A) 27.3 amp/sec
B) 27.8 amp/sec
C) 2.73 amp/sec
D) None of the above
Correct Answer: D
Solution :
The rate of increase of current \[=\frac{di}{dt}=\frac{d}{dt}{{i}_{0}}\left( 1-{{e}^{-Rt/L}} \right)=\frac{d}{dt}{{i}_{0}}-\frac{d}{dt}{{i}_{0}}{{e}^{-Rt/L}}\] \[=0-{{i}_{0}}{{e}^{-Rt/L}}.\frac{d}{dt}\left( -\frac{Rt}{L} \right)={{i}_{0}}\frac{R}{L}{{e}^{-Rt/L}}\] \[=\frac{50}{180}\times \frac{180}{5\times {{10}^{-3}}}\times {{e}^{-(180\times 0.001)/(5\times {{10}^{-3}})}}\]\[={{10}^{4}}\times {{e}^{-36}}A/\sec \]
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