Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 20, Problem 50P

(a)

To determine

The currents I1, I2, the potential difference across the resistors, power of the battery and induced emf in the inductor.

(a)

Expert Solution
Check Mark

Answer to Problem 50P

The current I1, I2 is respectively 1.7mA, zero, the potential difference across the resistors is respectively 45V, zero, the power supplied by the battery is 0.075W and the potential difference across the inductor is 45V.

Explanation of Solution

Write the expression to calculate I1.

  I1=εR

Here, ε is the voltage of battery and R is the resistance along the path of I1.

Substitute 45V for ε and 27kΩ for R in the above equation to calculate I1.

  I1=45V27kΩ(103Ω1)=1.7×103A(1mA103A)=1.7mA

At the instance of closing the switch the current through the inductor is zero and therefore I2 is zero.

The voltage across the resistance 3.0 is zero since the current through the loop including this resistor, that is I2 zero.

The whole voltage of the battery would fall across the resistor 27. Thus, the voltage across the 27 is 45V.

Write the expression for the power supplied by the battery.

  P=ε2R

Here, P is the power supplied by the battery.

Substitute 45V for ε and 27kΩ for R in the above equation to calculate P.

  P=(45V)227kΩ(103Ω1)=0.075W

Since the current through the inductor is zero (I2 is zero), the voltage across the inductor is 45V.

Conclusion:

Therefore, the current I1, I2 is respectively 1.7mA, zero, the potential difference across the resistors is respectively 45V, zero, the power supplied by the battery is 0.075W and the potential difference across the inductor is 45V.

(b)

To determine

The currents I1, I2, the potential difference across the resistors, power of the battery and induced emf in the inductor after the switch has been closed for a long time.

(b)

Expert Solution
Check Mark

Answer to Problem 50P

The current I1, I2 is respectively 1.7mA, 15mA, the potential difference across the resistors is respectively 45V, 45V, the power supplied by the battery is 0.75W and the potential difference across the inductor is 0V.

Explanation of Solution

Write the expression to calculate I1.

  I1=εR

Here, ε is the voltage of battery and R is the resistance along the path of I1.

Substitute 45V for ε and 27kΩ for R in the above equation to calculate I1.

  I1=45V27kΩ(103Ω1)=1.7×103A(1mA103A)=1.7mA

Write the expression to calculate I2.

  I2=εR

Here, ε is the voltage of battery and R is the resistance along the path of I2.

Substitute 45V for ε and 3.0kΩ for R in the above equation to calculate I2.

  I2=45V3.0kΩ(103Ω1)=15×103A(1mA103A)=15mA

The resistors come into parallel and hence the voltage across the two resistors is 45V.

Write the expression for the power supplied by the battery.

  P=ε2Req

Here, P is the power supplied by the battery and Req is the equivalent resistor.

Write the expression to calculate Req.

  Req=RRR+R

Rewrite the equation for P using the above expression.

  P=ε2RRR+R

Substitute 45V for ε, 27kΩ for R and 3.0kΩ for R in the above equation to calculate P.

  P=(45V)2(27kΩ)(3.0kΩ)27kΩ+3.0kΩ=(45V)22.7kΩ(103Ω1)=0.75W

After the switch has been closed for a long time the current reaches a constant value and the emf across the inductor is zero.

Conclusion:

Therefore, the current I1, I2 is respectively 1.7mA, 15mA, the potential difference across the resistors is respectively 45V, 45V, the power supplied by the battery is 0.75W and the potential difference across the inductor is 0V.

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Chapter 20 Solutions

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