Q 12-02-192JEE MainJEE Main 2019 (12 Apr, Shift 2)Easy
In the given circuit, the charge on the $4\ \mu$F capacitor will be:
Answer: (C) $24\ \mu$C
The top branch ($4\ \mu$F in series with $1 + 5 = 6\ \mu$F) is directly across the $10$ V cell; the $3\ \mu$F branch does not affect it.
$$C = \frac{4\times6}{4 + 6} = 2.4\ \mu\text{F},\qquad Q = 2.4\times10 = 24\ \mu\text{C}$$
In series, the $4\ \mu$F capacitor carries this charge.
Solution by Sreeraj P, M.Sc Physics