Five capacitors are connected across a 50 V battery as in the diagram. Four capacitors C₁ = C₂ = C₃ = C₄ = 10 μF form a series branch, while C₅ = 2.5 μF forms a parallel branch. Determine the equivalent capacitance and charge on each capacitor. Image
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Question

Five capacitors are connected across a 50 V battery as in the diagram. Four capacitors C₁ = C₂ = C₃ = C₄ = 10 μF form a series branch, while C₅ = 2.5 μF forms a parallel branch. Determine the equivalent capacitance and charge on each capacitor. Image

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G1 Physics · Accepted Answer

Step 1: The four 10 μF capacitors in series have 1/C(series) = 1/10 + 1/10 + 1/10 + 1/10 = 4/10 (μF)⁻¹. Thus C(series) = 2.5 μF. Step 2: C₅ = 2.5 μF is parallel with this series string, so C(equivalent) = C(series) + C₅ = 5 μF. Step 3: Each parallel branch has the full battery voltage 50 V. Charge on series branch Q(series) = C(series)V = 2.5 × 50 = 125 μC. Step 4: In a series connection, every capacitor carries the same charge; therefore Q₁ = Q₂ = Q₃ = Q₄ = 125 μC. Step 5: Q₅ = C₅V = 2.5 × 50…