WBJEE · Physics · Capacitance
A capacitor of capacitance \(C_{0}\) is charged to a potential \(V_{0}\) and is connected with another capacitor of capacitance \(C\) as shown. After closing the switch \(S,\) the common potential across the two capacitors becomes \(V\). The capacitance \(C\) is given by 
- A \(\frac{C_{0}\left(V_{0}-h\right)}{V_{0}}\)
- B \(\frac{C_{0}\left(V-V_{0}\right)}{V_{0}}\)
- C \(\frac{C_{0}\left(V+V_{0}\right)}{V}\)
- D \(\frac{C_{0}\left(V_{0}-V\right)}{V}\)
Answer & Solution
Correct Answer
(D) \(\frac{C_{0}\left(V_{0}-V\right)}{V}\)
Step-by-step Solution
Detailed explanation
When the switch \(S\) is closed, the common potential across the two capacitors becomes \(V\). so charge on isolated plates remains same.…
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