Question

A coil of inductance 8.4 mh and resistance 6 is connected to a 12 V battery. The current in the coil is 1.0 A approximately after time

• A. 500 ms
• B. 20 s
• C. 35 ms
• D. 1 ms

Answer 1

The correct option is D 1 ms

Given : $L=0.0084H$ $R=6\Omega$ $V=12$ volts
Current in the RL circuit at any time is given by $I_t=\frac{V}{R}(1-e^{\frac{-Rt}{L}})$
$\therefore$ $1=\frac{12}{6}(1-e^{\frac{-6t}{0.0084}})$
Or $1=2(1-e^{-714.3t})$
Or $e^{-714.3t}=0.5$
Taking natural log on both sides, we get
$-714.3t=\ln 0.5$
$-714.3t=-0.693$
$⟹t=9.7\times {10}^{-4}s\approx 1ms$