.47× 10−3 F

9: (2012-DEC) Determine the value of RL RC forwhich the circuit as shown in Figure 1.25 is resonatesat all frequencies.

40 Fμ 40 mH

LRCR


Solution:

The admittance of circuit is given by

fr =1

2π√LC

√R2L −

LC

R2C −

LC

The circuit will resonate at any frequency if

R2L = R2

C =L

C

RL = RC =

√L

C=

√40× 10−3

40× 10−6= 31.6 Ω



JAN 2017 CBCS Q 7 a) Prove that f0 =√f1f2 where

f1 and f2 are the two half power frequencies of aresonant circuits. 8 Marks

JAN 2017 CBCS Q 7 b) A series RLC circuit ofR = 100 Ω L=0.01H and C=0.01µF is connectedacross supply of 10 mV. Determine i) f0 ii) Qfactor, iii) BW iv) f1 and f2 iv) I0 8 Marks

Solution:i) LC = 0.01× 0.01× 10−6 = 1× 10−10

fo =1

2π√LC

=1

2π√

1× 10−10= 15.9× 103Hz

ii) Q = 1R

√LC = 2πfoL

R

Q =2πfoL

R=

2π × 15.9× 103 × 0.01

100= 10

iii) Bandwidth BW

B =R

L=

100

0.01= 10× 103rad/sec

Bandwidth in Hz

B =B

2π=

10× 103

2π= 1.59× 103Hz

iv) f1 and f2

R

2L=

100

2× 0.01= 5× 103

f1 =1

2π

− R

2L+

√(R

2L

)2

+1

LC

=

1

2π

[−5× 103 +

√(5× 103)2 +

1

1× 10−10

]=

1

2π

[−5× 103 +

√25× 106 + 1× 1010

]=

1

2π

[−5× 103 + 100× 103

]= 15.12× 103Hz

f2 =1

2π

R2L

+

√(R

2L

)2

+1

LC

=

1

2π

[5× 103 + 100× 103

]= 16.71× 103Hz

iv) I0

I0 =V

R=

10× 10−3

10= 1× 10−3Ampere

2017 JAN 5 a) A series R-L-C circuit is fed with50 V rms supply. At resonance the currentthrough circuit is 25 A and the voltage acrossinductor is 1250 volts. If C = 4µF determinethe values of R, L, Q, resonant frequency,bandwidth and half power frequencies. (12 M)

Solution:

R =V

I=

50

25= 2 Ω

At resonanceXL = XC

Quality factor Q is

Q =VL or VC

E=

1250

50= 25

Q =1

R

√L

C⇒

√L

C= QR

L = (QR)2C = (25× 2)2 × 4× 10−6 = 0.01H

fo =1

2π√LC

fo =1

2π√

0.01× 4× 10−6= 795.7 Hz

iii) Bandwidth BW

B =R

L=

2

0.01= 200rad/sec

Bandwidth in Hz

B =B

2π=

200

2π= 31.83Hz

iv) f1 and f2

R

2L=

2

2× 0.01= 100 LC = 0.01×4×10−6 = 4×10−8

f1 =1

2π

− R

2L+

√(R

2L

)2

+1

LC

=

1

2π

[−100 +

√(100)2 +

1

4× 10−8

]=

1

2π

[−100 +

√10× 103 + 25× 106

]=

1

2π

[−100 + 5× 103

]= 779.8 Hz

f2 =1

2π

R2L

+

√(R

2L

)2

+1

LC

=

1

2π

[100 + 5× 103

]= 811.69Hz



JULY 2016 Q 5 a) With respect to series resonantcircuit define resonant frequency fr and half powerfrequencies f1 and f2. Also show that the resonantfrequency is equal to the geometric mean of half powerfrequencies 10 Marks

JULY 2016 Q 5 B) ) A series circuit is energizedby constant voltage and constant frequencysupply. Resonance takes place due to variationof inductance and the supply frequency is 300Hz. The capacitance in the circuit is 10µF .Determine the value of resistance in the circuitif the quality factor is 5. Also find the valueof the inductance at half power frequencies. 10Marks

