A capacitor of capacitance $5\,\mu F$ is connected as shown in the figure. The internal resistance of the cell is $0.5\,\Omega $. The amount of charge on the capacitor plate is......$\mu C$
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(c) In steady state condition. No current flows through line $(1)$. Hence total current $i = \frac{{2.5}}{{(1 + 1 + 0.5)}} = 1\,A$
Potential difference a cross line $(2)$ $=$ potential difference a cross capacitor
$ = 1 \times 2 = 2\,Volt$
So, charge on capacitor = $5 × 2$ = $10 \,µC$
Potential difference a cross line $(2)$ $=$ potential difference a cross capacitor
$ = 1 \times 2 = 2\,Volt$
So, charge on capacitor = $5 × 2$ = $10 \,µC$
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