## 1. A 100Ω100Ω resistor is connected to a 220V220V, 50Hz50Hz ac supply.

a) What is the rms value of current in the circuit?

Ans: It is given that,

Resistance, R=100ΩR=100Ω

Voltage, V=220VV=220V

Frequency, f=50Hzf=50Hz

It is known that,

Irms=VrmsRIrms=VrmsR

⇒Irms=220100=2.2A⇒Irms=220100=2.2A

Therefore, the rms value of current in the circuit isIrms=2.2AIrms=2.2A.

## b) What is the net power consumed over a full cycle?

### Ans: It is known that,

Power=V×IPower=V×I

⇒Power=220×2.2⇒Power=220×2.2

⇒Power=484W⇒Power=484W

Therefore, the net power consumed over a full cycle is484W484W.

## 2.

a) The peak voltage of an ac supply is 300V300V. What is the rms voltage?

### Ans: It is given that,

Peak voltage of the ac supply, V0=300VV0=300V

It is known that,

Vrms=V02√Vrms=V02

⇒Vrms=3002√⇒Vrms=3002

⇒Vrms=212.1V⇒Vrms=212.1V

Therefore, the rms voltage is 212.1V212.1V.

## b) The rms value of current in an ac circuit is 10A10A. What is the peak current?

### Ans: It is given that,

Rms value of current in an ac circuit, Irms=10AIrms=10A

It is known that,

I0=2–√×IrmsI0=2×Irms

⇒I0=1.414×10⇒I0=1.414×10

⇒I0=14.14A⇒I0=14.14A

Therefore, the peak current is 14.14A14.14A.

## 3. A 44mH44mH inductor is connected to 220V220V,50Hz50Hz ac supply. Determine the rms value of the current in the circuit.

### Ans: It is known that,

Inductance, L=44mH=44×10−3HL=44mH=44×10−3H

Voltage,V=220VV=220V

Frequency, fL=50HzfL=50Hz

Angular frequency, ωL=2πfLωL=2πfL

It is known that,

Inductive reactance, XL=ωLL=2πfLLXL=ωLL=2πfLL

⇒XL=2×3.14×50×44×10−3Ω⇒XL=2×3.14×50×44×10−3Ω

⇒XL=13.8Ω⇒XL=13.8Ω

Irms=VXLIrms=VXL

⇒Irms=22013.82⇒Irms=22013.82

⇒Irms=15.92A⇒Irms=15.92A

Therefore, the rms value of the current in the circuit is 15.92A15.92A.

## 4. A 60μF60μF capacitor is connected to a 110V110V,60Hz60Hz ac supply. Determine the rms value of the current in the circuit.

### Ans: It is given that,

Capacitance, C=60μF=60×10−6FC=60μF=60×10−6F

Voltage, V=110VV=110V

Frequency, fC=60HzfC=60Hz

It is known that,

Irms=VXCIrms=VXC

XC=1ωCC=12πfCCXC=1ωCC=12πfCC

⇒XC=12×3.14×60×60×10−6⇒XC=12×3.14×60×60×10−6

⇒XC=44.248Ω⇒XC=44.248Ω

⇒Irms=11044.28⇒Irms=11044.28

⇒Irms=2.488A⇒Irms=2.488A

Therefore, the rms value of the current in the circuit is 2.488A2.488A.

## 5. In exercises 44 and 55 What is the net power absorbed by each circuit over a complete cycle? Explain your answer.

### Ans: From the inductive circuit,

Rms value of current, Irms=15.92AIrms=15.92A

Rms value of voltage, Vrms=220VVrms=220V

It is known that,

Net power absorbed, P=Vrms×IrmscosϕP=Vrms×Irmscosϕ

Where,

ϕϕ is the phase difference between voltage and current

For a pure inductive circuit, the phase difference between alternating voltage and current is 900900i.e., ϕ=900ϕ=900

⇒P=220×15.92cos900=0⇒P=220×15.92cos900=0

Therefore, net power absorbed is zero in a pure inductive circuit.

In a capacitive circuit,

Rms value of current, Irms=2.49AIrms=2.49A

Rms value of voltage, Vrms=110VVrms=110V

It is known that,

Net power absorbed, P=Vrms×IrmscosϕP=Vrms×Irmscosϕ

Where,

ϕϕ is the phase difference between voltage and current

For a pure capacitive circuit, the phase difference between alternating voltage and current is 900900i.e., ϕ=900ϕ=900

⇒P=110×2.49cos900=0⇒P=110×2.49cos900=0

Therefore, net power absorbed is zero in a pure capacitive circuit.

## 6. Obtain the resonant frequency ωrωr of a series LCR circuit with L=2.0HL=2.0H, C=32μFC=32μF and R=10ΩR=10Ω . What is the Q-value of this current?

### Ans: It is given that,

Inductance, L=2HL=2H

Capacitance,

C=32μF=32×10−6FC=32μF=32×10−6F

R=10ΩR=10Ω

It is known that,

Resonant frequency, ωr=1LC√ωr=1LC

⇒ωr=12×32×10−6√⇒ωr=12×32×10−6

⇒ωr=18×10−3⇒ωr=18×10−3

⇒ωr=125rad/s⇒ωr=125rad/s

Q−value=ωrLRQ−value=ωrLR

⇒Q=1RLC−−√⇒Q=1RLC

⇒Q=110232×10−6−−−−−−√⇒Q=110232×10−6

⇒Q=110×4×10−3⇒Q=110×4×10−3

⇒Q=25⇒Q=25

Therefore, the resonant frequency is 125rad/s125rad/s and Q-value is 2525.

