RL impedance calculation (numerical): A series RL circuit has resistance R = 60 Ω and inductive reactance XL = 92 Ω at a certain frequency. What is the magnitude of the total impedance Z?

Introduction / Context:In AC analysis, impedance combines resistance and reactance as perpendicular (orthogonal) components on the complex plane. For a series RL circuit, the magnitude of impedance is not the arithmetic sum R + XL but the vector sum. This question reinforces the correct computation for Z given R and XL.

Given Data / Assumptions:

• Series RL network with R = 60 Ω.
• Inductive reactance XL = 92 Ω at the operating frequency.
• Ideal components and sinusoidal steady state.

Concept / Approach:Because resistance (real axis) and inductive reactance (positive imaginary axis) are orthogonal, the magnitude of the series impedance is |Z| = sqrt(R^2 + XL^2). This follows directly from Pythagoras on the impedance triangle. The phase angle is theta = arctan(XL / R), but only the magnitude is requested here.

Step-by-Step Solution:

Verification / Alternative check:If you incorrectly add R + XL = 152 Ω, you overestimate the impedance because you ignored the quadrature relation. Phasor diagrams or complex arithmetic (Z = 60 + j92) confirm |Z| ≈ 110 Ω.

Why Other Options Are Wrong:102 Ω and 92 Ω underestimate the vector length; 152 Ω is the erroneous arithmetic sum, not the magnitude.

Common Pitfalls:Adding magnitudes linearly; forgetting units; mixing up series vs parallel combinations (parallel requires admittance addition, not impedance).

Final Answer:110 Ω