A 40 Ω resistor and a 50 mH inductor are connected in series to a 240 V, 60 Hz AC supply. The current is:
Correct answer: B. 5.8 A
- A. 4.8 A
- B. 5.8 A
- C. 6.8 A
- D. 7.8 A
Explanation
XL = 2π × 60 × 0.05 ≈ 18.85 Ω. Impedance Z = √(40² + 18.85²) ≈ √(1600 + 355.32) ≈ 41.2 Ω. Current I = 240/41.2 ≈ 5.83 A, closest to 5.8 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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