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A coil of inductance 0.01H is connected in series with a resistance $R=100 Ω$ and a battery, and the switch is closed. The time after which current in the circuit is $\frac{3}{4}$ times its maximum value is ______ miliseconds. [Take $\log_{e} = 2 = 0.7]$

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answer is 0.14.

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Detailed Solution

$i (t) = \frac{V}{R} (1 - e^{\frac{R}{L} t})$ and $i_{\max} = \frac{V}{R}$

$i (t) = \frac{3}{4} i_{\max} \Rightarrow t = \frac{2 L}{R} \log_{e} 2 = 1.4 \times 10^{- 4} s = 0.14 m s$

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