## 7. A charged 30μF30μF capacitor is connected to a 27mH27mH inductor. What is the angular frequency of free oscillations of the circuit?

### Ans: It is given that,

Capacitance, C=30μF=30×10−6FC=30μF=30×10−6F

Inductance, L=27mH=27×10−3HL=27mH=27×10−3H

It is known that,

Angular frequency of free oscillations, ωr=1LC√ωr=1LC

⇒ωr=127×10−3×30×10−6−−−−−−−−−−−−−−−−−−√⇒ωr=127×10−3×30×10−6

⇒ωr=19×10−4⇒ωr=19×10−4

⇒ωr=1.11×103rad/s⇒ωr=1.11×103rad/s

Therefore, the angular frequency of free oscillations of the circuit is

1.11×103rad/s1.11×103rad/s

## 8. Suppose the initial charge on the capacitor in exercise 77 is 6mC6mC . What is the total energy stored in the circuit initially? What is the total energy at a later time?

### Ans: It is known that,

Capacitance of the capacitor, C=30μF=30×10−6FC=30μF=30×10−6F

Inductance of the capacitor, L=27mH=27×10−3HL=27mH=27×10−3H

Charge on the capacitor, Q=6mC=6×10−3CQ=6mC=6×10−3C

It is known that,

Energy,E=12Q2CE=12Q2C

⇒E=12(6×10−3)230×10−6⇒E=12(6×10−3)230×10−6

⇒E=610=0.6J⇒E=610=0.6J

Therefore, the energy stored in the circuit initially is E=0.6JE=0.6J.

Total energy at later time will remain same as the initially stored i.e., 0.6J0.6J because energy is shared between the capacitor and the inductor.

## 9. A series LCR circuit with R=20ΩR=20Ω, L=1.5HL=1.5H and C=35μFC=35μF is connected to a variable frequency 200V200V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

### Ans: It is known that,

Resistance, R=20ΩR=20Ω

Inductance, L=1.5HL=1.5H

Capacitance, C=35μF=35×10−6FC=35μF=35×10−6F

Voltage, V=200VV=200V

It is known that,

Impedance,Z=R2+(XL−XC)2−−−−−−−−−−−−−−√Z=R2+(XL−XC)2

At resonance, XL=XCXL=XC

⇒Z=R=20Ω⇒Z=R=20Ω

I=VZ=20020I=VZ=20020

⇒I=10A⇒I=10A

Average power, P=I2RP=I2R

⇒P=102×20⇒P=102×20

⇒P=2000W⇒P=2000W

Therefore, the average power transferred is 2000W2000W.

## 10. A radio can tune over the frequency range of a portion of MWMW broadcast band: (800kHz800kHz to 1200kHz1200kHz). If its LC circuit has an effective inductance of 200μH200μH, what must be the range of its variable capacitor?

(Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radio wave.)

### Ans: It is given that,

The range of frequency(f)(f) of a radio is 800kHz800kHz to 1200kHz1200kHz.

Effective inductance of the circuit, L=200μH=200×10−6HL=200μH=200×10−6H

It is known that,

Capacitance of variable capacitor for f1f1 is C1=1ω12LC1=1ω12L

Where,

ω1ω1 is the angular frequency for capacitor for f1=2πf1f1=2πf1

⇒ω1=2×3.14×800×103rad/s⇒ω1=2×3.14×800×103rad/s

⇒C1=1(2×3.14×800×103)2×200×10−6⇒C1=1(2×3.14×800×103)2×200×10−6

⇒C1=1.9809×10−10F⇒C1=1.9809×10−10F

⇒C1=198.1pF⇒C1=198.1pF

C2=1ω22LC2=1ω22L

⇒C2=1(2×3.14×1200×103)2×200×10−6⇒C2=1(2×3.14×1200×103)2×200×10−6

⇒C2=0.8804×10−10F⇒C2=0.8804×10−10F

⇒C2=88.04pF⇒C2=88.04pF

Therefore, the range of the variable capacitor is from 88.04pF88.04pF to 198.1pF198.1pF.

## 11. Figure shows a series LCR circuit connected to a variable frequency 230V230V source. L=5.0HL=5.0H, C=80μFC=80μF, R=40ΩR=40Ω .

(Image will be Uploaded Soon)

a) Determine the source frequency which drives the circuit in resonance.

### Ans: It is given that,

Voltage, V=230VV=230V

Inductance, L=5.0HL=5.0H

Capacitance, C=80μF=80×10−6FC=80μF=80×10−6F

Resistance, R=40ΩR=40Ω

It is known that,

Source frequency at resonance=1LC√=1LC

⇒15×80×10−6√=50rad/s⇒15×80×10−6=50rad/s

Therefore, the source frequency of the circuit in resonance is 50rad/s50rad/s.

