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    				| 1. | A capacitor of `4 mu F` is connected as shown in the circuit. The internal resistance of the battery is `0.5 Omega`. The amount of charge on the capacitor plates will be . | 
| Answer» Correct Answer - D No current flows in upper arm of the circuit. Current in lower arm of the circuit, `I=(E)/(R+r)=(2.5)/(2+0.5)=1A` `:.` Terminal potential difference of battery, `V = E - I_( r) = 2.5 -1 xx 0.5 = 2V` So charge on the capacitor plates, `Q = CV` `= 4 mu F xx 2V=8 mu C`. | |