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A coil of inductance `0.20 H` is connected in series with a switch and a cell of emf `1.6 V`. The total resistance of the circuit is `4.0 Omega`. What is the initial rate of growth of the current when the switch is closed?A. `0.050 A s^(-1)`B. `0.40 A s^(-1)`C. `0.13 A s^(-1)`D. `8.0 A s^(-1)` |
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Answer» Correct Answer - D `V = RI + L(dI)/(dt)`, at `t =0, I = 0`, thus we have `(dI)/(dt) = (V)/(L) = (1.6)/(0.2) = 8 A//s` |
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