Question

# A coil of inductance 300mH and resistance 2Ω is connected to a source of voltage 2v. The current reaches half of its steady-state value in____seconds.

A

0.1

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B

0.3

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C

0.05

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D

0.105

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Solution

## The correct option is A 0.1Step 1: GivenInductance, L=300mH=$300×{10}^{-3}H$Resistance, R= 2ΩVoltage, V=2VStep 2: Use the formula of growth of currentFormula for the growth of current in inductor, $I={I}_{0}\left(1-{e}^{-\frac{Rt}{L}}\right)$Also, $I=\frac{{I}_{0}}{2}$ because the time requires is for the half of current steady-state value.Now, $\frac{{I}_{0}}{2}={I}_{0}\left(1-{e}^{-\frac{Rt}{L}}\right)\phantom{\rule{0ex}{0ex}}\frac{{I}_{0}}{2}={I}_{0}-{I}_{0}\left({e}^{-\frac{Rt}{L}}\right)\phantom{\rule{0ex}{0ex}}{I}_{0}\left({e}^{-\frac{Rt}{L}}\right)={I}_{0}-\frac{{I}_{0}}{2}\phantom{\rule{0ex}{0ex}}{I}_{0}\left({e}^{-\frac{Rt}{L}}\right)=\frac{{I}_{0}}{2}\phantom{\rule{0ex}{0ex}}{e}^{-\frac{Rt}{L}}=\frac{1}{2}$taking log on both sides-$-\frac{Rt}{L}=-\mathrm{log}2\phantom{\rule{0ex}{0ex}}\frac{Rt}{L}=\mathrm{log}2$Step 3: Calculate the time to reach half of its steady-state value$t=\frac{L}{R}\mathrm{log}2$Substitute values-$t=\frac{300×{10}^{-3}}{2}×0.693\phantom{\rule{0ex}{0ex}}t=150×{10}^{-3}×0.693\phantom{\rule{0ex}{0ex}}t=0.10395s\approx 0.1s$Thus, the time to reach half of its steady-state value is $0.10395s\approx 0.1s$.Hence, Option A is correct.

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