A capacitor of 50 μF is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is ______________μC.

Solution

Physics sir · 4 steps

At steady state no current enters the capacitor, so it takes the 2 V across the bottom resistor.

1. Steady state. After a long time the capacitor is fully charged and carries no current. All the current goes through the three 2 kΩ resistors in series, so the total resistance is 6 kΩ.
2. Find the current. I = (6 V)/(6 kΩ) = 1 mA. The drop across each resistor is 1 mA × 2 kΩ = 2 V.
3. Voltage on capacitor. The capacitor is in parallel with the bottom resistor only, so V_C = 2 V. The long plate of the battery is on top, so current flows down through the resistors and the upper plate sits at the higher potential.

   Q = CV_C = 50 μF × 2 V = 100 μC

4. So. The upper plate carries +100 μC, so the answer is 100.

The common slip: 300. Using the full battery voltage of 6 V across the capacitor gives 300. The capacitor only sees the 2 V across the bottom resistor.

The idea to keep. In a DC circuit at steady state, treat a capacitor as an open switch. Find the current through the resistors, then the capacitor takes the voltage of whatever it is in parallel with.

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