A) 0.3s
B) 0.15s
C) 0.1 s
D) 0.05s
Correct Answer: C
Solution :
[c] \[L=300\times {{10}^{-3}}H\] \[R=1\Omega \] \[I=\frac{{{I}_{0}}}{2}\] Using \[I={{I}_{0}}(1-{{e}^{-Rt/L}}),\]we get \[\frac{{{I}_{0}}}{2}={{I}_{0}}(1-{{e}^{-Rt/L}})\] \[\Rightarrow \frac{1}{2}=1-{{e}^{-Rt/L}}\Rightarrow {{e}^{-Rt/L}}=\frac{1}{2}\] \[\Rightarrow {{e}^{Rt/L}}=2\Rightarrow \frac{R}{L}t={{\log }_{e}}2=0.693\] \[\Rightarrow t=\frac{0.693\times 300\times {{10}^{-3}}}{2}s=0.1s\]You need to login to perform this action.
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