Imagine a L-R circuit with two switches. Switch 1 is closed while switch 2 is ke

Question

Imagine a L-R circuit with two switches. Switch 1 is closed while switch 2 is kept open. The inductance is L = 0.110H , and the resistance is R = 140 ohms.
Part A
When the current has reached its final value, the energy stored in the inductor is 0.260J . What is the emf of the battery?

ANSWER:327 V


I can't get PART B:

Does anyone know how to approach this problem. Thanks!!!!

Part B
After the current has reached its final value, S1 is opened and S2 is closed. How much time does it take for the energy stored in the inductor to decrease to a half of the original value?
t = sec

Explanation / Answer

I = V / R U = 1/2 L I^2 V^2 = 2 U R^2 / L V = 304.39 v

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