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
.

Options

A. 0
B. `4muC`
C. `16muC`
D. `8muC`

Answer

D

Explanation

Current flows through `2Omega` resistance from left to right, is given by
`I=V/(R+r) = (2.5V)/(2+0.5) =1A`
The potential difference across `2Omega` resistance `V =IR =1 xx2 =2V`
Since, capacitor is in parallel with `2Omega` resistance, so it also has 2V potential difference across it.
The charge on capacitor `q=CV=(4piF) xx2V =8muC`