Question

11(b) A 4.0 μF capacitor is connected in series with a 400 Ω resistor. The connection is placed across a source of root-mean-square value of 60V and alternating frequency (1500/π) Hz. Calculate: (i) the impedance of the circuit; (ii) the potential difference across the capacitor. [π = 3.14]

Accepted Answer

(i) Xc = 1/(2πfC) = 1/(2×(1500/π)×π×4×10⁻⁶) = 1/(2×1500×4×10⁻⁶) = 1/0.012 ≈ 83.3 Ω.
Impedance, Z = √(R²+Xc²) = √(400²+83.3²) = √166,938.9 ≈ 408.6 Ω.
(ii) Current, I = Vrms/Z = 60/408.6 ≈ 0.1469 A. Potential difference across capacitor, Vc = I×Xc = 0.1469×83.3 ≈ 12.24 V.

WAEC Physics 2025 Theory — Question 23

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