question_answer

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

A)  0.05 s                                   

B)  0.1 s                     

C)  0.15 s                   

D)         0.3 s

Correct Answer: B

Solution :

The current at any instant is given by                 \[I={{I}_{0}}(1-{{e}^{-Rt/L}})\]                 \[\frac{{{I}_{0}}}{2}={{I}_{0}}(1-{{e}^{-Rt/L}})\]                 \[\frac{1}{2}=(1-{{e}^{-Rt/L}})\]                 \[{{e}^{-Rt/L}}=1/2\]                 \[\frac{Rt}{L}=\ln 2\] \[\therefore \]  \[t=\frac{L}{R}\ln 2=\frac{300\times {{10}^{-3}}}{2}\times 0.693\]                 \[=150\times 0.693\times {{10}^{-3}}\]                 \[=0.10395s=0.1\,\,s\]