Physics - Capacitor Question with Solution | TestHub

In steady state, capacitors behave as open circuits; current flows only through the resistors.

From the circuit, potential difference across the 2 μF capacitor equals that across the 3 Ω resistor: ${\mathrm{V}}_{2\mu F}={\mathrm{V}}_{3\Omega }=3\mathrm{V}$

$q_{2\mu F}=2\times 3=6\mu C$ → (A) is correct.

Also ${\mathrm{V}}_{4\mu F}={\mathrm{V}}_{4\Omega }=4\mathrm{V}$ so $U=\frac{1}{2}\times 4\mu F\times (4V{)}^2=32\mu J$ → (B) is correct.

The 1 Ω and 3 Ω resistors lie in the same current path, so the current through them is equal → (C) is correct.

The 3 μF capacitor is in parallel with the 1 Ω resistor, so the potential differences are equal → (D) is correct.

Hence (A), (B), (C) and (D).