A capacitor of 4`muF` is connected as shwon in the figure. The internal resistance of the battery is `0.5Omega`. 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=1xx2=2V` ltbgt Since, capacitor is in parallel with `2Omega` resistance, so it also has 2V potential difference across it.
The charge on capacitor
`q=CV=(2muF)xx2V=8muC`.