A capacitor of capacitaance `5muF` is connected as shown in the figure.The internal resistane of the cell is `0.5 omega` The amount of charge on the cap[acitor plates is

Options

A. zero
B. `5muC`
C. `10muC`
D. `25muC`

Answer

C

Explanation

We know that capacitor offers infinite resistance for DC source. (battery is DC source)
Current taken from the cell
`i=(E)/(R+r)`
`=(2.5)/(1+1+0.5)=1`amp
Potential drop across capacitor
`V=E-ir`
`=2.5-1xx0.5=2V`
The current in `2Omega` resistor is zero.
Charge on capacitor
Q=CV
`=5xx2`
`=10muC`