Parallel RL at low power factor: What is the overall phase angle (in degrees) for a 24 Vac parallel RL circuit with R = 45 Ω and XL = 1100 Ω?

Introduction / Context:For parallel RL circuits, the source voltage is common, while currents through the resistor and inductor differ in phase. The overall current lags the voltage slightly when the inductor branch susceptance is small relative to conductance. Computing the phase angle from admittance is a standard AC analysis task.

Given Data / Assumptions:

• R = 45 Ω, XL = 1100 Ω at 24 Vac (RMS).
• Ideal components; frequency implicit in XL value.
• We seek the phase angle θ of total current relative to voltage.

Concept / Approach:Use admittance: Y = G + jB, where G = 1/R and B = −1/XL (inductive susceptance is negative). The phase angle of the total current is θ = arctan(B/G). For an inductive parallel circuit, θ is negative (current lags). We typically report the magnitude for comparison with given options.

Step-by-Step Solution:Compute conductance: G = 1/R = 1/45 ≈ 0.02222 S.Compute susceptance: B = −1/XL = −1/1100 ≈ −0.0009091 S.Find angle: θ = arctan(B/G) = arctan(−0.0009091 / 0.02222) ≈ −2.34 degrees.Magnitude ≈ 2.34 degrees → closest option is 2.300 degrees (lagging).

Verification / Alternative check:Compute power factor: pf = cos(|θ|) ≈ cos(2.34°) ≈ 0.9992, indicating a very small reactive component, consistent with a large XL in parallel.

Why Other Options Are Wrong:0.001° and 89.900° imply negligible or extreme phase shifts not matching B/G.

87.600° is characteristic of nearly pure reactance, not this case.

45.000° would require G = |B|, which is not true here.

Common Pitfalls:Confusing series and parallel formulas. In parallel, compute with admittances (G and B); in series, use impedances (R and X) and their phasor angle.

Final Answer:2.300 degrees