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An ideal inductor draws 10 A when connected to a 125 V, 50 Hz supply. A pure resistor across the same source draws 12.5 A. If the inductor and resistor are connected in series across a 100*sqrt(2) V, 40 Hz supply, find the current through the circuit.

1. 10 A
2. 12.5 A
3. 20 A
4. 25 A

Correct answer: 10 A

Solution

At 50 Hz the inductor reactance XL = 125/10 = 12.5 ohm and the resistance R = 125/12.5 = 10 ohm. Reactance scales with frequency, so at 40 Hz XL' = 12.5*(40/50) = 10 ohm. Series impedance Z = sqrt(R² + XL'²) = sqrt(10² + 10²) = 10*sqrt(2) ohm. With supply V = 100*sqrt(2) V, current I = V/Z = 100*sqrt(2)/(10*sqrt(2)) = 10 A.

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