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Q. An ideal coil of $10\, H$ is joined in series with a resistance of $5 \,\Omega$ and battery of $5\, V,\, 2 s$ after joining the current flowing (in ampere) in the circuit will be

AIIMSAIIMS 2011

Solution:

Given, $L=10\, H,\, R=5 \Omega,\, E=5\, V$
and $t=2\, s$.
Current in $R-L$ circuit
$I=I_{0}\left(1-e^{-R t / L}\right)$
$I=\frac{E}{R}\left(1-e^{-R t / L}\right)$
$=\frac{5}{5}\left(1-e^{-5 \times 2 / 10}\right)$
$I=\left(1-e^{-1}\right)$