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Question

Two capacitors of \( 15 \, \text{pF} \) each are connected in series. What is the equivalent
capacitance?

Options

Choose one · Correct answer highlighted

Explanation

**Parallel combination** has same voltage V across each, charges Q_i = C_i V, total Q = Σ Q_i = V Σ C_i, so C_eq = Σ C_i, sum of capacitances. For five 10 μF in series, 1/C=5/10=0.5, C_eq=2 μF, much smaller than individual, while parallel would be 50 μF. (1/C) = (1/15) + (1/15) = (2/15) . C = (15/2) = 7.5 pF . 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 7.5 pF follows, reflecting potential-capacitance relations.

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