A) 2 ´ 10?2 sec and 14.14 amp
B) 1 ´ 10?2 sec and 7.07 amp
C) 5 ´ 10?3 sec and 7.07 amp
D) 5 ´ 10?3 sec and 14.14 amp
Correct Answer: D
Solution :
Time taken by the current to reach the maximum value \[t=\frac{T}{4}=\frac{1}{4\nu }=\frac{1}{4\times 50}=5\times {{10}^{-3}}sec\] and \[{{i}_{o}}={{i}_{rms}}\sqrt{2}=10\sqrt{2}=14.14\]ampYou need to login to perform this action.
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