In a series RC circuit, as the frequency of the applied sinusoidal voltage increases, how does the circuit phase angle θ (of impedance) change?

Introduction / Context:
The phase angle θ of a series RC circuit indicates how far the current leads the applied voltage. Understanding how θ varies with frequency is central to filter design and Bode-plot intuition.

Given Data / Assumptions:

• Series RC with resistance R and capacitance C fixed.
• Frequency f increases.
• θ is defined from the impedance Z = R − jXc: θ = arctan(−Xc/R) (capacitive angle is negative).

Concept / Approach:

As frequency increases, capacitive reactance Xc = 1/(2 * π * f * C) decreases. With a smaller Xc relative to R, the magnitude |θ| shrinks toward 0°, meaning the circuit becomes more resistive and the phase lag (negative angle) reduces in magnitude.

Step-by-Step Solution:

Verification / Alternative check:

At very high f, Xc → 0 and θ → 0°. At very low f, Xc ≫ R and θ → −90°. This monotonic trend confirms decreasing |θ| with increasing f.

Why Other Options Are Wrong:

'Increases' contradicts the trend. 'Remains the same' is not true because Xc changes with f. 'Becomes erratic' does not describe linear, time-invariant behavior.

Common Pitfalls:

Ignoring the negative sign (capacitive lag) and comparing signed angles instead of magnitude; mixing RC with RL behavior.

Final Answer:

decreases