## b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

### Ans: It is known that,

At resonance, Impedance,Z=Z= Resistance,RR

⇒Z=R=40Ω⇒Z=R=40Ω

I=VZI=VZ

⇒I=23040=5.75A⇒I=23040=5.75A

Amplitude, I0=1.414×II0=1.414×I

⇒I0=1.414×5.75⇒I0=1.414×5.75

⇒I0=8.13A⇒I0=8.13A

Therefore, the impedance of the circuit is 40Ω40Ω and the amplitude of current at resonating frequency is 8.13A8.13A.

## c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

### Ans: It is known that,

Potential drop, V=IRV=IR

Across resistor, VR=IRVR=IR

⇒VR=5.75×40=230V⇒VR=5.75×40=230V

Across capacitor, VC=IXC=IωCVC=IXC=IωC

⇒VC=5.75×150×80×10−6⇒VC=5.75×150×80×10−6

⇒VC=1437.5V⇒VC=1437.5V

Across Inductor, VL=IXL=IωLVL=IXL=IωL

⇒VL=5.75×50×5⇒VL=5.75×50×5

⇒VL=1437.5V⇒VL=1437.5V

Across LC combination, VLC=I(XL−XC)VLC=I(XL−XC)

At resonance, XL=XCXL=XC

⇒VLC=0⇒VLC=0

Therefore, the rms potential drop across Resistor is 230V230V, Capacitor is 1437.5V1437.5V, Inductor is 1437.5V1437.5V and the potential drop across LC combination is zero at resonating frequency.

## 12. An LC circuit contains a 20mH20mH inductor and a 50μF50μF capacitor with an initial charge of 10mC10mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0t=0.

a) What is the total energy stored initially? Is it conserved during LC oscillations?

### Ans: It is given that,

Inductance of the inductor, L=20mH=20×10−3HL=20mH=20×10−3H

Capacitance of the capacitor, C=50μF=50×10−6FC=50μF=50×10−6F

Initial charge on the capacitor, Q=10mC=10×10−3CQ=10mC=10×10−3C

It is known that,

Total energy stored initially in the circuit, E=12Q2CE=12Q2C

⇒E=(10×10−3)22×50×10−6=1J⇒E=(10×10−3)22×50×10−6=1J

Therefore, the total energy stored in the LC circuit will be conserved because there is no resistor (R=0)(R=0) connected in the circuit.

## b) What is the natural frequency of the circuit?

### Ans: It is known that,

Natural frequency of the circuit, ν=12πLC√ν=12πLC

⇒ν=12π20×10−3×50×10−6√⇒ν=12π20×10−3×50×10−6

⇒ν=1032π=159.24Hz⇒ν=1032π=159.24Hz

Natural angular frequency, ωr=1LC√ωr=1LC

⇒ωr=120×10−3×50×10−6√⇒ωr=120×10−3×50×10−6

⇒ωr=110−6√=103rad/s⇒ωr=110−6=103rad/s

Therefore, the natural frequency is 159.24Hz159.24Hz and the natural angular frequency is 103rad/s103rad/s.

## c) At what time is the energy stored (i)(i) completely electrical (i.e., stored in the capacitor)? (ii)(ii) completely magnetic (i.e., stored in the inductor)?

### Ans:

(i) Completely electrical:

It is known that,

Time period for LC oscillations, T=1νT=1ν

⇒T=1159.24=6.28ms⇒T=1159.24=6.28ms

Total charge on the capacitor at time tt, Q′=Qcos(2πTt)Q′=Qcos(2πTt)

If energy stored is electrical, Q′=±QQ′=±Q

Therefore, it can be inferred that the energy stored in the capacitor is completely electrical at time, t=0,T2,T,3T2,…………..t=0,T2,T,3T2,………….. where, T=6.3msT=6.3ms.

(ii) Completely magnetic:

Magnetic energy is maximum, when electrical energy Q′Q′ is equal to 00.

Therefore, it can be inferred that the energy stored is completely magnetic at time, t=T4,3T4,5T4,………..t=T4,3T4,5T4,……….. where, T=6.3msT=6.3ms.

## d) At what times is the total energy shared equally between the inductor and the capacitor?

### Ans: Consider, Q′Q′ be the charge on capacitor when total energy is equally shared between the capacitor and the inductor at time tt.

When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor=1/2(=1/2(maximum energy)).

⇒12(Q′)2C=12(12Q2C)⇒12(Q′)2C=12(12Q2C)

⇒12(Q′)2C=14Q2C⇒12(Q′)2C=14Q2C

⇒Q′=Q2√⇒Q′=Q2

It is known that, Q′=Qcos2πTtQ′=Qcos2πTt

⇒Q2√=Qcos2πTt⇒Q2=Qcos2πTt

⇒cos2πTt=12√=cos(2n+1)π4;⇒cos2πTt=12=cos(2n+1)π4; n=0,1,2,3…..n=0,1,2,3…..

⇒t=(2n+1)T8⇒t=(2n+1)T8

Therefore, the total energy is equally shared between the inductor and the capacitor at the time,t=T8,3T8,5T8………..t=T8,3T8,5T8…………

## e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

### Ans: If a resistor is included in the circuit, then the total initial energy gets dissipated as heat energy in the circuit. The LC oscillation gets damped due to the resistance.

