Obtain the equivalent capacitance of the network shown in figure. F...

Question

Obtain the equivalent capacitance of the network shown in figure. For a 300 V supply, determine the charge on each capacitor.

Accepted Answer

📌 Standard Network (C₁ = C₃ = C₄ = 100 pF, C₂ = 200 pF):

Step 1 — Identify Series/Parallel Groups:
• C₂ and C₃ are in series:
1/C₂₃ = 1/200 + 1/100 = 3/200 → C₂₃ = 200/3 pF

• C₂₃ and C₄ are in parallel:
C₂₃₄ = 200/3 + 100 = 500/3 pF

• C₁ and C₂₃₄ are in series:
1/C_eq = 1/100 + 3/500 = 5/500 + 3/500 = 8/500
C_eq = 500/8 = 62.5 pF

Step 2 — Total Charge:
Q_total = C_eq × V = 62.5 × 10⁻¹² × 300 = 18750 pC ≈ 1.875 × 10⁻⁸ C

Step 3 — Charge Distribution:
• Q on C₁ = Q_total (series) = 1.875 × 10⁻⁸ C
• Voltage across C₂₃₄ = V − V_C₁ = 300 − (Q/C₁) = 300 − 187.5 = 112.5 V
• Q on C₄ = C₄ × 112.5 = 100×10⁻¹² × 112.5 = 1.125 × 10⁻⁸ C
• Q on C₂ = Q on C₃ = Q_total − Q_C₄ = 0.75 × 10⁻⁸ C

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