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

Any capacitor after getting fully charged behaves as an open circuit. In the given situation due to the presence of 4 `muF` capacitor, current will not flow through the branch containing it and 10 resistor. Thus, current flowing through 2 resistor will be as:
`I=(V)/(R +r)=(2.5)/(2+0.5)=1A`, Terminal voltage for a battery is given by, `V=E-Ir`

Here, `V=2.5-0.5xx1=2V`
As no current will be flowing through the 10 resistor, thus the terminal potential difference of the battery will appear as the potential difference between the plates of the capacitor.
Charge on plates of capacitor is given by,
`q=CV=4xx2=8muC`