CUET · PHYSICS · PYQ PAPER 2023
The capacities of two capacitors are \( C_1 \) and \( C_2 \) and their respective potentials are \( V_1 \) and \( V_2 \). If they are connected with a wire, then the loss of energy is given by :
- A \( \frac{C_1 C_2(V_1 + V_2)}{2(C_1 + C_2)} \)
- B \( \frac{C_1 C_2(V_1 - V_2)}{2(C_1 + C_2)} \)
- C \( \frac{C_1 C_2(V_1 - V_2)^2}{2(C_1 + C_2)} \)
- D \( \frac{(C_1 + C_2)(V_1 - V_2)}{C_1 C_2} \)
Answer & Solution
Correct Answer
(C) \( \frac{C_1 C_2(V_1 - V_2)^2}{2(C_1 + C_2)} \)
Step-by-step Solution
Detailed explanation
\( \Delta E = (\frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2) - \frac{(C_1 V_1 + C_2 V_2)^2}{2(C_1 + C_2)} \) \( \Delta E = \frac{(C_1 V_1^2 + C_2 V_2^2)(C_1 + C_2) - (C_1 V_1 + C_2 V_2)^2}{2(C_1 + C_2)} \)…
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