Question

a coil of self inductance 50H is joined to the terminals of a battery of e.m.f. 2V through a resistance of 10 ohm and a steady current is flowing through the circuit . if the battery is now disconnected then time in which the current will decay to 1/e of its steady value is
ans: 5 sec

Answer 1

decay of current in term of self inductance is given by, $i=i_0e^{-r\tau/L}$

where $i_0$ is value of current in steady state, r is resistance, L is self inductance and $\tau$ is time taken to decay of current.

here, $i=\frac{i_0}{e}$

r = 10 ohm , L = 50H

so, $\frac{i_0}{e}=i_0e^{-10\tau/50}$

or, $\frac{1}{e}=e^{-\tau/5}$

or, $e^{-1}=e^{-\tau/5}$

hence, $\tau=5$

time taken 5 sec in which the current will decay to 1/e of its steady value.

Answer 2

Answer:

here is the answer.

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