Calculate the amount of charge on capacitor of 4μF. The internal resistance of battery is 1Ω :

Solution

Physics sir · 4 steps

In steady state no current flows through the capacitor branch, so find the battery's terminal voltage first.

1. Steady state. Once the capacitors are fully charged, no current flows through the top branch. The only current loop is the battery with its 1 Ω internal resistance and the 4 Ω resistor.
2. Current and voltage. I = 5/(4 + 1) = 1 A. The terminal voltage is V = 5 - 1 × 1 = 4 V. The top branch is connected across the same two nodes, so it also sees 4 V. With no current, the 6 Ω resistor has no voltage drop.
3. Combine capacitors. The two 2 μF capacitors in parallel give 4 μF. That is in series with the 4 μF capacitor, so the branch capacitance is C = (4 × 4)/(4 + 4) = 2 μF.

   Q = C V = 2 μF × 4 V = 8 μC

4. So. Capacitors in series carry the same charge, so the 4 μF capacitor holds 8 μC, which is option (a).

The idea to keep. In a DC circuit at steady state, a capacitor branch carries no current. Get the voltage across that branch from the resistive part of the circuit, and include internal resistance. Then series capacitors share one charge.

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