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 \[I={{I}_{0}}(1-{{e}^{-Rt/L}})\] \[\frac{{{I}_{0}}}{2}={{I}_{0}}(1-{{e}^{-Rt/L}})\] \[\frac{1}{2}=(1-{{e}^{-Rt/L}})\] \[{{e}^{-Rt/L}}=1/2\] \[\frac{Rt}{L}=\ln \,2\] \[\therefore \] \[t=\frac{L}{R}\ln \,2=\frac{300\times {{10}^{-3}}}{2}\times 0.693\] \[=150\times 0.693\times {{10}^{-3}}\] \[=0.10395\,s=0.1\,s\]You need to login to perform this action.
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