In an electrical circuit drawn below the amount of charge stored in the capacitor is _______ μ C.

Solution

Physics sir · 4 steps

In steady state no current flows through the capacitor, so drop that branch and find the voltage across it.

1. Steady state. Once the capacitor is fully charged, no current flows through its branch. So R₂ carries no current and has no voltage drop across it.
2. Find the current. The current then flows only around the outer loop: battery, R₁ and R₃ in series. That gives I = 10/(4 + 6) = 1 A.
3. Voltage across C. The capacitor branch joins the same two nodes as R₃, so it is in parallel with R₃. With no drop across R₂, V_C = I R₃ = 1 × 6 = 6 V.

   Q = C V_C = 10 μF × 6 V = 60 μC

4. So. The charge stored in the capacitor is 60 μC.

The common slip: 100. Taking the full battery voltage of 10 V across the capacitor gives 100, not 60. Only the 6 V across R₃ appears across C.

The idea to keep. For a DC circuit in steady state, treat a capacitor branch as an open circuit. Its voltage equals the voltage across whatever it sits in parallel with, because the resistor in series with it drops nothing.

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