Two identical capacitors C₁ and C₂ of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor C ₁ using battery of emf V volt. Now, disconnecting a and b the terminals b and c are connected. Due to this, what will be the percentage loss of energy?

Step-by-step solution

When C ₁ is connected to voltage source, it is charged to a potential V and this will be stored as a potential energy in the capacitor given by U = 1 2 CV ^2 When key is disconnected from battery and band c are connected, the charge will be transformed from the capacitor C ₁ to capacitor C ₂ , then The loss of energy due to redistribution of charge is given by aligned & aligned U & = C ₁ C ₂ 2 ( C ₁+ C ₂ ) ( V ₁- V ₂ )^2 & = C C 2( C + C ) ( V -0)^2= 1 4 CV ^2 [ C ₁= C ₂ ] aligned & Percentage loss = U U 100= 1 4 C