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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