Series RLC at 5 kHz: A 3 kΩ resistor, a 0.05 µF capacitor, and a 120 mH inductor are connected in series to a 20 V (rms) source operating at 5 kHz. What is the circuit impedance expressed in polar form (magnitude and implied angle)?

Introduction / Context:This problem checks phasor-based AC analysis for a series RLC network, including calculating inductive and capacitive reactances at a specified frequency, forming the complex impedance, and interpreting the result in polar form. Such computations are routine in filter design, resonance analysis, and impedance matching.

Given Data / Assumptions:

• R = 3 kΩ = 3000 Ω.
• L = 120 mH = 0.12 H; C = 0.05 µF = 5 × 10^-8 F.
• Frequency f = 5 kHz (sinusoidal steady state).
• Series connection; ideal components (no parasitics beyond L and C).

Concept / Approach:For a series RLC circuit, the impedance is Z = R + j(XL − XC). Compute XL = 2π f L and XC = 1 / (2π f C). Then find magnitude |Z| = √(R^2 + (XL − XC)^2) and the phase angle θ = arctan((XL − XC)/R).

Step-by-Step Solution:

Verification / Alternative check:Because XL ≫ XC and both are comparable to R, the magnitude must exceed R and be in the several-kilohm range. A quick check with vector addition confirms a hypotenuse bigger than 3 kΩ, close to 4.3 kΩ as computed.

Why Other Options Are Wrong:

• 636 Ω: This is the capacitor’s reactance magnitude, not the total series impedance.
• 3,769 Ω: This is the inductor’s reactance magnitude, not |Z|.
• 433 Ω: An order-of-magnitude error; series contributions make |Z| much larger.

Common Pitfalls:

• Forgetting to subtract XC from XL (sign matters).
• Mixing units (µF, kHz) without converting to SI before calculation.

Final Answer:4,337 Ω (corresponding to approximately 4.337 kΩ ∠ +46.8°).