Question

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 1

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`.