At steady state the charge on the capacitor, as shown in the circuit below, is _________ μC.

Solution

Physics sir · 4 steps

At steady state the capacitor passes no current, so find the current first, then read the voltage across 10 Ω.

1. Capacitor branch. Once the capacitor is fully charged, no current flows through its branch. The whole current goes through the 10 Ω and 15 Ω resistors, which are in series with the battery.
2. Find the current. Total resistance is 10 + 15 = 25 Ω, so the current is I = 5/25 = 0.2 A.
3. Voltage across capacitor. The capacitor is connected across the ends of the 10 Ω resistor, so it has the same voltage: V = 0.2 × 10 = 2 V.

   Q = CV = 8 μF × 2 V = 16 μC

4. So. The charge on the capacitor at steady state is 16 μC.

The common slip: 40. Taking the full battery voltage of 5 V across the capacitor gives 8 × 5 = 40 μC. The 15 Ω resistor takes 3 V of it, which leaves only 2 V across the capacitor.

The idea to keep. In a DC circuit at steady state, treat a capacitor as an open switch. Its voltage equals the voltage across whatever it is connected in parallel with.

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