9. In the figure below, switch S1 is closed while switch S2 is kept open. The in

Question

9. In the figure below, switch S1 is closed while switch S2 is kept open. The inductance is L = 0.160 H, and the resistance is R = 125 ?. When the current has reached its final value, the energy stored in the inductor is UL = 0.32 J.

a) What is the maximum value of the current?

b) What is the emf of the battery?

c) 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?

Explanation / Answer

We know that Energy,U=1/2(LI2)

L=0.160

U=0.32

So,I2=2U/L=2*0.32/0.16=4

I=2 A

b)EMF,V=Current *resistance=2*125=250 V

c)

The charge Q as function of time is given by

Q = Q0[e-(t/RC)]

Energy = LI2/2

So energy as a function of time is given by LI2/2 = L0I2/2 [e-(Rt/L)] or

E = Eo [e[-t125/.16] or E = Eo e-781t or

781t = ln (Eo/E) = ln 2 So the required time, T is given by

T = (ln 2)/781 = 8.8*10-4 s

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