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 38P

(a)

To determine

The peak current in the circuit.

(a)

Expert Solution
Check Mark

Answer to Problem 38P

The required peak current is 7.94 A.

Explanation of Solution

`Given:

Inductance of inductor is L=36mH

Resistance of resistor is R=40Ω

Ideal voltage output is ε=(345V)×cos(150πt)

Formula used:

Expression for peak current is

  εIZ=0I=εZ

  I=εR2+ ( ωL )2.......... (1)

Here, ε,R,Landω are voltage, resistance, angular frequency and inductance of inductor.

Calculation:

Substitute (345V)×cos(150πt) for ε , 40Ω for R , 150π for ω , 36mH for L in equation (1)

  I=345×cos( 150π×t) ( 40Ω ) 2 + ( 150π×36× 10 3 ) 2 I=(7.94A)×cos150πt

Conclusion:

Hence, the requiredpeak current is 7.94 A.

(b)

To determine

The peak and RMS voltage across the inductor.

(b)

Expert Solution
Check Mark

Answer to Problem 38P

The required peak voltage and RMS voltage are Vpeak=345V and Vrms=244V .

Explanation of Solution

Given:

Inductance of inductor is L=36mH

Resistance of resistor is R=40Ω

Ideal voltage output is ε=(345V)×cos(150πt)

Formula used:

RMS current can be obtained as

  Vrms=Vpeak2.......... (1)

Here, Vpeak is the peak voltage.

Calculation:

Expression for output voltage is

  ε=(345V)×cos(150πt)

From above expression, peak voltage is found to be 345V i.e. Vpeak=345V .

Substitute 345V for Vpeak in equation (1)

  Vrms=3452V rms=244V

Conclusion:

Hence, the required peak voltage and RMS voltage are Vpeak=345V and Vrms=244V .

(c)

To determine

The average power dissipated.

(c)

Expert Solution
Check Mark

Answer to Problem 38P

  Pavg=1.3kW

Explanation of Solution

Given:

Inductance of inductor is L=36mH

Resistance of resistor is R=40Ω

Ideal voltage output is ε=(345V)×cos(150πt)

Peak current is Ipeak=7.94A

Formula used:

Average power can be obtained as

  Pavg=I2rms×RPavg=( I peak 2 )2×R

  Pavg=12I2peak×R.......... (1)

Here, Ipeak is the peak current and R is resistance.

Calculation:

Substitute 7.94A for Ipeak and 40Ω for R in equation (1)

  Pavg=12×(7.94A)2×(40Ω)P avg=1.3kW

Conclusion:

Hence, the required average power is Pavg=1.3kW.

(d)

To determine

The peak and average magnetic energy stored in inductor.

(d)

Expert Solution
Check Mark

Answer to Problem 38P

  Upeak=1.1J and Uavg=0.57J

Explanation of Solution

Given:

Inductance of inductor is L=36mH

Resistance of resistor is R=40Ω

Ideal voltage output is ε=(345V)×cos(150πt)

  Ipeak=7.94A

Formula used:

Expression for peak magnetic energy stored in inductor is

  Upeak=12LI2peak.......... (1)

Expression for average magnetic energy stored in inductor is

  Uavg=14LI2peak.......... (2)

Here, L is inductance of inductor and Ipeak is the peak current.

Calculation:

Substitute 36×103H for L and 7.94A for Ipeak in equation (1)

  Upeak=12×(36× 10 3)×(7.94)2U peak=1.1J

Average energy is calculated as

  Uavg=14×(36× 10 3)×(7.94)2U avg=0.57J

Conclusion:

Hence, the required peak and average magnetic energy is Upeak=1.1J and Uavg=0.57J

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