A 100 V RMS source is connected across two parallel branches: a 20 Ω resistor and an ideal inductor with Xʟ = 25 Ω. Determine the magnitude of the source current.
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A 100 V RMS source is connected across two parallel branches: a 20 Ω resistor and an ideal inductor with Xʟ = 25 Ω. Determine the magnitude of the source current.

G1 Physics · Accepted Answer

The resistor current is Iᵣ=100/20=5 A in phase with the voltage. The inductor current is Iʟ=100/25=4 A lagging by 90°. Therefore source current magnitude = √(5²+4²)=√41 A.