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A `mu F` capacitor is charged by a `400 V` supply through `0.1 M Omega` resistance. The time taken by the capacitor to develop a potential difference of `300 V` is : (Given `log_(10) 4 = 0.602`)A. `2.2 sec`B. `1.1 sec`C. `0.55 sec`D. `0.48 "secs"` |
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Answer» Correct Answer - D `tau = (L)/(R ) = RC` `R = sqrt(L//C) = sqrt(8//2 xx 10^(-6)), R = 2 K Omega` |
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