In the given circuit, $\mathrm{C}_{1}=2 \mu \mathrm{~F}, \mathrm{C}_{2}=0.2 \mu \mathrm{~F}, \mathrm{C}_{3}=2 \mu \mathrm{~F}, \mathrm{C}_{4}=4 \mu \mathrm{~F}$, $\mathrm{C}_{5}=2 \mu \mathrm{~F}, \mathrm{C}_{6}=2 \mu \mathrm{~F}$, the charge stored on capacitor $\mathrm{C}_{4}$ is $\_\_\_\_$ $\mu \mathrm{C}$.

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Capacitance

In the given circuit,

C_{1} = 2 μ F, C_{2} = 0.2 μ F, C_{3} = 2 μ F, C_{4} = 4 μ F

C_{5} = 2 μ F, C_{6} = 2 μ F

, the charge stored on capacitor

C_{4}

____

μ C

Answer: 4

With the

10 V

source across the two outer corners, label the four internal junction potentials and apply charge conservation at each junction using

Q = C Δ V

for every capacitor. Solving the node equations for the top-middle, top-right, bottom-middle and bottom-right junctions gives potentials of

7.5 V

5.5 V

2.5 V

and

4.5 V

respectively, with the source plates held at

10 V

and

0 V

. Capacitor

C_{4}

bridges the top-right and bottom-right junctions, so the potential difference across it is

5.5 - 4.5 = 1 V

. The stored charge is

Q_{4} = C_{4} Δ V = 4 μ F \times 1 V = 4 μ C

. The charge stored on

C_{4}

4 μ C

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Integer answer