Phase angle of a series RC network with specified reactance ratio A sinusoid drives a series RC circuit where the capacitive reactance magnitude is four-thirds of the resistance: |XC| : R = 4 : 3. What is the phase angle between the source voltage and the circuit current (current leads negative angle; report the magnitude)?

Introduction / Context:
For a series RC circuit, the current leads the total voltage by an angle whose magnitude depends on the ratio of capacitive reactance to resistance. Designers use this relationship to set filter corner slopes, timing responses, and phase-shift networks. Here we compute the phase given a specific |XC| to R ratio.

Given Data / Assumptions:

• Topology: series RC.
• Ratio: |XC| : R = 4 : 3 (so |XC| = 4k, R = 3k for some common factor k).
• Steady-state sinusoidal operation with ideal components.
• Phase angle φ defined as the angle between total voltage and current; for RC, current leads voltage by |φ|. We report the magnitude.

Concept / Approach:
In a series RC, the impedance is Z = R − j|XC|. The phase angle between V and I is φ = arctan(−|XC|/R); current leads by |φ|. The magnitude is |φ| = arctan(|XC|/R). With |XC|/R known, we can directly compute the arctangent.

Step-by-Step Solution:

Verification / Alternative check:
Construct a right triangle with adjacent = R = 3 and opposite = |XC| = 4. The hypotenuse is 5. The angle opposite the “4” side is arcsin(4/5) = 53.13°, matching the arctan result. Impedance magnitude is 5k (arbitrary k), consistent with the triangle.

Why Other Options Are Wrong:

Common Pitfalls:
Mixing sign and magnitude conventions; forgetting that in series RC, current leads voltage; confusing |XC|/R with R/|XC|.

Final Answer:
53.13°