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A coil of inductance `300mh` and resistance `2Omega` is connected to a source of voltage `2V`. The current reaches half of its steady state value inA. `0.3 s`B. `0.15 s`C. `0.1 s`D. `0.05 s` |
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Answer» Correct Answer - C `L = 300 xx 10^(-3) H` `R = 10 Omega` `I = (I_(0))/(2)` Using `I = I_(0)(1-e^(-Rt//L))`, we get `(I_(0))/(2) = I_(0) (1-e^(-Rt//L))` `rArr (1)/(2) = 1 - e^(-Rt//L) rArr e^(-Rt//L) = (1)/(2)` `rArr e^(Rt//L) = 2 rArr (R )/(L)t = log_(e) 2 = 0.693` `rArr t = (0693 xx 300 xx 10^(-3))/(2)s = 0.1 s` |
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