An ideal cell of emf $10\, V$ is connected in circuit shown in figure. Each resistance is $2\, \Omega .$ The potential difference (in $V$) across the capacitor when it is fully charged is
JEE MAIN 2020, Diffcult
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$R _{1}$ to $R _{5} \rightarrow$ each $2 \Omega$

Cap. is fully charged

So no current is there in branch $ADB$

Effective circuit of current flow

$R_{e q}=\left(\frac{4 \times 2}{4+2}\right)+2$

$R _{ eq }=\frac{4}{3}+2=\frac{10}{3} \Omega$

$i=\frac{10}{10 / 3}=3 A$

So potential different across $AEB$

$\Rightarrow 2 \times 1+2 \times 3=8 V$

Hence potential difference across

Capacitor $=\Delta V = V _{ AEB }=8 V$

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