Compute impedance of a given series RL network (numerical): A 7.5 kΩ resistor is in series with an inductive reactance of 10 kΩ. What is the magnitude of the impedance Z of the series RL circuit?

Introduction / Context:
Many quick design and exam questions focus on computing the magnitude of impedance for first-order RL series networks. The result determines current draw and power factor at a given frequency. The calculation uses the impedance triangle, not simple arithmetic addition.

Given Data / Assumptions:

• Series elements: R = 7.5 kΩ and XL = 10 kΩ.
• Ideal components and sinusoidal steady state.
• Asked for magnitude |Z| only.

Concept / Approach:
For series RL, Z = R + jXL. Therefore |Z| = sqrt(R^2 + XL^2). With values in kilo-ohms, you can keep the kΩ factor outside the square-root arithmetic for convenience, then apply it back to the result.

Step-by-Step Solution:

Verification / Alternative check:
Using rectangular to polar conversion: Z = 7.5 + j10 (kΩ). Magnitude √(7.5^2 + 10^2) = 12.5 kΩ; angle theta = arctan(10/7.5) ≈ 53.1 degrees (not asked but confirms consistency).

Why Other Options Are Wrong:
17.5 kΩ is the incorrect arithmetic sum R + XL. 10.6 kΩ results from an arithmetic/rounding mistake. 7.5 kΩ ignores reactance entirely.

Common Pitfalls:
Switching units mid-calculation; forgetting to use vector sum; mistaking reactance for resistance when estimating power.

Final Answer:
12.5 kΩ