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Q.

Two capacitors having capacitance C₁and C₂respectively are connected as shown in figure. Initially, capacitor C₁is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C₁ is now connected to uncharged capacitor C₂ by closing the switch S. The amount of charge on the capacitor C₂, after equilibrium is:

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a

$\frac{C_{1} C_{2}}{(C_{1} + C_{2})} V$

b

$\frac{(C_{1} + C_{2})}{C_{3} C_{3}} V$

c

$(C_{1} + C_{2}) V$

d

$(C_{1} - C_{2}) V$

answer is A.

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Detailed Solution

Initially the total charge on the C₁ which is C₁V,

Now this charge is distributed among both the capacitors.

Common potential is $\frac{C_{1} V}{C_{1} + C_{2}}$

So charge on capacitor C₂is given by $\frac{C_{1} C_{2}}{C_{1} + C_{2}} V$

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Two capacitors having capacitance C1 and C2 respectively are connected as shown in figure. Initially, capacitor C1 is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C1 is now connected to uncharged capacitor C2 by closing the switch S. The amount of charge on the capacitor C2, after equilibrium is: