A 24 V(rms) parallel RL circuit has R = 45 Ω and inductive reactance XL = 1100 Ω. What is the magnitude of the phase angle between the source voltage and the total current?

Introduction:
In a parallel RL circuit, voltage is common to both branches, while currents split through the resistive and inductive paths. The overall current leads or lags the voltage by a small angle determined by the ratio of susceptance to conductance. This problem targets computing the magnitude of that phase angle from component values.

Given Data / Assumptions:

• Parallel RL circuit with ideal components.
• R = 45 Ω, XL = 1100 Ω.
• Source voltage magnitude is 24 V(rms); however, the phase angle depends on R and XL, not directly on V.

Concept / Approach:
Work with admittances. For parallel RL: G = 1 / R and B = −1 / XL (inductive). The magnitude of the phase angle φ between voltage and total current is |φ| = arctan(|B| / G) = arctan( (1 / XL) / (1 / R) ) = arctan(R / XL ).

Step-by-Step Solution:
1) Compute G = 1 / 45 ≈ 0.02222 S.2) Compute |B| = 1 / 1100 ≈ 0.0009091 S.3) Ratio |B| / G = (1 / 1100) / (1 / 45) = 45 / 1100 ≈ 0.04091.4) Angle magnitude: |φ| = arctan(0.04091) ≈ 2.35°, which rounds to about 2.3°.

Verification / Alternative check:
Using |φ| = arctan(R / XL) gives the same result directly: arctan(45 / 1100) ≈ 2.35° → 2.3° (rounded).

Why Other Options Are Wrong:
0.001°: unrealistically small; would require XL ≫≫ R.87.6° and 89.9°: imply an almost purely reactive or extremely unbalanced circuit; not supported by R/XL ≈ 0.041.

Common Pitfalls:
Using series formulas tan φ = XL / R (which applies to series RL) instead of the admittance ratio for parallel RL. Always switch to conductance/susceptance for parallel analysis.

Final Answer:
2.3°