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An inductor of inductance L = 400 mH and resistors of resistance R_(1) = 2 Omega and R_(2) = 2Omega are connected to a battery of emf 12V as shown in the figure.The internal resistance of the battery is negligible.The switch S is closed at t = 0.The potential drop across L as a function of time is |
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Answer» `(12)/(t)e^(-3t)V` `i = (E)/(R_(2))[1 - e^(-R_(2)t/L] Rightarrow (di)/(dt) = (E)/(R_(2)).(R_(2))/(L) e^(-R_(2)t/L) = (E)/(L)e ^(-(R_(2)t))/(L)]` Hence, potential drop across `L = ((E)/(L)e^(-R_(2)t/L))L = Ee^(-R_(2)t/L) = 12 e -(2t)/(400 xx 10^(-3)) = 12e^(-5tV)` |
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