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In the circuit shown in the figure, `E = 50.0 V, R = 250 Omega` and `C = 0.500 muF.` The switch `S` is closed for a long time, and no voltage is measured across the capacitor. After the switch is opened, the voltage across the capacitor reaches a maximum value of `150 V`. What is the inductance `L`? |
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Answer» Correct Answer - B In steady state when switch was closed, `i_0=E//R=(1//5)A=0.2A` After switch is opened it becomes `L-C` circuit in which peak value current is `0.2 A`. `:. 1/2 Li_0^2=1/2CV_0^2` or `L=V_0^2/i_0^2.C` `=((150)^2)/((0.2)^2)xx0.5xx10^-6` `=0.28H` |
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