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 42P

(a)

To determine

The average power delivered to each resistor if the driving frequency is 100 Hz.

(a)

Expert Solution
Check Mark

Answer to Problem 42P

The required average powers are P1=45.6W and P2=55.6W .

Explanation of Solution

Given:

Ideal AC voltage source is ε1=(20V)cos(2πft)

An ideal battery (whose EMF is ε2 ) is ε2=16V

Resistances of two resistor are R1=10Ω and R2=8Ω

Inductance of inductor is L=6mH

Driving frequency is f=100Hz

Formula used:

Total power delivered to R1 and R2 is expressed as

P1=P1,dc+P1,ac…… (1)

And

P2=P2,dc+P2,ac…… (2)

Here, Pdc and Pdc are dc and ac power respectively

Expression for power is

P=ε2R…… (3)

Here, R and ε are resistance and voltage.

Calculation:

Dc powers delivered to the resistor whose resistances are R1andR2 are

P1,dc=ε22R1….. (4)

And

P2,dc=ε22R2…… (5)

Average AC power is

P1,dc=ε21,rmsR1=ε21,peak2R1

P1,dc=ε21,peak2R1…… (6)

Apply Kirchhoff’s loop law,

R1I1Z2I2=0…… (7)

Solve equation (7)

I2=R1Z2I1I2=R1ε 1,peakZ2R1

I2=ε1,peakZ2…… (8)

Average ac power delivered is calculated as

P2,ac=12I22,rmsR2P2,ac=12( ε 1,peak Z 2 )2R2

P2,ac=ε21,peakR22Z22…… (9)

Substitute ε22R1 for P1,dc , ε22R2 for P2,dc , ε21,peak2R1 for P1,dc , ε21,peakR22Z22 for P2,ac in equation (1) and (2)

P1=ε22R1+ε21,peak2R1…… (10)

And

P2=ε22R2+ε21,peakR22Z22…… (11)

Substitute 16 V for ε2 ,20 V for ε1,peak , 10Ω for R1 , 8Ω for R2 , 100 Hz for f and 6mH for L ;in equation (10) and (11)

P1= 16210+ 2022×10P1=45.6W

And

P2= 1628+ 202×82×[ 8 2+2×π×100×6× 10 3]P2=55.6W

Conclusion:

Hence, the requiredaverage powers are P1=45.6W and P2=55.6W .

(b)

To determine

The average power delivered to each resistor if the driving frequency is 200 Hz.

(b)

Expert Solution
Check Mark

Answer to Problem 42P

The required average power are P1=45.6W and P2=54.37W .

Explanation of Solution

Given:

Ideal AC voltage source is ε1=(20V)cos(2πft)

An ideal battery (whose EMF is ε2 ) is ε2=16V

Resistances of two resistor are R1=10Ω and R2=8Ω

Inductance of inductor is L=6mH

Driving frequency is f=200Hz

Formula used:

Total power delivered to R1 and R2 is expressed as

P1=P1,dc+P1,ac…… (1)

And

P2=P2,dc+P2,ac…… (2)

Here, Pdc and Pdc are dc and ac power respectively

Expression for power is

P=ε2R…… (3)

Here, R and ε are resistance and voltage.

Calculation:

Dc powers delivered to the resistor whose resistances are R1andR2 are

P1,dc=ε22R1….. (4)

And

P2,dc=ε22R2…… (5)

Average AC power is

P1,dc=ε21,rmsR1=ε21,peak2R1

P1,dc=ε21,peak2R1…… (6)

Apply Kirchhoff’s loop law,

R1I1Z2I2=0…… (7)

Solve equation (7)

I2=R1Z2I1I2=R1ε 1,peakZ2R1

I2=ε1,peakZ2…… (8)

Average ac power delivered is calculated as

P2,ac=12I22,rmsR2P2,ac=12( ε 1,peak Z 2 )2R2

P2,ac=ε21,peakR22Z22…… (9)

Substitute ε22R1 for P1,dc , ε22R2 for P2,dc , ε21,peak2R1 for P1,dc , ε21,peakR22Z22 for P2,ac in equation (1) and (2)

P1=ε22R1+ε21,peak2R1…… (10)

And

P2=ε22R2+ε21,peakR22Z22…… (11)

Substitute 16 V for ε2 , 20 V for ε1,peak , 10Ω for R1 , 8Ω for R2 , 200 Hz for f and 6mH for L in equation (10) and (11)

P1= 16210+ 2022×10P1=45.6W

And

P2= 1628+ 202×82×[ 8 2+2×π×200×6× 10 3]P2=54.37W

Conclusion:

Hence, the required average powers are P1=45.6W and P2=54.37W .

(c)

To determine

The average power delivered to each resistor if the driving frequency is 800 Hz.

(c)

Expert Solution
Check Mark

Answer to Problem 42P

The required average powerare P1=45.6W and P2=48.99W .

Explanation of Solution

Given:

Ideal AC voltage source is ε1=(20V)cos(2πft)

An ideal battery (whose EMF is ε2 ) is ε2=16V

Resistances of two resistor are R1=10Ω and R2=8Ω

Inductance of inductor is L=6mH

Driving frequency is f=800Hz

Formula used:

Total power delivered to R1 and R2 is expressed as

P1=P1,dc+P1,ac…… (1)

And

P2=P2,dc+P2,ac…… (2)

Here, Pdc and Pdc are dc and ac power respectively

Expression for power is

P=ε2R…… (3)

Here, R and ε are resistance and voltage.

Calculation:

Dc powers delivered to the resistor whose resistances are R1andR2 are

P1,dc=ε22R1….. (4)

And

P2,dc=ε22R2…… (5)

Average AC power is

P1,dc=ε21,rmsR1=ε21,peak2R1

P1,dc=ε21,peak2R1…… (6)

Apply Kirchhoff’s loop law,

R1I1Z2I2=0…… (7)

Solve equation (7)

I2=R1Z2I1I2=R1ε 1,peakZ2R1

I2=ε1,peakZ2…… (8)

Average ac power delivered is calculated as

P2,ac=12I22,rmsR2P2,ac=12( ε 1,peak Z 2 )2R2

P2,ac=ε21,peakR22Z22…… (9)

Substitute ε22R1 for P1,dc , ε22R2 for P2,dc , ε21,peak2R1 for P1,dc , ε21,peakR22Z22 for P2,ac in equation (1) and (2)

P1=ε22R1+ε21,peak2R1…… (10)

And

P2=ε22R2+ε21,peakR22Z22…… (11)

Substitute 16 V for ε2 , 20 V for ε1,peak , 10Ω for R1 , 8Ω for R2 , 800 Hz for f and 6mH for L in equation (10) and (11)

P1= 16210+ 2022×10P1=45.6W

And

P2= 1628+ 202×82×[ 8 2+2×π×800×6× 10 3]P2=48.99W

Conclusion:

Hence, the required average powers are P1=45.6W and P2=48.99W .

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