A capacitor of 4 microfarad is connected as shown in the circuit figure.The internal resistance of the battery is 0.5 ohm the amount of charge on the capacitor plates will be
Current in the lower arm of the circuit, I=2?+0.5?2.5V?=1A Potential difference across the internal resistance of cell =(0.5?)(1A)=0.5V and potential difference across the?4?F?capacitor 2.5V?0.5V=2V Charge on the capacitor plates,?Q=CV=(4?F(2V)=8?C
