In this figure the resistance of the coil of galvanometer G is 2 Ω. The emf of the cell is 4 V. The ratio of potential difference across C₁ and C₂ is:

Solution

Physics sir · 4 steps

In steady state the capacitors block current, so it flows A to B to G to C to D through all three resistances.

1. Steady state. Once the capacitors are fully charged, no current flows through C₁ or C₂. The only path left is A, then 6 Ω, then B, then G (2 Ω), then C, then 8 Ω, then D.
2. Find the current. The three resistances are in series: 6 + 2 + 8 = 16 Ω. So I = 4/16 = 0.25 A.
3. Node potentials. Take V_D = 0 and V_A = 4 V. Then V_B = 4 - 0.25 × 6 = 2.5 V and V_C = 0.25 × 8 = 2 V.

   V_(C₁) = V_A - V_C = 2 V, V_(C₂) = V_B - V_D = 2.5 V

4. So. V_(C₁)/V_(C₂) = 2/2.5 = 4/5, which is option (d).

The idea to keep. In a DC circuit at steady state, treat every capacitor as a broken wire. Find the current in the resistor path, then read each capacitor's voltage as the potential difference between its two ends.

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