In this figure the resistance of the coil of galvanometer G is $2 \Omega$. The emf of the cell is 4 V. The ratio of potential difference across $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ is:

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Capacitance

In this figure the resistance of the coil of galvanometer G is

2Ω

. The emf of the cell is 4 V. The ratio of potential difference across

C_{1}

and

C_{2}

is:

(A)1
(B) $\frac{4}{5}$ ✓ Correct
(C) $\frac{3}{4}$
(D) $\frac{5}{4}$

Answer: (B)

Solution

At steady state, capacitors block current. The circuit current is

i = \frac{4}{6 + 2 + 8} = \frac{1}{4}

A. Voltage across

C_{1}

(between A and C) is

V_{1} = i (6 + 2) = 2

V. Voltage across

C_{2}

(between B and D) is

V_{2} = i (2 + 8) = 2.5

V. Therefore

\frac{V _{1}}{V _{2}} = \frac{2}{2.5} = \frac{4}{5}

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option

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