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Q. A coil of inductance 300 mH and resistance 2 $\Omega$ is connected to a source of voltage 2 V. The current reaches half of its steady state value in

JIPMERJIPMER 2011Alternating Current

Solution:

The charging of inductance is given by
$I = I_0 [1 - e^{-Rt/L} ]$
$\therefore \: \frac{I_0}{2} = I_0 [ 1 - e^{-RT/L} ]$
or $e^{-Rt/L} = \frac{1}{2} $ or $\frac{Rt}{L} =$ 1n 2
or $t = \frac{0.693 \times 300 \times 10^{-3}}{2}$
= 0.1 s