Series RLC at 200 Hz: For a circuit with R = 12 Ω, L = 10 mH, and C = 80 µF connected to a 200 Hz, 15 V AC source, determine the total impedance in polar form (magnitude ∠ angle, in ohms).

Introduction / Context:Analyzing impedance in a series RLC circuit is fundamental to AC circuit design. By computing the resistive and reactive parts at a given frequency, we can express the total impedance in polar form, which clearly shows magnitude and phase angle for phasor calculations.

Given Data / Assumptions:

• Frequency f = 200 Hz, source 15 V (voltage value is not required to find Z).
• R = 12 Ω.
• L = 10 mH.
• C = 80 µF.
• Ideal components; steady-state sinusoidal operation.

Concept / Approach:

Compute inductive reactance XL = 2 * π * f * L, capacitive reactance Xc = 1 / (2 * π * f * C). For series RLC, the net reactance is X = XL − Xc and the impedance is Z = R + jX. Convert rectangular Z to polar form: |Z| = √(R^2 + X^2), angle θ = arctan(X / R).

Step-by-Step Solution:

Verification / Alternative check:

Because XL is slightly larger than Xc, the circuit is slightly inductive, so a small positive phase angle is expected. The magnitude is close to R because X is small compared to R, which matches the computed 12.28 Ω.

Why Other Options Are Wrong:

12.57∠12.34° Ω and 12.62∠12.34° Ω overstate the magnitude. 9.95∠12.34° Ω wrongly uses Xc as the magnitude of Z.

Common Pitfalls:

Using X = Xc − XL (sign error), forgetting unit conversions (mH, µF), or mixing degrees and radians in the arctangent.

Final Answer:

12.28∠12.34° Ω