## 13. A coil of inductance 0.5H0.5H and resistance 100Ω100Ω is connected to a 240V,50Hz240V,50Hz ac supply.

a) What is the maximum current in the coil?

### Ans: It is given that,

Inductance of the inductor, L=0.5HL=0.5H

Resistance of the resistor, R=100ΩR=100Ω

Potential of the supply voltage, V=240VV=240V

Frequency of the supply, ν=50Hzν=50Hz

It is known that,

Peak voltage, V0=2–√VV0=2V

⇒V0=2–√×240⇒V0=2×240

⇒V0=339.41V⇒V0=339.41V

Angular frequency of the supply, ω=2πνω=2πν

⇒ω=2π×50=100πrad/s⇒ω=2π×50=100πrad/s

Maximum current in the circuit, I0=V0R2+ω2L2√I0=V0R2+ω2L2

⇒I0=339.41(100)2+(100π)2(0.50)2√=1.82A⇒I0=339.41(100)2+(100π)2(0.50)2=1.82A

Therefore, the maximum current in the coil is 1.82A1.82A.

## b) What is the time lag between the voltage maximum and the current maximum?

### Ans: It is known that,

Equation for voltage, V=V0cosωtV=V0cosωt

Equation for current, I=I0cos(ωt−ϕ)I=I0cos(ωt−ϕ)

Where,

ϕϕ is the phase difference between voltage and current.

At time t=0t=0, V=V0V=V0 (voltage is maximum)

If ωt−ϕ=0ωt−ϕ=0 i.e., at t=ϕωt=ϕω , I=I0I=I0 (current is maximum)

Therefore, the time lag between maximum voltage and maximum current is ϕωϕω .

⇒tanϕ=ωLR⇒tanϕ=ωLR

⇒tanϕ=2π×50×0.5100=1.57⇒tanϕ=2π×50×0.5100=1.57

⇒ϕ=tan−1(1.57)⇒ϕ=tan−1(1.57)

⇒ϕ=57.5∘=57.5π180rad⇒ϕ=57.5∘=57.5π180rad

Time lag, t=ϕωt=ϕω

⇒t=57.5π180×2π×50⇒t=57.5π180×2π×50

⇒t=3.19×10−3s⇒t=3.19×10−3s

⇒t=3.2ms⇒t=3.2ms

Therefore, the time lag between the maximum voltage and maximum current is 3.2ms3.2ms.

## 14. Obtain the answers (a)(a) to (b)(b) in Exercise 1313 if the circuit is connected to a high frequency supply (240V,10kHz)(240V,10kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

### Ans: It is given that,

Inductance of the inductor, L=0.5HL=0.5H

Resistance of the resistor, R=100ΩR=100Ω

Potential of the supply voltage, V=240VV=240V

Frequency of the supply, ν=10kHz=104Hzν=10kHz=104Hz

Angular frequency, ω=2πν=2π×104rad/sω=2πν=2π×104rad/s

Peak Voltage, V0=V2–√=1102–√VV0=V2=1102V

1. Maximum current, I0=V0R2+ω2C2√I0=V0R2+ω2C2

⇒I0=2402√(100)2+(2π×104)2×(0.5)2√=1.1×10−2A⇒I0=2402(100)2+(2π×104)2×(0.5)2=1.1×10−2A

Therefore, the maximum current in the coil is 1.1×10−2A1.1×10−2A.

2. The time lag between maximum voltage and maximum current is ϕωϕω .

For phase difference ϕϕ:tanϕ=ωLRtanϕ=ωLR

⇒tanϕ=2π×104×0.5100=100π⇒tanϕ=2π×104×0.5100=100π

⇒ϕ=tan−1(100π)⇒ϕ=tan−1(100π)

⇒ϕ=89.82∘=89.82π180rad⇒ϕ=89.82∘=89.82π180rad

Time lag, t=ϕωt=ϕω

⇒t=89.82π180×2π×104⇒t=89.82π180×2π×104

⇒t=25×10−6s⇒t=25×10−6s

⇒t=25μs⇒t=25μs

Therefore, the time lag between the maximum voltage and maximum current is 25μs25μs.

It can be observed that I0I0 is very small in this case.

Thus, at high frequencies, the inductor amounts to an open circuit.

In a dc circuit, after a steady state is achieved, ω=0ω=0. Thus, inductor LL behaves like a pure conducting object.

## 15. A 100μF100μF capacitor in series with a 40Ω40Ω resistance is connected to a 110V,60Hz110V,60Hz supply.

a) What is the maximum current in the circuit?

### Ans: It is given that,

Capacitance of the capacitor, C=100μF=100×10−6FC=100μF=100×10−6F

Resistance of the resistor, R=40ΩR=40Ω

Supply voltage, V=110VV=110V

Frequency oscillations, ν=60Hzν=60Hz

Angular frequency, ω=2πν=2π×60rad/sω=2πν=2π×60rad/s

It is known that,

For a RC circuit, Impedance: Z=R2+1ω2C2−−−−−−−−√Z=R2+1ω2C2

Peak Voltage, V0=V2–√=1102–√VV0=V2=1102V

Maximum current; I0=V0ZI0=V0Z

⇒I0=V0R2+1ω2C2√⇒I0=V0R2+1ω2C2

⇒I0=1102√(40)2+1(120π)2(10−4)2√⇒I0=1102(40)2+1(120π)2(10−4)2

⇒I0=1102–√1600+1(120π)2(10−4)2−−−−−−−−−−−−−−−√=3.24A⇒I0=11021600+1(120π)2(10−4)2=3.24A

Therefore, the maximum current in the circuit is 3.24A3.24A.

