QJEEM20Phb515: JEE Main 2020 Physics - Class 12 JEEM20Phb515 Solution: An ideal cell of emf 10 V is connected in circuit shown in…

Capacitor is fully charged

So no current is there in branch ADB

$$\therefore$$ Effective circuit of current flow :


R_eq = $$\left(\frac{4\times 2}{4+2}\right)+2$$

R_eq = $$\frac{4}{3}+2$$ = $$\frac{10}{3}\mathrm{\Omega }$$

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

So potential different across AEB

= 2 × 1 + 2 × 3 = 8V

Hence potential difference across Capacitor = 8V