JEE MAIN - Physics (2022 - 26th June Morning Shift - No. 9)
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 :
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$${{{C_1}{C_2}} \over {({C_1} + {C_2})}}V$$
$${{({C_1} + {C_2})} \over {{C_1}{C_2}}}V$$
$$({C_1} + {C_2})V$$
$$({C_1} - {C_2})V$$
Explanation
$$
V_{\text {common }}=\frac{C_1 V}{C_1+C_2}
$$
$\Rightarrow$ Charge on capacitor $C_2$
$\Rightarrow$ Charge on capacitor $C_2$
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