In the given circuit, C₁=2 μ F, C₂=0.2 μ F, C₃=2 μ F, C₄=4 μ F, C₅=2 μ F, C₆=2 μ F, The charge stored on capacitor C₄ is ____________ μ C.

Solution

Physics sir · 5 steps

Collapse the circuit from the far right, find the voltage across C₂, then come back out to C₄.

1. Right loop. C₃, C₄ and C₅ form one path with no junction in between, so they are in series: 1/C = 1/2 + 1/4 + 1/2 = 5/4, which gives C₃₄₅ = 0.8 μF.
2. Add C2. This branch and C₂ connect the same top and bottom nodes, so they are in parallel: Cₚ = 0.8 + 0.2 = 1 μF.
3. Whole circuit. C₁, Cₚ and C₆ are in series with the battery: 1/C_(eq) = 1/2 + 1/1 + 1/2 = 2, so C_(eq) = 0.5 μF. The battery delivers Q = 0.5 × 10 = 5 μC.

   Vₚ = Q/Cₚ = (5 μC)/(1 μF) = 5 V

4. Back to C4. The same 5 V sits across the C₃–C₄–C₅ branch, so its charge is 0.8 × 5 = 4 μC. Series capacitors carry equal charge, so C₄ holds all of it.
5. So. The charge stored on C₄ is 4 μC.

The common slip: 5. Taking the battery's full charge of 5 μC as the charge on C₄ gives 5. That charge splits at the parallel junction, and 1 μC of it goes to C₂.

The idea to keep. For a ladder of capacitors, reduce from the end farthest from the battery, find the total charge, then unfold step by step: parallel parts share voltage, series parts share charge.

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