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 C1 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 ?

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 C1 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 ?

(a) 25%

(b) 75%

(c) 0%

(d) 50%

Answer :  (d) 50%

\( U_i = \frac{1}{2} C V^2 \)

\( U_f = \frac{1}{2} C \left( \frac{V}{2} \right)^2 \times 2 \quad \text{(Voltage split between two capacitors)} \)

% loss of Energy \( = \frac{U_i - U_f}{U_i} \times 100 \)

% loss of Energy

\( = \frac{\frac{1}{2} C V^2 \left[ 1 - \frac{1}{2} \right]}{\frac{1}{2} C V^2} \times 100 \)

\( = 50\% \)