Effective capacitance of parallel combination of two capacitors C₁ and C₂ is 10 μF. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor C₂ is 4 times that of C₁. If these capacitors are connected in series, their effective capacitance will be :

Solution

Physics sir · 3 steps

Same 1 V on both, so energy tracks capacitance: 4 times the energy means 4 times the capacitance.

1. Energy to capacitance. Each capacitor gets the same V = 1 V, and U = 1/2CV². With V fixed, U is proportional to C, so U₂ = 4U₁ gives C₂ = 4C₁.
2. Use the parallel value. In parallel, C₁ + C₂ = 10 μF. So C₁ + 4C₁ = 10, which gives C₁ = 2 μF and C₂ = 8 μF.

   Cₛₑᵣᵢₑₛ = (C₁ C₂)/(C₁ + C₂) = (2 × 8)/(2 + 8) = 1.6 μF

3. So. The series combination has an effective capacitance of 1.6 μF, which is option (c).

The idea to keep. Pick the energy formula that holds the common quantity fixed: same voltage means U = 1/2CV², same charge means U = Q²/2C. Then the energy ratio gives the capacitance ratio directly.

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