NTA Abhyas JEE Main2020PhysicsElectrostaticsPractice
A 10 μ F capacitor and a 20 μ F capacitor are connected in series across a 200 V supply line. The charged capacitors are then disconnected from the line and reconnected such that those same polarities are connected to each other and no external voltage is applied. The potential difference across capacitors is
Options
- A400 9 V
- B800 3 V
- C400 V
- D200 V
Correct answer
A. 400 9 V
Step-by-step solution
  C s = 10 × 20 10 + 20 = 200 30 = 20 3   μ F   Q   = C s V   Q   = 20 3   μ F × 200   V   Q   = 4000 3   μC Now,   V   =   4000   μC 3 × 30   μF = 4000 90   V = 400 9   V