In the circuit shown, the cell is ideal, with $emf$ $=$ $15$ $V$. Each resistance is of $3 $ $\Omega$ . The potential difference across the capacitor is.....$V$
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As a fully charged capacitor draws no current so no current flow through resistor on right The current flow through remaining resistors as shown in figure.
Applying Kirchhoff's law
For loop $ABEFGA:$ $R i_{1}+R\left(i_{1}+i_{2}\right)=15$
$\Longrightarrow 2 i_{1}+i_{2}=\frac{15}{R}=\frac{15}{3}=5 \ldots .(1)$
For loop $ABCDEFGA:$ $R i_{2}+R i_{2}+R\left(i_{1}+i_{2}\right)=15$
$\Longrightarrow i_{1}+3 i_{2}=\frac{15}{R}=\frac{15}{3}=5 \ldots . .(2)$
From $(1)$ and $( 2 ):$
$i_{1}+3\left(5-2 i_{1}\right)=5$
$\Rightarrow 5 i_{1}=10$
$\Rightarrow i_{1}=2 A$
Putting, $i_{1}=2$ in $(1)$
$\Rightarrow i_{2}=5-2(2)=1 A$
Thus, Potential across capacitor is: $V_{c}=R i_{2}+R\left(i_{1}+i_{2}\right)=(3 \times 1)+3(2+1)=3+$
$9=12 V$
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