Three capacitors C₁ = 2μF, C₂ = 6μF and C₃ = 12μF are connected as shown in figure. Find the ratio of the charges on capacitors C₁, C₂ and C₃ respectively :

Solution

Physics sir · 4 steps

C₂ and C₃ share one charge in series, and that branch sits in parallel with C₁ across the same V.

1. Spot the branches. Between A and B there are two paths. The top path has C₂ and C₃ in series through D. The bottom path has C₁ alone. Both paths get the full battery voltage V.
2. Series branch. In series, both capacitors carry the same charge. Their equivalent capacitance is (6 × 12)/(6 + 12) = 4 μF, so Q₂ = Q₃ = 4V μC.
3. Lone capacitor. C₁ sits directly across A and B, so Q₁ = C₁ V = 2V μC.

   Q₁ : Q₂ : Q₃ = 2V : 4V : 4V = 1 : 2 : 2

4. So. The charges on C₁, C₂ and C₃ are in the ratio 1 : 2 : 2, option (c).

The idea to keep. Capacitors in series always carry equal charge. So find the series equivalent first, then multiply by the voltage across that whole branch.

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