Thank you for reporting, we will resolve it shortly
Q.
A capacitor of capacitance $ 5 \mu F $ is connected as shown in the figure. The internal resistance of the cell is $ 0.5 \Omega$ The amount of charge on the capacitor plates is
In steady state, there will be no current in the capacitor branch
Net resistance of the circuit $ R = 1 + 1 + 0.5 = 2.5 \Omega $
Current drawn from the cell. $ i = \frac{V}{R} = \frac{2.5}{2.5} = 1A $
PotentiaI drop across two parallel branches
$ V = E - ir = 2.5 - 1 \times 0.5 $
$ = 2.5 - 0.5 $
$ = 2.0V $
So, charge on the capacitor plates
$ q = CV = 5 \times 2 $
$ = 10 \mu C $
