JEE PYQ: Capacitance Question 5
Question 5 - 2021 (17 Mar Shift 2)
A $2\mu$F capacitor $C_1$ is first charged to a potential difference of 10 V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor $C_2$ of $8\mu$F. The charge in $C_2$ on equilibrium condition is ____$\mu$C. (Round off to the Nearest Integer)
Show Answer
Answer: 16
Solution
$20 = (C_1 + C_2)V \Rightarrow V = 2$ volt. $Q_2 = C_2 V = 8 \times 2 = 16\mu$C.
BETA
