JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
Two capacitors having capacitance \(C _{1}\) and \(C _{2}\) respectively are connected as shown in figure. Initially, capacitor \(C _{1}\) is charged to a potential difference \(V\) volt by a battery. The battery is then removed and the charged capacitor \(C_{1}\) is now connected to uncharged capacitor \(C _{2}\) by closing the switch \(S\). The amount of charge on the capacitor \(C _{2}\), after equilibrium is
- A \(\frac{ C _{1} C _{2}}{\left( C _{1}+ C _{2}\right)} V\)
- B \(\frac{\left( C _{1}+ C _{2}\right)}{ C _{1} C _{2}} V\)
- C \(\left( C _{1}+ C _{2}\right) V\)
- D \(\left( C _{1}- C _{2}\right) V\)
Answer & Solution
Correct Answer
(A) \(\frac{ C _{1} C _{2}}{\left( C _{1}+ C _{2}\right)} V\)
Step-by-step Solution
Detailed explanation
Charge on capacitor \(C _{2}\) \(=\frac{ C _{2} \times Q _{\text {total }}}{ C _{\text {total }}}=\frac{ C _{2}\left[ C _{1} V \right]}{ C _{1}+ C _{2}}=\frac{ C _{1} C _{2} V }{ C _{1}+ C _{2}}\)
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