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
1.1. SERIES RESONANCE Chapter 1. Resonance
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 3
1.1. SERIES RESONANCE Chapter 1. Resonance
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.
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 4
1.1. SERIES RESONANCE Chapter 1. Resonance
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
The resonant frequency ωo is
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 5
1.1. SERIES RESONANCE Chapter 1. Resonance
Figure 1.9: Example
LC = 312× 10−3 × 1.25× 10−12 = 0.39× 10−12
The resonant frequency ωo is
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
The resonant frequency ωo is
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
Figure 1.11: Example
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 6
1.1. SERIES RESONANCE Chapter 1. Resonance
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 7
1.1. SERIES RESONANCE Chapter 1. Resonance
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
Solution: (a) Resonant frequency
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.
Solution: (a) Resonant frequency
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 8
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
Consider a parallel circuit consists of resistor andinductor in one branch and capacitor in another branchwhich is as shown in Figure 1.13.
R
CL
CI LII
Figure 1.13: General parallel resonant circuit
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 9
1.4. PARALLEL RESONANT CIRCUIT BY CONSIDERING CAPACITANCE RESISTANCEChapter 1. Resonance
1.4 Parallel Resonant circuit byconsidering capacitance resis-tance
Consider a parallel circuit consists of resistor andinductor in one branch and capacitor in another branchwhich is as shown in Figure 1.14.
CR
C L
CI LII
LR
Figure 1.14: General parallel resonant circuit
The impedance of the inductor branch is
ZL = RL + jωL
The admittance of the inductor branch is
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
The admittance of the inductor branch is
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 10
1.4. PARALLEL RESONANT CIRCUIT BY CONSIDERING CAPACITANCE RESISTANCEChapter 1. Resonance
Calculate the resonant frequency of the circuitshown in Figure 1.16
AC
AL
AR
Figure 1.16: General parallel resonant circuit
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
Figure 1.17: General parallel resonant circuit
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
Figure 1.18: General parallel resonant circuit
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 11
1.4. PARALLEL RESONANT CIRCUIT BY CONSIDERING CAPACITANCE RESISTANCEChapter 1. Resonance
Figure 1.19: General parallel resonant circuit
Solution: The admittance of circuit is given by
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
Figure 1.20: General parallel resonant circuit
Solution: The admittance of circuit is given by
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
Figure 1.21: General parallel resonant circuit
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 12
1.4. PARALLEL RESONANT CIRCUIT BY CONSIDERING CAPACITANCE RESISTANCEChapter 1. Resonance
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
Figure 1.22: General parallel resonant circuit
Solution: The admittance of circuit is given by
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
Figure 1.23: General parallel resonant circuit
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)
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 13
1.4. PARALLEL RESONANT CIRCUIT BY CONSIDERING CAPACITANCE RESISTANCEChapter 1. Resonance
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
Figure 1.24: General parallel resonant circuit
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
Figure 1.25: General parallel resonant circuit
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 14
1.4. PARALLEL RESONANT CIRCUIT BY CONSIDERING CAPACITANCE RESISTANCEChapter 1. Resonance
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 15
1.4. PARALLEL RESONANT CIRCUIT BY CONSIDERING CAPACITANCE RESISTANCEChapter 1. Resonance
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 16
1.4. PARALLEL RESONANT CIRCUIT BY CONSIDERING CAPACITANCE RESISTANCEChapter 1. Resonance
√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
Figure 1.26: Example
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 17
1.4. PARALLEL RESONANT CIRCUIT BY CONSIDERING CAPACITANCE RESISTANCEChapter 1. Resonance
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 18
1.4. PARALLEL RESONANT CIRCUIT BY CONSIDERING CAPACITANCE RESISTANCEChapter 1. Resonance
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
Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 19