In the given figure, the charge stored in 6 μ F capacitor, when points A and B are joined by a connecting wire is __________ μ C.

Solution

Physics sir · 4 steps

Once the capacitors are charged, no current flows through them: find the potential at A–B, then use Q = CV.

1. Steady state. After a long time, a capacitor passes no current. The only path from 9 V to ground is top → 6 Ω → A → wire → B → 3 Ω → ground.
2. Current and potential. The two resistors are in series, so I = 9/(6 + 3) = 1 A. The drop across 6 Ω is 6 V, so A and B both sit at 9 - 6 = 3 V.
3. Voltage on 6 μF. The 6 μF capacitor joins the top wire at 9 V to B at 3 V. The potential difference across it is 9 - 3 = 6 V.

   Q = CV = 6 μF × 6 V = 36 μC

4. So. The charge stored in the 6 μF capacitor is 36 μC.

The common slip: 18. Using the 3 V potential at B as the voltage across the 6 μF capacitor gives 18. The capacitor sees the difference between its plates, 9 - 3 = 6 V.

The idea to keep. In a DC circuit at steady state, treat every capacitor as a break. Find the node potentials from the resistor current, then the charge on each capacitor is C times the difference across its two plates.

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