Solution: LC = 0.01 × 0.01 × 10−6 = 1 × 10−10

The resonant frequency fr is

fr =1

2π

1√LC

√LC =

1

2π

1

fr=

1

2π

1

300= 5.3× 10−4

LC = 2.8× 10−7

L =2.8× 10−7

10× 10−6= 28.1× 10−3Henry

Q =2πfrL

R

R =2πfrL

Q=

2π × 300× 0.0281

5= 10.6 Ω

The value of the inductance at half power frequenciesis

ωr = 2πfr = 2× π × 300 = 1.88× 103

L1 =1

ω2C− R

ω

=1

(1.88× 103)2.10× 10−6− 10.6

1.88× 103

= 0.02829− 5.638× 10−3 = 22.65× 10−3 H

L2 =1

ω2C+R

ω

=1

(1.88× 103)2.10× 10−6+

10.6

1.88× 103

= 0.02829 + 5.638× 10−3 = 33.93× 10−3 H

DEC/JAN 2015 5 a) What is resonance? Derivean expression for cut-off frequencies 8 Marks

DEC/JAN 2015 5 b) Calculate the half powerfrequencies of series resonant circuit where theresonance frequency is 150 KHz and bandwidthis 75 KHz 4 Marks

Solution:Frequency in radians is

ωr = 2πfr = 2× π × 150× 103 = 942.5× 103 radians

Bandwidth in radians is

B = 2πB = 2× π × 75× 103 = 471.2× 103 radians

ω1 = −B2

+

√(B

2

)2

+ ω2r

= −235.6× 103 +√

55.5× 109 + 888.3× 109

= −235.6× 103 + 971.5× 103

= 736× 103 Radians

ω2 =B

2+

√(B

2

)2

+ ω2r

= 235.6× 103 + 971.5× 103 = 1.2× 106

= 1.2× 106 Radians

Frequency in Hertz

f1 =ω1

2π=

736× 103

2× π= 117.13× 103 Hz

f2 =ω1

2π=

1.2× 106

2× π= 191× 103 Hz

DEC 2015 5 a) It is required that a series RLCcircuit should resonate at 500 KHz. Determinethe values of R, L, and C if the bandwidth ofthe circuit is 10 KHz and its impedance is 100Ω at resonance. Also find the voltages acrossL and C resonance if the applied voltage is 75volts 10 Marks

Solution:At resonance the circuit acts as a pure resistive hencethe value of R=100 Ω

B =R

2πL

L =R

2πB=

100

2× π × 10× 103= 1.59× 10−3H

f0 =1

2π√LC



√LC =

1

2πf0=

1

2π × 500× 103= 0.318× 10−6

LC = (0.318× 10−6)2 = 0.1× 10−12

C =0.1× 10−12

1.59× 10−3= 62.83× 10−12F

XL = 2πf0L = 2π × 500× 103 × 1.59× 10−3

= 5× 103Ω

At resonance current I is

I =V

R=

75

100= 0.75A

XC =1

2πf0C=

1

2π × 500× 106 × 62.9× 10−12

= 5× 103Ω

At resonance

XL = XC

Voltage across capacitor/inductor is

VC = VL = IXC = 0.75× 5× 103 = 3750V

JULY 2014 5 a) Define the following terms i)Resonance ii) Q factor iii) Selectivity of seriesRLC circuit iv) Bandwidth 4 Marks

JULY 2014 5 b) Prove that f0 =√f1f2 where f1

and f2 are the two half power frequencies ofresonant circuit 8 Marks

JULY 2014 5 C) A series RLC circuit has R=4Ω L=1 mH and C= 10 µF . Calculate the Qfactor, band width resonant frequency and thehalf power frequencies f1 and f2 8 Marks

Solution:

fr =1

2π√LC

=1

2π√

1× 10−3 × 10× 10−6

= 1591.5 Hz

B =R

2πL=

4

2π × 1× 10−3= 636.62Hz

R

2L=

4

2× 1× 10−3= 2× 103

f1 =1

2π

−R2L

+

√(R

2L

)2

+1

LC

=

1

2π

[−2× 103 +

√4× 106 + 0.1× 109

]=

1

2π

[−2× 103 + 10198

]= 1305 Hz

f2 =1

2π

R2L

+

√(R

2L

)2

+1

LC

=

1

2π

[2× 103 + 10198

]= 1941 Hz

JULY 2013 5 a) For the series RLC circuit shownin Figure find the resonant frequency, halfpower frequencies, band width Quality factorand 10 Marks

( )iv t+

-100 Ω

0.4 F 0.5 H


Solution:Resonant frequency is

fr =1

2π√LC

=1

2π√

0.4× 0.5= 0.356 Hz

Given data may wrongly printed because frequencyis always more than 1 Hz

JULY 2012 5 a) Define the following terms i)Resonance ii) Q factor iii) Selectivity of seriesRLC circuit iv) Bandwidth 6 Marks

