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 ?

Question

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) 75% (B) 0% (C) 50% (D) 25%

Tardigrade Admin · Accepted Answer

U textinitial = (1/2)CV2 Loss = (C.C/2(C + C))(V - 0)2 = (1/4) CV2 % Loss = ((1/4) CV2/(1)2CV2) × 100 = 50 %

75%

0%

50%

25%

Uinitial​=21​CV2 Loss =2(C+C)C.C​(V−0)2 =41​CV2 % Loss =21​CV241​CV2​×100=50%

$U_{initial} = \frac{1}{2} C V^{2}$
Loss $= \frac{C . C}{2 ( C + C )} (V - 0)^{2}$
$= \frac{1}{4} C V^{2}$
$%$ Loss $= \frac{\frac{1}{4} C V ^{2}}{\frac{1}{2} C V ^{2}} \times 100 = 50%$