## b) What is the time lag between the current maximum and the voltage maximum?

### Ans: It is known that,

In a capacitor circuit, the voltage lags behind the current by a phase angle of ϕϕ.

tanϕ=1ωCR=1ωCRtanϕ=1ωCR=1ωCR

⇒tanϕ=1120π×10−4×40=0.6635⇒tanϕ=1120π×10−4×40=0.6635

⇒ϕ=tan−1(0.6635)⇒ϕ=tan−1(0.6635)

⇒ϕ=33.56∘=33.56π180rad⇒ϕ=33.56∘=33.56π180rad

It is known that,

Time lag, t=ϕωt=ϕω

⇒t=33.56π180×120π⇒t=33.56π180×120π

⇒t=1.55×10−3s⇒t=1.55×10−3s

⇒t=1.55ms⇒t=1.55ms

Therefore, the time lag between maximum current and maximum voltage is 1.55ms1.55ms.

## 16. Obtain the answers to (a)(a) and (b)(b) in Exercise 1515 if the circuit is connected to a

110V,12kHz110V,12kHz

supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.

### Ans: It is given that,

Capacitance of the capacitor, C=100μF=100×10−6FC=100μF=100×10−6F

Resistance of the resistor, R=40ΩR=40Ω

Supply voltage, V=110VV=110V

Frequency oscillations, ν=12kHz=12×103Hzν=12kHz=12×103Hz

Angular frequency, ω=2πν=2π×12×103rad/s=24π×103rad/sω=2πν=2π×12×103rad/s=24π×103rad/s

Peak Voltage, V0=V2–√=1102–√VV0=V2=1102V

1. It is known that,

For a RC circuit, Impedance: Z=R2+1ω2C2−−−−−−−−√Z=R2+1ω2C2

Maximum current; I0=V0ZI0=V0Z

⇒I0=V0R2+1ω2C2√⇒I0=V0R2+1ω2C2

⇒I0=1102√(40)2+1(24π×103)2(10−4)2√⇒I0=1102(40)2+1(24π×103)2(10−4)2

⇒I0=1102–√1600+(1024π)2−−−−−−−−−−−√=3.9A⇒I0=11021600+(1024π)2=3.9A

Therefore, the maximum current in the circuit is 3.9A3.9A.

2. It is known that,

In a capacitor circuit, the voltage lags behind the current by a phase angle of ϕϕ.

tanϕ=1ωCR=1ωCRtanϕ=1ωCR=1ωCR

⇒tanϕ=124π×103×10−4×40=196π⇒tanϕ=124π×103×10−4×40=196π

⇒ϕ=tan−1(196π)⇒ϕ=tan−1(196π)

⇒ϕ=0.2∘=0.2π180rad⇒ϕ=0.2∘=0.2π180rad

It is known that,

Time lag, t=ϕωt=ϕω

⇒t=0.2π180×24π×103⇒t=0.2π180×24π×103

⇒t=0.04×10−6s⇒t=0.04×10−6s

⇒t=0.04μs⇒t=0.04μs

Therefore, the time lag between maximum current and maximum voltage is 0.04μs0.04μs.

It can be concluded that ϕϕ tends to become zero at high frequencies. At a high frequency, capacitor CC acts as a conductor.

In a dc circuit, after the steady state is achieved, ω=0ω=0. Therefore, capacitor CC amounts to an open circuit.

## 17. Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if three elements, L,C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 11 for this frequency.

### Ans: It is given that,

An inductor (L)(L), a capacitor (C)(C) and a resistor (R)(R) is connected in parallel with each other in a circuit where,

Inductance, L=5.0HL=5.0H

Capacitance, C=80μF=80×10−6FC=80μF=80×10−6F

Resistance, R=40ΩR=40Ω

Potential of the voltage source, V=230VV=230V

It is known that,

Impedance (Z)(Z) of the given LCR circuit is given as:

1Z=1R2+(1ωL−ωC)2−−−−−−−−−−−−−−√1Z=1R2+(1ωL−ωC)2

Where,

ωω is the angular frequency

At resonance: 1ωL−ωC=01ωL−ωC=0

⇒ω=1LC√⇒ω=1LC

⇒ω=15×80×10−6√=50rad/s⇒ω=15×80×10−6=50rad/s

Therefore, the magnitude of ZZ is maximum at 50rad/s50rad/s and the total current is minimum.

Rms current flowing through inductor LL: IL=VωLIL=VωL

⇒IL=23050×5=0.92A⇒IL=23050×5=0.92A

Rms current flowing through capacitor CC: IC=V1ωC=ωCVIC=V1ωC=ωCV

⇒IC=50×80×10−6×230=0.92A⇒IC=50×80×10−6×230=0.92A

Rms current flowing through resistor RR: IR=VRIR=VR

⇒IR=23040=5.75A⇒IR=23040=5.75A

The current RMS value in the inductor is 0.92A0.92A, in the capacitor is 0.92A0.92A and in the resistor is 5.75A5.75A.

