CUET · PHYSICS · PYQ PAPER 2023
A capacitor of capacitance \(C_1\) is charged upto \(V\) volt and then connected to an uncharged capacitor of capacity \(C_2\). The final potential difference across each will be :
- A \(\frac{V}{\left(C_1+C_2\right)}\)
- B V
- C \(\frac{C_1 V}{\left(C_1+C_2\right)}\)
- D \(\frac{V}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{C_1 V}{\left(C_1+C_2\right)}\)
Step-by-step Solution
Detailed explanation
Initial charge on \(C_1\): \(Q = C_1 V\) Final potential difference: \(V_{final} = \frac{Q}{C_1 + C_2} = \frac{C_1 V}{C_1 + C_2}\)
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