A 10 Ω resistor and a 40 μF capacitor are connected in parallel to a 110 V, 50 Hz AC supply. The total current is:
Correct answer: C. 11.5 A
- A. 9.5 A
- B. 10.5 A
- C. 11.5 A
- D. 12.5 A
Explanation
For resistor, IR = 110/10 = 11 A. For capacitor, Xc = 1/(2π × 50 × 40 × 10⁻⁶) ≈ 79.58 Ω, I = 110/79.58 ≈ 1.38 A. Total current I = √(11² + 1.38²) ≈ √(121 + 1.90) ≈ 11.1 A, closest to 11.5 A.
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About Alternating Current
Alternating current varies periodically, so phase relationships become essential when current passes through a resistor, capacitor or inductor. The chapter distinguishes resistance from capacitive and inductive reactance, explains phase lead and lag, and connects changing electric and magnetic fields with the production and properties of electromagnetic waves.
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