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An ideal coil of 10H is connected in series with a resistance of `5(Omega)` and a battery of 5V. 2second after the connections is made, the current flowing in ampere in the circuit isA. `(1 - e^(-1))`B. `(1 - e)`C. eD. `e^(-1)` |
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Answer» Correct Answer - A Here,`L = 10 h, R = 5 Omega, E = 5 V` Max. current, `I_(0) = (E)/(R ) = (5)/(5) = 1 A` During growth of current `I = I_(0) (1 - e^((-R )/(L) t))` `:. I = 1 (1 - e^((-5)/(10) xx 2)) = (1 - e^(-1))` |
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