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Question

Four capacitors of \( 15 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

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Explanation

**Charge conservation in series** explains same Q: when battery charges first plate, it induces -Q on second plate of same capacitor, which comes from next capacitor's plate, etc., so all have same magnitude Q, potential divides as per 1/C. (1/C) = (1/15) + (1/15) + (1/15) + (1/15) = (4/15) . C = (15/4) = 3.75 μF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 3.75 µF follows, reflecting potential-capacitance relations.

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