Series RC driven by AC: A 12 kΩ resistor is in series with a 0.02 µF capacitor across a 1.2 kHz source. If the current is I = 0.3∠0° mA (polar form reference), what is the source voltage in polar form (magnitude and phase)?

Introduction / Context:In AC steady-state analysis, the source voltage phasor equals the product of current phasor and total impedance. For a series resistor–capacitor (RC) network, the impedance is Z = R − jXc with Xc = 1 / (2π f C). Translating between rectangular and polar forms is a core skill used in phasor diagrams and impedance matching.

Given Data / Assumptions:

• R = 12 kΩ; C = 0.02 µF; f = 1.2 kHz.
• Current phasor: I = 0.3∠0° mA = 0.0003∠0° A.
• Assume ideal components (no parasitics).

Concept / Approach:Compute capacitive reactance Xc = 1 / (2π f C). Then Z_total = R − jXc. The source voltage phasor is V = I * Z. Magnitude |V| = |I| * |Z| and angle ∠V = ∠I + ∠Z. Although the options list magnitudes, we will also state the phase for completeness.

Step-by-Step Solution:

Verification / Alternative check:Rectangular multiplication: V ≈ I * (R − jXc) = 0.0003 * (12000 − j6629.7) ≈ (3.6 − j1.989) V. Polar magnitude √(3.6^2 + 1.989^2) ≈ 4.11 V, angle arctan(−1.989/3.6) ≈ −28.9°, matching the polar computation.

Why Other Options Are Wrong:

• 45.6 V and 411.3 V: Orders of magnitude too high given a 0.3 mA current through a ~13.7 kΩ impedance.
• 0.411 V: Off by a factor of 10, likely from milliampere mishandling.

Common Pitfalls:

• Mishandling microfarads and kilohms; always convert to SI before computing Xc.
• Forgetting that a capacitor introduces a negative imaginary component in impedance.

Final Answer:4.11 V (with phase approximately −29°).