In the circuit shown in the figure, the total charge is 750 μC and the voltage across capacitor C₂ is 20 V. Then the charge on capacitor C₂ is :

Solution

Physics sir · 3 steps

All 750 μC passes through C1 and then splits between C2 and C3, so find C3's share and subtract.

1. Read the circuit. C₂ and C₃ are in parallel, and that pair is in series with C₁. So the full 750 μC sits on C₁, and the same total is shared between C₂ and C₃.
2. Same voltage. Parallel capacitors have the same voltage across them. So C₃ also has 20 V across it, and its charge is Q₃ = C₃ V = 8 μF × 20 V = 160 μC.

   Q₂ = Qₜₒₜₐₗ - Q₃ = 750 μC - 160 μC

3. So. Q₂ = 590 μC, which is option (c).

The idea to keep. In series, the charge is the same on each part; in parallel, the charge splits but the voltage is shared. Use the shared voltage to find one branch, then subtract from the total.

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