Parallel RC phase behavior — as the capacitance C increases in a parallel RC circuit excited by a sinusoid, how does the overall phase angle between the source voltage and total current change?

Introduction / Context:
Phase angle in parallel circuits depends on the relative contributions of conductance and susceptance. In a parallel RC network, changing the capacitor value alters the reactive branch current dramatically, which changes the total current phasor and the angle it makes relative to the impressed voltage.

Given Data / Assumptions:

• Sinusoidal steady-state excitation.
• Parallel RC: one branch is a resistor R (conductance G = 1/R), the other is a capacitor C (susceptance Bc = omega * C).
• Phase angle theta is the angle of total current relative to voltage (voltage is the reference in admittance analysis).

Concept / Approach:
Total admittance is Y = G + jBc, where Bc = omega * C. As C increases, Bc increases proportionally, so the reactive component of current grows relative to the resistive component. The phase angle of the total current is theta = arctan(Bc / G). Therefore, increasing C increases Bc, which increases |theta| toward +90 degrees (capacitive behavior).

Step-by-Step Solution:

Verification / Alternative check:
Consider extreme cases. If C is very small, Bc ≈ 0 and the circuit is nearly resistive (theta ≈ 0). If C is very large, Bc dominates and theta approaches +90 degrees, indicating strongly capacitive current.

Why Other Options Are Wrong:

• “More inductive” contradicts the nature of a capacitor.
• “No change” ignores the dependence of Bc on C.
• “Reverses sign / pure resistor” has no basis; adding capacitance increases capacitive character.

Common Pitfalls:
Confusing series and parallel behavior or using reactance Xc = 1/(omega*C) directly in parallel when admittance is more convenient.

Final Answer:
Correct — increasing C increases the capacitive phase angle magnitude.