Q 12-03-265JEE MainJEE Main 2019 (9 Apr, Shift 1)Medium
Determine the charge on the capacitor in the following circuit:
Answer: (D) $200\ \mu\text{C}$
In steady state no current flows through the capacitor branch.
$2\ \Omega$ and $10\ \Omega$ in series ($12\ \Omega$) are in parallel with $4\ \Omega$: $\dfrac{12\times4}{16} = 3\ \Omega$. Total resistance $6 + 3 = 9\ \Omega$, so $I = 72/9 = 8\ \text{A}$.
Voltage across the $4\ \Omega$ resistor: $72 - 8\times6 = 24\ \text{V}$. Current through the $2\ \Omega$–$10\ \Omega$ branch: $24/12 = 2\ \text{A}$, so the capacitor (in parallel with $10\ \Omega$) has $V = 20\ \text{V}$:
$$Q = CV = 10\ \mu\text{F}\times20\ \text{V} = 200\ \mu\text{C}$$
Solution by Sreeraj P, M.Sc Physics