JEE Main 2024

Question

The charge accumulated on the capacitor connected in the following circuit is ______ $\mu$C. (Given $C = 150\ \mu	ext{F}$)

![10V source with R1=1Ω, R2=2Ω, R3=6Ω, R4=4Ω and capacitor C between A and B](https://www.edvaya.com/jee/physics/q54-capacitor-circuit-charge-2024.png) A). 200 B). 400 C). 300 D). 100

Edvaya · Accepted Answer

The correct answer is 400. At steady state, no current flows through the capacitor branch. The circuit reduces to a Wheatstone-bridge-like voltage divider.
The potential at the nodes across the capacitor are:
$V_A = 10 \left(\dfrac{R_2}{R_1+R_2}ight) = 10 \left(\dfrac{2}{1+2}ight) = \dfrac{20}{3}\ 	ext{V}$
$V_B = 10 \left(\dfrac{R_4}{R_3+R_4}ight) = 10 \left(\dfrac{4}{6+4}ight) = 4\ 	ext{V}$
Voltage across capacitor:
$V_C = V_A - V_B = \dfrac{20}{3} - 4 = \dfrac{8}{3}\ 	ext{V}$
Charge: $Q = CV_C = 150 	imes \dfrac{8}{3} = 400\ \mu	ext{C}$

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