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A two-branch parallel AC circuit has Branch A with R_A = 20 ohm and X_L = 15 ohm, and Branch B with R_B = 6 ohm and X_C = 8 ohm. The total current from the source is I_total = sqrt(29) A. Find the current through the inductor branch (Branch A).

1. Zero
2. 4 A
3. 2 A
4. 5 A

Correct answer: 2 A

Solution

Z_A = 25 ohm, Z_B = 10 ohm. I_A = V/25, I_B = V/10. Phase angle of branch A: phi_A = arctan(15/20) = arctan(3/4) (lagging). Phase of branch B: phi_B = arctan(-8/6) = arctan(-4/3) (leading). Phasor sum must have magnitude sqrt(29). With V as reference, I_A = V/25 at -arctan(3/4) and I_B = V/10 at +arctan(4/3). Setting |I_total| = sqrt(29): solve for V then find I_A = V/25.

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