In the following figure R = 15.1 Ohm and the battery emf is 6.49 V With switch S

Question

In the following figure R = 15.1 Ohm and the battery emf is 6.49 V With switch S2 open, switch S1 is closed. After several minutes. S1 is opened and S2 is closed. At 2.15 ms after S1 is opened, the current has decayed to 0.328 A . Calculate the inductance of the coil. Express your answer to three significant figures and include the appropriate units. How long after S1 is opened will the current reach 1.00% of its original value? Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

in LR series

with battery connected

i = (e/R)*(1-e^-(t/RL))

0.328 = (6.49/15.1)*(1-e^-(0.00215/(15.1*L)))

L = 9.88*10^-5 H

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with s1 opened

i = I*e^-(t/RL)

i = 0.01 I

0.01 = e^-(t/(15.1*9.88*10^-5))

t = 0.0068s = 6.8 ms

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