# An ideal cell of emf 10V is connected in circuit shown in figure. Each resistance is . The potential difference (in V) across the capacitor when it is fully charged is ......... Option: 1 2 Option: 2 4 Option: 3 6 Option: 4 8

when the capacitor is fully charged it behaves as an open circuit  (voltage drop with no current)

So the new circuit diagram is given below.

So

$\begin{array}{l} \mathrm{i}=\frac{10}{\frac{4}{3}+2}=\frac{10 \times 3}{10}=3 \mathrm{Amp} \\ \Rightarrow \mathrm{i}_{1}=2 \mathrm{~A} \ \& \ \mathrm{i}_{2}=1 \mathrm{~A} \\ \Rightarrow \mathrm{~V}_{\mathrm{AB}}=1 \times 2+3 \times 2=8 \mathrm{~V} \end{array}$

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