Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 29, Problem 41P

(a)

To determine

The RMS current in all branch of the circuit if frequency is 500 Hz.

(a)

Expert Solution
Check Mark

Answer to Problem 41P

Irms=7.07A , IR,rms=3.17A and IRL,rms=6.31A .

Explanation of Solution

Given:

Load resistance is RL=20Ω

Inductance of inductor is L=3.20mH

Resistance of resistor is R=4Ω

Frequency is f=500Hz

Ideal voltage output is ε=(100V)×cos(2πft)

Formula used:

Phase angle is expressed as

δ=tan1(RL2πfL)…… (1)

Here, RL,f,L are load resistance, frequency and inductance of inductor.

Resultant reactance for parallel combination is is

Z=1R 2L+X 2L…… (2)

Here, XL is inductance of inductor.

Peak current is

Ipeak=εpeakZ…… (3)

Calculation:

Substitute 20Ω for RL , 500 Hz for f , 3.20 mH for L in equation (1)

δ=tan1( 20 2π×500×3.20× 10 3 )δ=63.31°

XL=2πfLXL=2×π×500×3.20×103XL=10.05Ω

Substitute 20Ω for RL , 10.05 Ω for XL in equation (2)

Z2=1 20 2 + ( 10.05 ) 2 Z2=8.97A

Therefore, final value of Z is calculated as

Z=42+ ( 8.97 )2+2×4×cos( 63.31)Z=10Ω

Substitute 100 V for Vpeak and 10Ω for Z in equation (3)

Ipeak=10010Ipeak=10A

RMS current is

Irms=102I rms=7.07A

V2,peak=IpeakZ2=10×8.97V2,peak=89.7V

And

V2,rms=12V2,peakV2,rms=12×89.7V2,rms=63.43V

The current for resistor and inductor are

IR,rms=V 2,rmsRLIR,rms=63.43V20I R,rms=3.17A

And

IRL,rms=V 2,rmsXLIRL,rms=63.4310.05I R L ,rms=6.31A

Conclusion:

Hence, the requiredcurrent across each branch are Irms=7.07A , IR,rms=3.17A and IRL,rms=6.31A .

(b)

To determine

RMS current in all branch of the circuit if frequency is 2000 Hz.

(b)

Expert Solution
Check Mark

Answer to Problem 41P

The required current across each branch are Irms=3.84A , IR,rms=3.44A and IRL,rms=1.71A .

Explanation of Solution

Given:

Load resistance is RL=20Ω

Inductance of inductor is L=3.20mH

Resistance of resistor is R=4Ω

Frequency is f=2000Hz

Ideal voltage output is ε=(100V)×cos(2πft)

Formula used:

Phase angle is expressed as

δ=tan1(RL2πfL)…… (1)

Here, RL,f,L are load resistance, frequency and inductance of inductor.

Resultant inductance is

Z=1R 2L+X 2L…… (2)

Here, XL is inductance of inductor.

Peak current is

Ipeak=εpeakZ…… (3)

Calculation:

Substitute 20Ω for RL , 2000 Hz for f , 3.20 mH for L in equation (1)

δ=tan1( 20 2π×2000×3.20× 10 3 )δ=26.4°

XL=2πfLXL=2×π×2000×3.20×103XL=40.20Ω

Substitute 20Ω for RL , 40.20 Ω for XL in equation (2)

Z2=1 20 2 + ( 40.20 ) 2 Z2=17.9Ω

Therefore, final value of Z is calculated as

Z=42+ ( 17.9 )2+2×4×cos( 26.4)Z=18.40Ω

Substitute 100 V for Vpeak and 18.40Ω for Z in equation (3)

RMS current is

Irms=5.432I rms=3.84A

V2,peak=IpeakZ2=5.43×17.90V2,peak=97.19V

And

V2,rms=12V2,peakV2,rms=12×97.19V2,rms=68.72V

The current for resistor and inductor are

IR,rms=V 2,rmsRLIR,rms=68.72V20I R,rms=3.44A

And

IRL,rms=V 2,rmsXLIRL,rms=68.72V40.20I R L ,rms=1.71A

Conclusion:

Hence, the required current across each branch are Irms=3.84A , IR,rms=3.44A and IRL,rms=1.71A .

(c)

To determine

The fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is 500 Hz.

(c)

Expert Solution
Check Mark

Answer to Problem 41P

  P R L P R L +PR=80%

Explanation of Solution

Given:

Load resistance is RL=20Ω

Inductance of inductor is L=3.20mH

Resistance of resistor is R=4Ω

Frequency is f=500Hz

RMS current is Irms=7.07A

RMS current across RL is IRL=6.31A

Ideal voltage output is ε=(100V)×cos(2πft)

Formula used:

The fraction of the total power delivered by the source that is dissipated in load resistor can be expressed as

  PRLPRL+PR=(1+PRPR L )

PRLPRL+PR=(1+ I 2 rms I 2 R L ,rms R L )1…… (1)

Here, PRL,Irms,PR,RL and R are power dissipated across load resistor, RMS current, power dissipated across resistor, resistance across inductor and resistance of resistor respectively.

Calculation:

Substitute 500 Hz for f , 4Ω for R , 20Ω for RL , 7.07A for Irms and 6.31A for IRL in equation (1)

  P R L P R L +PR=(1+ ( 7.07 ) 2 ×4 ( 6.31 ) 2 ×20)1P R L P R L +PR=0.80 P R L P R L + P R =80%

Conclusion:

Hence, the required the fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is 500 Hz is P R L P R L +PR=80% .

(d)

To determine

The fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is 2000 Hz.

(d)

Expert Solution
Check Mark

Answer to Problem 41P

  P R L P R L +PR=50%

Explanation of Solution

Given:

Load resistance is RL=20Ω

Inductance of inductor is L=3.20mH

Resistance of resistor is R=4Ω

Frequency is f=2000Hz

RMS current is Irms=3.84A

RMS current across RL is IRL=1.71A

Ideal voltage output is ε=(100V)×cos(2πft)

Formula used:

The fraction of the total power delivered by the source that is dissipated in load resistor can be expressed as

  PRLPRL+PR=(1+PRPR L )

PRLPRL+PR=(1+ I 2 rms I 2 R L ,rms R L )1…… (1)

Here, PRL,Irms,PR,RL and R are power dissipated across load resistor, RMS current, power dissipated across resistor, resistance across inductor and resistance of resistor respectively.

Calculation:

Substitute 2000 Hz for f , 4Ω for R , 20Ω for RL , 3.84A for Irms and 1.71A for IRL in equation (1)

  P R L P R L +PR=(1+ ( 3.84 ) 2 ×4 ( 1.71 ) 2 ×20)1P R L P R L +PR=0.50 P R L P R L + P R =50%

Conclusion:

Hence, the required the fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is 2000 Hz is P R L P R L +PR=50% .

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