A galvanometer (G) of 2 Ω resistance is connected in the given circuit. The ratio of charge stored in C₁ and C₂ is :

Solution

Physics sir · 4 steps

At steady state the capacitors block current, so it flows through 4 Ω, G and 6 Ω only; read the potentials.

1. Steady current path. Once the capacitors are charged, no current passes through C₁ or C₂. The only path left is 4 Ω → G (2 Ω) → 6 Ω, all in series across the 6 V cell.

   I = (6 V)/((4 + 2 + 6) Ω) = 0.5 A

2. Node potentials. Take the left node at 6 V and the right node at 0 V. The 4 Ω drops 2 V, so the top node is at 4 V. The galvanometer drops 1 V, so the bottom node is at 3 V. The 6 Ω drops the remaining 3 V.
3. Charge on each. C₁ joins the left node to the bottom node, so it sees 6 - 3 = 3 V and holds Q₁ = 4 × 3 = 12 μC. C₂ joins the top node to the right node, so it sees 4 - 0 = 4 V and holds Q₂ = 6 × 4 = 24 μC.
4. So. Q₁ : Q₂ = 12 : 24 = 1/2, which is option (d).

The idea to keep. In a DC circuit at steady state, treat every capacitor branch as open. Find the steady current through the resistors, get the node potentials, then use Q = CV across each capacitor.

Answer from the NTA final answer key. Solution by PrepWiser.