Capacitors

JEE Main level, single correct, medium

165 of 180

Question 165 of 180

Two capacitors having capacitance C1{C}_{1} and C2{C}_{2} respectively are connected as shown in figure. Initially, capacitor C1{C}_{1} is charged to a potential difference VV volt by a battery. The battery is then removed and the charged capacitor C1{C}_{1} is now connected to uncharged capacitor C2{C}_{2} by closing the switch SS. The amount of charge on the capacitor C2{C}_{2}, after equilibrium, is

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Answer: (A) C1C2(C1+C2)V\dfrac{{C}_{1}{C}_{2}}{\left({C}_{1}+{C}_{2}\right)}V

The initial charge on the capacitor C1{C}_{1} will be Q=C1VQ={C}_{1}V. If after closing the switch the charge flown through the circuit is qq, the charge on C2{C}_{2} will be qq and C1{C}_{1} will be Q−qQ-q. The potential across the capacitors will be same. Hence,
Q−qC1=qC2⇒q=C2QC1+C2⇒q=C1C2C1+C2V\dfrac{Q-q}{{C}_{1}}=\dfrac{q}{{C}_{2}} \Rightarrow q=\dfrac{{C}_{2}Q}{{C}_{1}+{C}_{2}} \Rightarrow q=\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}V

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