A single-phase circuit has R = 20 Ω and an inductor with X_L = 40 Ω. If the supply is 120 V RMS, what is the current magnitude?

A single-phase circuit has R = 20 Ω and an inductor with X_L = 40 Ω. If the supply is 120 V RMS, what is the current magnitude?

• A. 2.12 A
• B. 2.68 A
• C. 3.18 A
• D. 4.00 A

Answer: (B)

In AC circuits, current magnitude is determined by the impedance, which combines resistance and inductive reactance. For a resistor and inductor in series, the impedance magnitude is Z = sqrt(R^2 + X_L^2). The current magnitude is I = V_RMS / Z. With R = 20 Ω and X_L = 40 Ω, Z = sqrt(20^2 + 40^2) = sqrt(2000) ≈ 44.72 Ω. So I ≈ 120 / 44.72 ≈ 2.68 A. The voltage and current are out of phase by φ where tan φ = X_L / R = 40/20 = 2, giving φ ≈ 63.4 degrees (current lags voltage).

In AC circuits, current magnitude is determined by the impedance, which combines resistance and inductive reactance. For a resistor and inductor in series, the impedance magnitude is Z = sqrt(R^2 + X_L^2). The current magnitude is I = V_RMS / Z. With R = 20 Ω and X_L = 40 Ω, Z = sqrt(20^2 + 40^2) = sqrt(2000) ≈ 44.72 Ω. So I ≈ 120 / 44.72 ≈ 2.68 A. The voltage and current are out of phase by φ where tan φ = X_L / R = 40/20 = 2, giving φ ≈ 63.4 degrees (current lags voltage).