The charge accumulated on the capacitor connected in the following circuit is _______ μ C (Given C=150 μ F)

Solution

Physics sir · 4 steps

Once charged, the capacitor carries no current, so find the potentials at A and B from the two resistor arms.

1. Steady state. After a long time no current flows through the capacitor branch. The circuit becomes two separate arms across the 10 V battery: R₁ in series with R₂ through A, and R₃ in series with R₄ through B.
2. Potential at A. Take the battery's negative terminal (bottom) as 0 V. Arm R₁R₂ has 3 Ω, so the current is 10/3 A. Then V_A = 10/3 × 2 = 20/3 V.
3. Potential at B. Arm R₃R₄ has 10 Ω, so the current is 1 A. Then V_B = 1 × 4 = 4 V.

   Q = C (V_A - V_B) = 150 μF × (20/3 - 4) V = 150 × 8/3 μC

4. So. The capacitor sees 8/3 V, and its charge is Q = 400 μC.

The common slip: 1500. Putting the full 10 V across the capacitor gives 1500. The capacitor only sees the difference between the potentials at A and B.

The idea to keep. In a DC circuit at steady state a capacitor acts like an open switch. Find the current in the resistors without it, then use the potential difference across its two ends.

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