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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Similar questions

Q. A coil of self inductance

50 H

is joined to the terminals of a battery of emf

2 V

through a resistance of

10 Ω

and a steady current is flowing through the circuit. If the battery is now disconnected then time in which the current will decay to

\frac{1}{e}

of its steady value is

A uniformly wound solenoidal coil of self-inductance $1.8 \times 10^{- 4}$ henry and resistance 6 ohm is broken up into two coils having lengths in the ratio 1 : 2. These identical coils are then connected in parallel across a 12 volt battery of negligible resistance. The steady state current through the battery is___ ampere.