As show in the figure, in steady state, the charge stored in the capacitor is ____________ × 10⁻⁶ C.

Solution

Physics sir · 4 steps

Once the capacitor is full, its branch carries no current, so it just copies the voltage across R.

1. Steady state. A charged capacitor blocks DC, so no current flows in the bottom branch. With zero current, R' = 200 Ω has zero voltage drop across it.
2. Current in the loop. The only current path is the battery, r and R in series. So I = 10/(10 + 100) = 1/11 A.
3. Voltage across C. The capacitor branch is connected across the same two points as R, so V_C = IR = 1/11 × 100 = 100/11 V.

   Q = CV_C = 1.1 × 10⁻⁶ × 100/11 = 10 × 10⁻⁶ C

4. So. The charge stored is 10 μC, so the answer is 10.

The common slip: 11. Taking the full EMF of 10 V across the capacitor ignores the drop across the internal resistance and gives Q = 11 μC, not 10.

The idea to keep. In DC steady state, treat a capacitor branch as an open circuit. Any resistor in series with it drops no voltage, so the capacitor's voltage equals the voltage across whatever it sits in parallel with.

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