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

Options

A. `80muC`
B. `40muC`
C. `20muC`
D. `10muC`

Answer

d

Explanation

In steady state, there will be no current in the capacitro branch .Net resistacne of the circuit `R=1+1+0.5=2.5omega`
Current drawn from the cell `i=(V)/(R )=(2.5)/(2.5)=1A`
Potential drop across two parallel bracnhes
`V=E-ir=2.5-1xx0.5=2.5-0.5=2.0V`
So, charge on the capacito plates
`q=CV=5xx2=10muC`