A) 1 s
B) 2 s
C) e second
D) 2 e second
Correct Answer: B
Solution :
Given,\[L=60\,\,H\] \[R=30\Omega\] \[V=100V\] In\[L\text{-}R\]circuit current to reach 63.2% of its final value in time constant. Equation of current in\[L\text{-}R\]circuit \[I={{I}_{0}}(1-{{e}^{-Rt/L}})\] Required time = time constant \[=\frac{L}{R}=\frac{60}{30}=2s\]You need to login to perform this action.
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