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In the circuit shown in figure `L=10H, R=5Omega, E=15V.` The switch `S` is closed at `t=0`. At `t=2s` the current in the circuit is A. `3(1-1/e)A`B. `3(1-1/e^2)A`C. `3(1/e)A`D. `3(1/e^2)A` |
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Answer» Correct Answer - A `tau_L=L/R=2s` `i_0=E/R=3A, t=2s` `i=i_0(1-e^(-t//tau_L))` Substituting the given values, we can find i. |
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