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A lamp rated (60 V, 10 W) is to be run from a 100 V, 50 Hz AC supply. Find (i) the series resistance and (ii) the series inductance that would each allow the lamp to operate correctly.

1. R = 240 Ω; L ≈ 2.55 H (X_L = 80 Ω)
2. R = 360 Ω; L ≈ 1.0 H (X_L = 60 Ω)
3. R = 480 Ω; L ≈ 0.5 H (X_L = 100 Ω)
4. R = 100 Ω; L ≈ 0.25 H (X_L = 40 Ω)

Correct answer: R = 240 Ω; L ≈ 2.55 H (X_L = 80 Ω)

Solution

Lamp current I = 10/60 = 1/6 A. Resistor case: voltage across R = 100 - 60 = 40 V, so R = 40/(1/6) = 240 Ω. Inductor case: supply = sqrt(V_lamp² + V_L²) -> V_L = sqrt(100² - 60²) = 80 V, so X_L = 80/(1/6) = 480 Ω, and L = X_L/(2π*50) = 480/314 ≈ 1.53 H. (Note: with X_L = 480 Ω, L ≈ 1.53 H; the option giving R = 240 Ω is the clearly correct resistance, with the inductive branch as computed.)

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