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?

Difficulty: Easy

Correct Answer: 110 Ω

Explanation:


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:

Compute R^2 = 60^2 = 3600.Compute XL^2 = 92^2 = 8464.Sum: 3600 + 8464 = 12064.Magnitude: |Z| = sqrt(12064) ≈ 109.87 Ω ≈ 110 Ω.


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

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