Q 12-02-148JEE MainJEE Main 2020 (5 Sep, Shift 2)Medium
In the circuit shown, the charge on the $5\ \mu\text{F}$ capacitor is:
Answer: (D) $5.45\ \mu\text{C}$
Take the potential of O as zero. In the figure both cells have their longer (positive) plate on the right, so the left end of the circuit is at $-6$ V and the right end is at $+6$ V.
Let the common top junction be at potential $V$. The total charge on the three plates joined at this junction is zero:
$$2(V+6) + 4(V-6) + 5(V-0) = 0$$
$$11V = 12 \Rightarrow V = \frac{12}{11}\ \text{V}$$
$$Q_5 = 5\times\frac{12}{11} = \frac{60}{11} \approx 5.45\ \mu\text{C}$$
Solution by Sreeraj P, M.Sc Physics