A) 0.5 amp/sec
B) 2.0 amp/sec
C) 2.5 amp/sec
D) 0.25 amp/sec
Correct Answer: C
Solution :
\[i={{i}_{0}}\left( 1-{{e}^{\frac{-Rt}{L}}} \right)\Rightarrow \frac{di}{dt}=\frac{d}{dt}{{i}_{0}}-\frac{d}{dt}{{i}_{0}}{{e}^{\frac{-Rt}{L}}}\] \[\Rightarrow \frac{di}{dt}=0-{{i}_{0}}\left( -\frac{R}{L} \right)\ {{e}^{-\frac{Rt}{L}}}=\frac{{{i}_{0}}R}{L}{{e}^{-\frac{Rt}{L}}}\] Initially, \[t=0\Rightarrow \frac{di}{dt}=\frac{{{i}_{0}}\times R}{L}=\frac{E}{L}=\frac{5}{2}=2.5\ amp/sec.\]
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