Exams › JEE Advanced › Physics
A circuit contains two capacitors C and 2C. Initially, capacitor C is connected to a battery of EMF V through switch S1 (S1 closed, S2 open), fully charging it to voltage V. Then S1 is opened and S2 is closed, connecting C in parallel with uncharged capacitor 2C. After charge redistribution, find the ratio of charge on capacitor 2C to charge on capacitor C (Q₂C / Q_C).
- 1/2
- 2
- 1
- 3/2
Correct answer: 2
Solution
Initially C is charged to V, so Q0 = CV. When S2 closes (S1 open), C and 2C are connected in parallel. Total charge Q0 = CV redistributes to equalize voltage. Final voltage V_f = CV / (C+2C) = V/3. Charge on 2C: Q₂C = 2C * (V/3) = 2CV/3. Charge on C: Q_C = C * (V/3) = CV/3. Ratio Q₂C / Q_C = (2CV/3) / (CV/3) = 2.
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