Series RL impedance magnitude — calculation practice: In a series RL circuit used in basic electronics, the impedance magnitude |Z| is found from the resistor and inductive reactance as |Z| = sqrt(R^2 + X_L^2). If R = 15 kΩ and X_L = 10 kΩ, what is the approximate value of the total circuit impedance?

Introduction / Context:
Series RL circuits appear in filters, chokes, and transient-limiting networks. The combined opposition to current is not a simple sum because the resistor (R) and the inductor's reactance (X_L) act at right angles in the impedance plane. This question checks your ability to compute the impedance magnitude |Z| from given R and X_L values and to pick the correct approximate answer from options.

Given Data / Assumptions:

• Resistor R = 15 kΩ.
• Inductive reactance X_L = 10 kΩ (at the operating frequency).
• Series combination; standard phasor relationships apply.
• We need the magnitude |Z|, not the angle.

Concept / Approach:
For a series RL circuit, impedance is Z = R + jX_L. The magnitude is |Z| = sqrt(R^2 + X_L^2). Because R and X_L are perpendicular components (real and imaginary), the Pythagorean relationship applies. Units must be consistent (kΩ with kΩ).

Step-by-Step Solution:

Verification / Alternative check:
A quick bound check helps: if R = 15 and X_L = 10, |Z| must be larger than the larger component (15 kΩ) but smaller than their sum (25 kΩ). The value 18 kΩ lies within this reasonable range, confirming the calculation.

Why Other Options Are Wrong:

• 5 kΩ / 6 kΩ: too small; magnitude cannot be less than the largest individual component in series.
• 25 kΩ: that is the arithmetic sum, not the vector magnitude.
• 20 kΩ: plausible-sounding estimate but higher than the correct Pythagorean result.

Common Pitfalls:
Adding R and X_L arithmetically; forgetting that reactance is orthogonal to resistance; mixing Ω with kΩ; rounding prematurely before comparing with options.

Final Answer:
18 kΩ