Series RC impedance in rectangular form: A 47 Ω resistor is in series with a capacitor whose reactance is 150 Ω. What is the total impedance Z expressed in rectangular form?

Introduction / Context:
Representing impedances in rectangular form (R ± jX) is central to AC circuit analysis. For capacitors, the reactance is negative imaginary; for inductors, it is positive imaginary. This sign convention enables straightforward phasor addition of series elements.

Given Data / Assumptions:

• Resistor R = 47 Ω (real, in-phase with voltage).
• Capacitive reactance magnitude X_C = 150 Ω.
• Series connection; sinusoidal steady-state.

Concept / Approach:
Impedances in series add directly as complex numbers. For a capacitor, X_C is represented as −jX_C. Therefore, Z_total = R + (−jX_C) = R − jX_C. Avoid converting to polar unless requested; the problem asks specifically for rectangular form.

Step-by-Step Solution:

Verification / Alternative check:
Polar magnitude: |Z| = √(47^2 + 150^2) ≈ √(2209 + 22500) ≈ √24709 ≈ 157.2 Ω; angle ≈ arctan(−150/47) ≈ −72.7°. Converting back to rectangular reproduces 47 − j150 Ω, confirming correctness.

Why Other Options Are Wrong:

• Z = 47 Ω + j150 Ω: Wrong sign for capacitive reactance (that would be inductive).
• Z = 197 Ω or Z = 103 Ω: These look like real-only magnitudes; impedance has both real and imaginary parts here.

Common Pitfalls:

• Using +j for capacitors instead of −j.
• Treating reactance magnitudes as simple arithmetic with resistance without complex notation.

Final Answer:
Z = 47 Ω – j150 Ω