## 18. A circuit containing an 80mH80mH inductor and a 60μF60μF capacitor in series is connected to a

230V,50Hz230V,50Hz

supply. The resistance of the circuit is negligible.

a) Obtain the current amplitude and rms values.

### Ans: It is given that,

Inductance, L=80mH=80×10−3HL=80mH=80×10−3H

Capacitance, C=60μF=60×10−6FC=60μF=60×10−6F

Supply voltage, V=230VV=230V

Frequency, ν=50Hzν=50Hz

Angular frequency, ω=2πν=100πrad/sω=2πν=100πrad/s

Peak voltage, V0=V2–√=2302–√VV0=V2=2302V

It is known that,

Maximum current: I0=V0(ωL−1ωC)I0=V0(ωL−1ωC)

⇒I0=2303–√(100π×80×10−3−1100π×60×10−6)⇒I0=2303(100π×80×10−3−1100π×60×10−6)

⇒I0=2303–√(8π−10006π)=−11.63A⇒I0=2303(8π−10006π)=−11.63A

The negative sign is because ωL<1ωCωL<1ωC

Amplitude of maximum current, |I0|=11.63A|I0|=11.63A

⇒I=I02√=−11.632√⇒I=I02=−11.632

⇒I=−8.22A⇒I=−8.22A, which is the rms value of current.

## b) Obtain the rms values of potential drops across each element.

### Ans: It is known that,

Potential difference across the inductor, VL=I×ωLVL=I×ωL

⇒VL=8.22×100π×80×10−3⇒VL=8.22×100π×80×10−3

⇒VL=206.61V⇒VL=206.61V

Potential difference across the capacitor, VC=I×1ωCVC=I×1ωC

⇒VC=8.22×1100π×60×10−6⇒VC=8.22×1100π×60×10−6

⇒VC=436.3V⇒VC=436.3V, which is the rms value of potential drop.

## c) What is the average power transferred to the inductor?

### Ans: Average power transferred to the inductor is zero as actual voltage leads the current by π2π2.

## d) What is the average power transferred to the capacitor?

### Ans: Average power transferred to the capacitor is zero as actual voltage lags the current by π2π2.

## e) What is the total average power absorbed by the circuit? (‘Average’ implies ‘averaged over one cycle’.)

### Ans: The total average power absorbed (averaged over one cycle) is zero.

## 19. Suppose the circuit in Exercise 1818 has a resistance of 15Ω15Ω . Obtain the average power transferred to each element of the circuit, and the total power absorbed.

### Ans: It is given that,

Average power transferred to the resistor

=788.44W=788.44W

Average power transferred to the capacitor =0W=0W

Total power absorbed by the circuit

=788.44W=788.44W

Inductance of inductor,

L=80mH=80×10−3HL=80mH=80×10−3H

Capacitance of capacitor,

C=60μF=60×10−6FC=60μF=60×10−6F

Resistance of resistor,

R=15ΩR=15Ω

Potential of voltage supply,

V=230VV=230V

Frequency of signal,

ν=50Hzν=50Hz

Angular frequency of signal,

ω=2πν=2π×(50)=100πrad/sω=2πν=2π×(50)=100πrad/s

It is known that,

Impedance, Z=R2+(ωL−1ωC)2−−−−−−−−−−−−−−√Z=R2+(ωL−1ωC)2

⇒Z=(15)2+(100π(80×10−3)−1(100π×60×10−6))2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√⇒Z=(15)2+(100π(80×10−3)−1(100π×60×10−6))2

⇒Z=(15)2+(25.12−53.08)2−−−−−−−−−−−−−−−−−−−√=31.728Ω⇒Z=(15)2+(25.12−53.08)2=31.728Ω

Now,

I=VZI=VZ

⇒I=23031.728=7.25A⇒I=23031.728=7.25A

The elements are connected in series to each other. Therefore, impedance of the circuit is given as current flowing in the circuit,

Average power transferred to resistance is given as: PR=I2RPR=I2R

⇒PR=(7.25)2×15=788.44W⇒PR=(7.25)2×15=788.44W

Average power transferred to capacitor,PC=PC= Average power transferred to inductor, PL=0PL=0

Total power absorbed by the circuit: PT=PR+PC+PLPT=PR+PC+PL

PT=788.44+0+0=788.44WPT=788.44+0+0=788.44W

Therefore, the total power absorbed by the circuit is

788.44W788.44W

.

## 20. A series LCR circuit with

L=0.12H,C=480nF,R=23 ΩL=0.12H,C=480nF,R=23 Ω

is connected to a 230V230V variable frequency supply.

a) What is the source frequency for which current amplitude is maximum? Obtain this maximum value.

### Ans: It is given that,

Inductance, L=0.12HL=0.12H

Capacitance, C=480nF=480×10−9FC=480nF=480×10−9F

Resistance, R=23ΩR=23Ω

Supply voltage, V=230VV=230V

Peak voltage, V0=230×2–√=325.22VV0=230×2=325.22V

It is known that,

Current flowing in the circuit, I0=V0R2+(ωL−1ωC)2√I0=V0R2+(ωL−1ωC)2

Where,

I0I0 is maximum at resonance.

