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 \[I={{I}_{0}}\,(1-{{e}^{-Rt/L}})\] \[\frac{{{I}_{0}}}{2}\,={{I}_{0}}\,(1-{{e}^{-Rt/L}})\Rightarrow \,\frac{1}{2}\,=(1-{{e}^{-Rt/L}})\] \[{{e}^{-Rt/L}}\,=\frac{1}{2}\Rightarrow \,\frac{Rt}{L}\,=\ln 2\] \[\therefore \,\,t=\frac{L}{2}\,\ln 2=\frac{300\times {{10}^{-3}}}{2}\times 0.693\] \[=150\times 0.693\,\times {{10}^{-3}}\] \[t=0.10395s=0.1s\]
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