DEC 2012 5 b) A series RLC circuit has R=10 ΩL=0.01 H and C= 0.01 µF and it is connectedacross 10 mV supply. Calculate i)f0, ii)Q0

iii)band width iv) f1 and f2 v) I0 10 Marks

Solution:

fo =1

2π√LC

=1

2π√

0.01× 0.01× 10−6

= 15915.5 Hz

Qo =2πf0L

R=

2π × 15915.5× 0.01

10= 100

B =R

2πL=

10

2π × 0.01= 160Hz

R

2L=

10

2× 10× 10−3= 500



f1 =1

2π

−R2L

+

√(R

2L

)2

+1

LC

=

1

2π

[−500 +

√250× 103 + 1× 1010

]=

1

2π

[−500 + 100× 103

]= 15836 Hz

f2 =1

2π

R2L

+

√(R

2L

)2

+1

LC

=

1

2π

[500 + 100× 103

]= 15995 Hz

Io =V

R=

10× 10−3

10= 1× 10−3A

DEC 2011 5 a) Define the following terms i)Resonance ii) Q factor iii) Selectivity of seriesRLC circuit iv) Bandwidth 6 Marks

JULY 2010 5 b) Derive for resonant circuit f0 =√f1f2 where f1 and f2 are the two half power

frequencies 8 Marks

DEC 2010 5 C) An RLC series circuit has R =1KΩ L=100 mH and C= 10 µF . If a voltageof 100 V is applied across series combinationdetermine i)Resonant frequency ii) Q factor,and iii) Half power frequencies 4 Marks

Solution:

fr =1

2π√LC

=1

2π√

100× 10−3 × 10× 10−6

= 159.15 Hz

ii) Q factor

Q =2πfrL

R=

2π × 159.15× 100× 10−3

1× 103= 0.1

B =R

2πL=

1× 103

2π × 100× 10−3= 1592.35Hz

iii) Half power frequencies

R

2L=

1× 103

2× 100× 10−3= 5× 103

f1 =1

2π

−R2L

+

√(R

2L

)2

+1

LC

=

1

2π

[−5× 103 +

√25× 106 + 1× 106

]=

1

2π

[−5× 103 + 5099

]= 15.75 Hz

f2 =1

2π

R2L

+

√(R

2L

)2

+1

LC

=

1

2π

[5× 103 + 5099

]= 1607 Hz

May/June 2010 5 C) Determine the i) line currentii) the power factor and iii) voltage across coilwhen a coil of resistance 40Ω and inductance of0.75 H forms a part of series circuit, for whichthe resonant frequency is 55 Hz If the supplyvoltage is 250 V 50 Hz 8 Marks

Solution:

ωrL = 2πfrL =1

ωrC= 2π × 55× 0.75 = 259.05

XL at 50 Hz

XL = ωL = 2πfL = 2π × 50× 0.75 = 235.5

XC = 259.05× 55

50= 284.96

Circuit Impedance at 50 Hz is

Z = 40 + J(XL −XC) = 40 + J(235.5− 284.96)

= 40− J49.46 = 63.616 − 51.04

i) line current

I =V

R=

250

63.61= 3.93 A

ii) Power factor is

I = cosφ = cos51.04 = 0.629leading

iii) voltage across coil

V = IZ = I√R2 +X2

L = 3.93√

402 + 235.52 = 938.77



June/July 2009 5 C) A series circuit RLC consistsof R = 1KΩ and an inductance of 100 mHin series with capacitance of 10 nF . Ifa voltage of 100 V is applied across seriescombination determine i)Resonant frequencythe ii) maximum current in the circuit iii) Qfactor, and iii) Half power frequencies 8 Marks

Solution:

fr =1

2π√LC

=1

2π√

100× 10−3 × 10× 10−9

= 5 KHz

ii) Maximum current in the circuit

I =V

R=

100

1× 103= 0.1A

iii) Q factor

Q =2πfrL

R=

2π × 5× 103 × 100× 10−3

1× 103= 3.14

iv) Half power frequencies

R

2L=

1× 103

2× 100× 10−3= 5× 103

f1 =1

2π

−R2L

+

√(R

2L

)2

+1

LC

=

1

2π

[−5× 103 +

√25× 106 + 1× 109

]=

1

2π

[−5× 103 + 32× 103

]= 27 KHz

f2 =1

2π

R2L

+

√(R

2L

)2

+1

LC

=

1

2π

[5× 103 + 32× 103

]= 37 KHz


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