At resonance: ωRL−1ωRC=0ωRL−1ωRC=0

Where,

ωRωRis the resonance angular frequency

ωR=1LC√ωR=1LC

⇒ωR=10.12×480×10−9√⇒ωR=10.12×480×10−9

⇒ωR=4166.67rad/s⇒ωR=4166.67rad/s

Resonant frequency, νR=ωR2πνR=ωR2π

⇒νR=4166.672×3.14=663.48Hz⇒νR=4166.672×3.14=663.48Hz

Maximum current, (I0)Max=V0R(I0)Max=V0R

⇒(I0)Max=325.2223=14.14A⇒(I0)Max=325.2223=14.14A

## b) What is the source frequency for which average power absorbed by the circuit is maximum? Obtain the value of this maximum power.

### Ans: It is known that,

Maximum average power absorbed by the circuit; (PV)Max=12(I0)2MaxR(PV)Max=12(I0)Max2R

⇒(PV)Max=12×(14.14)2×23⇒(PV)Max=12×(14.14)2×23

⇒(PV)Max=2299.3W⇒(PV)Max=2299.3W

Therefore, the resonant frequency, νR=663.48HzνR=663.48Hz

## c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?

### Ans: It is known that,

The power transferred to the circuit is half the power at resonant frequency.

Frequencies at which power transferred is half, =ωR±Δω=2π(νR±Δν)=ωR±Δω=2π(νR±Δν)

Where,

Δω=R2LΔω=R2L

⇒Δω=232×0.12=95.83rad/s⇒Δω=232×0.12=95.83rad/s

Therefore, the change in frequency, Δν=12πΔωΔν=12πΔω

Δν=95.832π=15.26HzΔν=95.832π=15.26Hz

νR+Δν=663.48+15.26=678.74HzνR+Δν=663.48+15.26=678.74Hz

νR−Δν=663.48−15.26=648.22HzνR−Δν=663.48−15.26=648.22Hz

Therefore, at 648.22Hz648.22Hz and 678.74Hz678.74Hz frequencies, the power transferred is half.

At these frequencies, current amplitude:I′=12√×(I0)MaxI′=12×(I0)Max

⇒I′=14.142√=10A⇒I′=14.142=10A

Therefore, the current amplitude is 10A10A.

## d) What is the Q-factor of the given circuit?

### Ans: It is known that,

Q-factor of the given circuit, Q=ωrLRQ=ωrLR

⇒Q=4166.67×0.1223=21.74⇒Q=4166.67×0.1223=21.74

Therefore, the Q-factor of the given circuit is 21.7421.74.

## 21. Obtain the resonant frequency and Q-factor of a series LCR circuit with L=3.0HL=3.0H ,C=27μFC=27μF and R=7.4ΩR=7.4Ω. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 22. Suggest a suitable way.

### Ans: It is given that,

Inductance, L=3.0HL=3.0H

Capacitance, C=27μF=27×10−6FC=27μF=27×10−6F

Resistance,R=7.4ΩR=7.4Ω

It is known that,

At resonance, angular frequency of the source for the given LCR series circuit is ωr=1LC√ωr=1LC

⇒ωr=13×27×10−6√⇒ωr=13×27×10−6

⇒ωr=1039=111.11rad/s⇒ωr=1039=111.11rad/s

Therefore, the resonant frequency is 111.11rad/s111.11rad/s.

Q-factor of the series, Q=ωrLRQ=ωrLR

⇒Q=111.11×37.4=45.0446⇒Q=111.11×37.4=45.0446

Therefore, the Q-factor is 45.044645.0446.

To improve the sharpness of the resonance by reducing ‘full width at half maximum’ by a factor of 22without changing ωrωr , reduce the resistance to half.

⇒R=7.42=3.7Ω⇒R=7.42=3.7Ω

Therefore, required resistance is 3.7Ω3.7Ω.

## 22. Answer the following questions:

a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?

### Ans: Yes, in any ac circuit, the applied instantaneous voltage is equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit.

The same is not true for rms voltage because voltages across different elements may not be in phase.

## b) A capacitor is used in the primary circuit of an induction coil.

### Ans: Yes, a capacitor is used in the primary circuit of an induction coil.

This is because, when the circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparks.

## c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.

### Ans: The dc signal will appear across capacitor CC because for dc signals, the impedance of an inductor LL is negligible while the impedance of a capacitor CC is very high (almost infinite).

Therefore, a dc signal appears acrossCC.

For an ac signal of high frequency, the impedance of LLis high and that of CC is very low.

Thus, an ac signal of high frequency appears across LL.

## d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.

### Ans: When an iron core is inserted in the choke coil (which is in series with a lamp connected to an ac line), the lamp will glow dimly.

This is because the choke coil and the iron core increase the impedance of the circuit.

## e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?

### Ans: As the choke coil reduces the voltage across the tube without wasting much power, it is used in the fluorescent tubes with ac mains. An ordinary resistor cannot be used instead of choke coil because it wastes power in the form of heat.

## 23. A power transmission line feeds input power at 2300V2300V to a step-down transformer with its primary windings having 40004000 turns. What should be the number of turns in the secondary in order to get output power at 230V230V?

