Parallel RL phase behavior: In a parallel RL circuit excited by a sinusoidal voltage, what happens to the phase angle between the total current and the applied voltage if the resistance R is increased (with L and frequency fixed)?

Introduction / Context:
Parallel RL networks are often analyzed more cleanly in the admittance domain. The total current leads the applied voltage by an angle set by the ratio of susceptance to conductance. Changing the resistance changes conductance, which directly affects the phase angle between total current and voltage.

Given Data / Assumptions:

• Parallel branches: a resistor R and an inductor L.
• Frequency and inductance are fixed; excitation is sinusoidal.
• Interest is the angle between total current and applied voltage.

Concept / Approach:
Work with admittance: Y = G − jB, where G = 1 / R and B = 1 / XL = 1 / (2 * pi * f * L). The phase of the total current relative to voltage is given by tan(phi) = B / G. Increasing R decreases G. With B fixed, B / G therefore increases, so |phi| increases. In a parallel RL, current leads voltage (inductive susceptance), so the lead increases in magnitude as R increases.

Step-by-Step Solution:

Verification / Alternative check:
Vector diagrams of branch currents show the resistive branch current shrinking with higher R, leaving the inductive branch current more dominant, so the phasor sum tilts further toward the inductive axis.

Why Other Options Are Wrong:
“Angle decreases” contradicts tan(phi) = R / XL. “Unchanged” ignores that G depends on R. “Resonant” behavior requires a capacitor; an RL parallel circuit cannot exhibit resonance.

Common Pitfalls:
Applying series RL intuition to a parallel network; mixing impedance and admittance phase formulas.

Final Answer:
The magnitude of the phase angle increases (more inductive current lead over conductance)