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hidemi

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- Homework Statement
- (1) An inductance L, resistance R, and ideal battery of emf ε are wired in series and the circuit is allowed to come to equilibrium. A switch in the circuit is opened at time t = 0, at which time the current is ε/R. At any later time t the potential difference across the resistor is given by:

A.ε (1−e^−Lt/R)

B.εe^−Lt/R

C.ε (1 +e^−Rt/L)

D.εe^−Rt/L

E. ε (1−e^−Rt/L) The answer is D.

(2) An inductance L, resistance R, and ideal battery of emf ε are wired in series. A switch in the circuit is closed at time t = 0, at which time the current is zero. At any later time t the emf of the inductor is given by:

A.ε (1−e^−Lt/R)

B.εe^−Lt/R

C.ε (1 +e^−Rt/L)

D.εe^−Rt/L

E. ε (1−e^−Rt/L) The answer is D.

- Relevant Equations
- Potential Difference = ε*e^(-Rt/L)

I would like to share my understanding of both of these questions.

For Question (1), the inductance L is releasing its stored energy to be dissipated by the resistance R. As time passes, the voltage across L is decreasing and thus the potential difference across the resistor will be close to 0.

For Question (2), the inductance L is being charged, as the current is given zero initially and thus voltage is at its maximum (because the inductance has a maximum reactance in the beginning). As time passes, the emf of the inductance drops (less reactance), so the emf of the inductor is close to 0.

Let me know if my thought is correct.

For Question (1), the inductance L is releasing its stored energy to be dissipated by the resistance R. As time passes, the voltage across L is decreasing and thus the potential difference across the resistor will be close to 0.

For Question (2), the inductance L is being charged, as the current is given zero initially and thus voltage is at its maximum (because the inductance has a maximum reactance in the beginning). As time passes, the emf of the inductance drops (less reactance), so the emf of the inductor is close to 0.

Let me know if my thought is correct.