### Ans: It is given that,

Input voltage, V1=2300VV1=2300V

Number of turns in primary coil, n1=4000n1=4000

Output voltage, V2=230VV2=230V

Number of turns in secondary coil ,n2=?n2=?

It is known that,

Voltage is related to number of terms: V1V2=n1n2V1V2=n1n2

⇒2300230=4000n2⇒2300230=4000n2

⇒n2=4000×2302300=400⇒n2=4000×2302300=400

Therefore, the number of turns in the second winding is 400400.

## 24. At a hydroelectric power plant, the water pressure head is at a height of 300m300m and the water flow available is 100m3/s100m3/s . If the turbine generator efficiency is 6060 , estimate the electric power available from the plant (g=9.8m/s2g=9.8m/s2 ).

### Ans: It is known that,

Height of water pressure head,

h=300mh=300m

Volume of water flow per second, V=100m3/sV=100m3/s

Efficiency of turbine generator,

n=60n=60

Acceleration due to gravity, g=9.8m/s2g=9.8m/s2

Density of water,

ρ=103kg/m3ρ=103kg/m3

It is known that,

Electric power available from the plant=η×hρgV=η×hρgV

⇒P=0.6×300×103×9.8×100⇒P=0.6×300×103×9.8×100

⇒P=176.4×106W⇒P=176.4×106W

⇒P=176.4MW⇒P=176.4MW

Therefore, the estimated electric power available from the plant is 176.4MW176.4MW.

## 25. A small town with a demand of 800kW800kW of electric power at 220V220V is situated 15km15km away from an electric plant generating power at 440V440V . The resistance of the two-wire line carrying power is 0.5Ωperkm0.5Ωperkm. The town gets power from the line through a 400−220V400−220V step-down transformer at a substation in the town.

a) Estimate the line power loss in the form of heat.

### Ans: It is given that,

Total electric power required,

P=800kW=800×103WP=800kW=800×103W

Supply voltage,

V=220VV=220V

Voltage at which electric plant is generating power, V′=440VV′=440V

Distance between the town and power generating station, d=15kmd=15km

Resistance of the two wire lines carrying power=0.5Ω/km=0.5Ω/km

Total resistance of the wires,

R=(15+15)0.5=15 ΩR=(15+15)0.5=15 Ω

A step-down transformer of rating

4000−220V4000−220V

is used in the sub-station.

Input voltage, V1=4000VV1=4000V

Output voltage, V2=220VV2=220V

It is known that,

Rms current in the wire lines: I=PV1I=PV1

⇒I=800×1034000=200A⇒I=800×1034000=200A

Line power loss=I2R=I2R

⇒(200)2×15⇒(200)2×15

⇒600×103W=600kW⇒600×103W=600kW

Therefore, the line power loss is 600kW600kW.

## b) How much power must the plant supply, assuming there is negligible power loss due to leakage?

### Ans: Assuming that there is negligible power loss due to leakage of the current:

Total power supplied by the plant=800kW+600kW=1400kW=800kW+600kW=1400kW

Therefore, the plant must supply 1400kW1400kW of power.

## c) Characterise the step up transformer at the plant.

### Ans: It is known that,

Voltage drop in the power line=IR=IR

⇒V=200×15=3000V⇒V=200×15=3000V

Total voltage transmitted from the plant=3000+4000=7000V=3000+4000=7000V

The power generated is 440V440V.

Therefore, the rating of the step-up transformer situated at the power plant is 440V−7000V440V−7000V.

## 26.Do the same exercise as above with the replacement of the earlier transformer by a 40,000−220V40,000−220V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?

### Ans: It is given that,

Total electric power required,

P=800kW=800×103WP=800kW=800×103W

Supply voltage,

V=220VV=220V

Voltage at which electric plant is generating power, V′=440VV′=440V

Distance between the town and power generating station, d=15kmd=15km

Resistance of the two wire lines carrying power=0.5Ω/km=0.5Ω/km

Total resistance of the wires,

R=(15+15)0.5=15 ΩR=(15+15)0.5=15 Ω

The rating of a step-down transformer is

40000V−220V40000V−220V

.

Input voltage, V1=40000VV1=40000V

Output voltage, V2=220VV2=220V

1. It is known that,

Rms current in the wire lines: I=PV1I=PV1

⇒I=800×10340000=20A⇒I=800×10340000=20A

Line power loss=I2R=I2R

⇒(20)2×15⇒(20)2×15

⇒6×103W=6kW⇒6×103W=6kW

Therefore, the line power loss is 6kW6kW.

2. Assume that there is negligible power loss due to leakage of the current:

Total power supplied by the plant=800kW+6kW=806kW=800kW+6kW=806kW

Therefore, the plant must supply 806kW806kW of power.

3. It is known that,

Voltage drop in the power line=IR=IR

⇒V=20×15=300V⇒V=20×15=300V

Total voltage transmitted from the plant=300+40000=40300V=300+40000=40300V

The power generated in the plant is generated at 440V440V.

Therefore, the rating of the step-up transformer situated at the power plant is 440V−40300V440V−40300V.

Power loss during transmission=6001400×100=42.8=6001400×100=42.8

In previous exercise the power loss due to the same reason is =6806×100=0.744=6806×100=0.744

As the power loss is less for a high voltage transmission, High voltage transmissions are preferred for this purpose.