Two identical capacitors C_1 and C_2 of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charg - PHYSICS MCQ NEET 2019 2026 | ExamTest.live | ExamTest.live

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Two identical capacitors C_{1} and C_{2} 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 a battery of emf V volt. Now disconnecting a and b terminals, terminals b and c are connected. Due to this, what will be the percentage loss of energy?

A $75%$
B $0%$
C $50%$
D $25%$

Solution & Step-by-step Explanation

Initial energy stored in C_{1} : U_{i} = \frac{1}{2} C V^{2} .When C_{1} is connected to C_{2} (which is uncharged), they share the charge until they reach common potential V^{'} = \frac{C _{1} V + C _{2} ( 0 )}{C _{1} + C _{2}} = \frac{C V}{2 C} = \frac{V}{2} .Final energy U_{f} = \frac{1}{2} (C_{1} + C_{2}) (V^{'})^{2} = \frac{1}{2} (2 C) (\frac{V}{2})^{2} = \frac{1}{4} C V^{2} .Loss of energy Δ U = U_{i} - U_{f} = \frac{1}{2} C V^{2} - \frac{1}{4} C V^{2} = \frac{1}{4} C V^{2} .Percentage loss = \frac{Δ U}{U _{i}} \times 100 = \frac{1/4}{1/2} \times 100 = 50% .

Practice this question

Try it yourself before checking the explanation above.

Two identical capacitors C_{1} and C_{2} 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 a battery of emf V volt. Now disconnecting a and b terminals, 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%

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