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An inductor `(L = 100 mH)`, a resistor `(R = 100 Omega)`, and a battery `(W = 100 V)` are initially connected in series as shown by the figure. After a long time the battery is disconnected after short circuiting the points `A` and `B`. The currents in the circuit `1 mm` after the short circuit is A. `0.1 A`B. `1 A`C. `(1)/(e) A`D. `e A` |
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Answer» Correct Answer - C `I = I_(0)e^(-Rt//L)` `= (E)/(R )e^(-Rt//L)` `= (100)/(100) e^(-) (100)/(100 xx 10^(-3)) xx 10^(-3)` `= (1)/(e)A` |
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