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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?

Asked in Odisha NEET 2019 · Redistribution of charge

Figure: Redistribution of charge
Answer: (3) 50%

Step-by-step solution

Let C₁=C₂=C. After charging: Q=CV, Uᵢ=1/2CV².

Key to b–c: C₁ shares its charge with C₂ in parallel; common voltage (CV)/(2C)=V/2.

U_f=1/2(2C)(V/2)²=1/4CV²=(Uᵢ)/2.

Loss =(Uᵢ-U_f)/(Uᵢ)×100=50% (dissipated as heat and radiation in the connecting wires).

Why the other options are wrong

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