JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Capacitance

A galvanometer ( G ) of

2Ω

resistance is connected in the given circuit. The ratio of charge stored in

C_{1}

and

C_{2}

is :

(A) $\frac{2}{3}$
(B) $\frac{3}{2}$
(C)1
(D) $\frac{1}{2}$ ✓ Correct

Answer: (D)

Solution

In steady state no current flows through either capacitor branch, so current flows only around the two resistive arms of the bridge driven by the 6 V source. With the given resistances the equivalent resistance is

12 Ω

, giving a current

I = \frac{6}{12} = 0.5

A. This produces a potential difference of 3 V across the branch holding

C_{1}

and 4 V across the branch holding

C_{2}

. The stored charges are

q_{1} = C_{1} V_{1} = 4 \times 3 = 12 μ C

and

q_{2} = C_{2} V_{2} = 6 \times 4 = 24 μ C

, so

\frac{q _{1}}{q _{2}} = \frac{12}{24} = \frac{1}{2}

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option

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