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In the circuit shown in figure the switch S is closed at t = 0. A long time after closing the switch A. voltage drop across the capacitor is EB. current through the battery is `(E)/(R_(1)+R_(2)+R_(3))`C. energy stored in the capacitor is `(1)/(2)C (((R_(2)+R_(3))E)/(R_(1)+R_(2)+R_(3)))^(2)`D. current through the resistance `R_(4)` becomes zero |
| Answer» Correct Answer - B::C::D | |