In the network shown below, the charge accumulated in the capacitor in steady state will be:

Solution

Physics sir · 4 steps

In steady state the capacitor branch carries no current, so it just copies the voltage of the 6 Ω beside it.

1. Capacitor branch. Once the capacitor is fully charged, no current flows through it. So no current flows in the middle branch either, and its 6 Ω has no voltage drop across it.
2. Find the current. The current only goes round the outer loop: the battery, the 4 Ω and the bottom 6 Ω are in series. That gives I = 3/(4 + 6) = 0.3 A.
3. Capacitor voltage. The middle branch and the bottom 6 Ω join the same left and right wires, so they have the same voltage across them. That voltage is 0.3 × 6 = 1.8 V, and all of it sits across the capacitor.

   Q = CV = 4 μF × 1.8 V = 7.2 μC

4. So. The steady-state charge on the capacitor is 7.2 μC, which is option (b).

The idea to keep. At steady state, remove the capacitor branch to find the currents. Then put it back and read its voltage from the parts it is in parallel with. A resistor in series with it drops no voltage.

Answer from the NTA final answer key. Solution by PrepWiser.