| The time at which current in \[{{R}_{2}}\] becomes half of the steady value of current when the circuit is switched on at t = 0: |
 |
A) 1.03 ms
B) 1.40 ms
C) 2.09 ms
D) 3.19 ms
Correct Answer: B
Solution :
| \[i=\frac{E}{R}\left( 1-{{e}^{-\,\,\frac{Rt}{L}}} \right)\] |
| Where \[R=2+3=5\Omega ,\] \[L=10\times {{10}^{-3}}\] Henry at time\[{{t}_{1}}\,i=\frac{1}{2}\times \] (steady value of current) |
| \[i=\frac{1}{2}\times \frac{E}{R}\Rightarrow \frac{E}{2R}\] |
| \[i=\frac{E}{R}\left[ 1-{{e}^{-\frac{Rt}{L}}} \right]\]\[\Rightarrow \frac{E}{2R}=\frac{E}{R}\left[ 1-{{e}^{-\,\,\frac{Rt}{L}}} \right]\] |
| \[{{e}^{\frac{-Rt}{L}}}=\frac{1}{2}\]\[\Rightarrow \frac{tR}{L}=In2\] |
| \[t=\frac{L}{R}In2=\frac{10\times {{10}^{-3}}}{5}\times 0.7=1.4\,\,milli\,\,\text{second}\] |
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