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| 1. |
The capacitor of capacitance C in the circuit shown in fully charged initially. Resistance is R.– After the switch S is closed, the time taken to reduce the stored energy in the capacitor to half its initial value isA. `RC//2`B. `2RC` in 2C. RC ln 2D. `(RC ln 2)/(2)` |
| Answer» Correct Answer - D | |