A coil of self inductance 50 henry is joined to the terminals of a battery of emf 2 volts through a resistance of 10 ohm and a steady current is flowing through the circuit. If the battery is now disconnected, the time in which the current will decay to 1/e of steady value is

500s

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50s

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15s

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Solution

The correct option is C 5s
the decay current in L-R circuit is
$i = i_{0} e^{- \frac{R}{L}} t$
$i = i_{0} e^{- \frac{10}{50}} t$
$e^{- 1} = e^{- \frac{t}{5}}$
$\frac{t}{5} = 1$
$t = 5 s e c$

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