JEE Main & Advanced
Sample Paper
JEE Main - Mock Test - 14
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 time t is
A)0.05s
B)0.1s
C)0.15s
D)0.3s
Correct Answer:
B
Solution :
[b]: The current at any instant is given by \[\Rightarrow \]\[I={{I}_{0}}(1-{{e}^{-Rt/L}})\] \[\Rightarrow \]\[\frac{{{I}_{0}}}{2}={{I}_{0}}(1-{{e}^{-Rt/L}})\]or\[\frac{1}{2}=(1-{{e}^{-Rt/L}})\] or\[{{e}^{-Rt/L}}=\frac{1}{2}\]or\[\frac{Rt}{L}=\ln 2\] \[\therefore \]\[t=\frac{L}{R}\ln 2=\frac{300\times {{10}^{-3}}}{2}\times 0.693\] \[=150\times 0.693\times {{10}^{-3}}=0.10395s=0.1s\]