An ideal cell of emf 10 V is connected in circuit shown in figure… | JEE Main 2020 (2 Sep, Shift 2)

Q32JEE Main 2020 · 2 SEP · SHIFT 2Medium

Capacitors in DC Circuits & RC

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 ______.

Numerical answer type. Work it out, then open the answer.

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Answer: 8

Steady state: no current through $C$ and $R_5$ . $R_1+R_2 = 4\ \Omega$ in parallel with $R_3$ gives $\dfrac{4}{3}\ \Omega$ ; plus $R_4$ : $\dfrac{10}{3}\ \Omega$ .
$I = \dfrac{10}{10/3} = 3\text{ A}$ ; drop on $R_4 = 6\text{ V}$ , so 4 V across the parallel part and $1\text{ A}$ in $R_1$ .
Capacitor is between the $R_1$ – $R_2$ junction and the negative terminal: $V_C = 10 - 1\times2 = 8\text{ V}$ .

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