Total opposition in a series RL circuit: For a series AC circuit containing both resistance and inductive reactance, how should the total opposition to current (impedance magnitude) be found?

Introduction / Context:
Impedance combines resistance (real part) and reactance (imaginary part). In a series RL circuit, these components are orthogonal on the complex plane, so the total opposition to AC current is obtained by vector addition, not simple arithmetic addition. This is foundational when computing currents, voltages, and power factors.

Given Data / Assumptions:

• Series connection of R and L with sinusoidal excitation.
• Reactance is inductive only: XL = 2 * pi * f * L.
• Goal: impedance magnitude for current calculation via Ohm’s law I = V / |Z|.

Concept / Approach:
Represent the impedance as Z = R + jXL. The magnitude is |Z| = sqrt(R^2 + XL^2) from the impedance triangle. Simple sums (R + XL) ignore the 90-degree phase difference and produce systematic errors, especially at higher reactance ratios. The correct approach preserves orthogonality via Pythagorean addition.

Step-by-Step Solution:

Verification / Alternative check:
Phasor diagrams show the resistive and reactive drops at right angles; the hypotenuse length equals |Z|. Experimental measurements with an LCR meter corroborate the vector relationship across frequencies.

Why Other Options Are Wrong:
Arithmetic sum and average ignore phase. Subtraction could apply to reactive cancellation (XL − XC) but not to R with XL in magnitude computation.

Common Pitfalls:
Adding R and XL directly; forgetting that only reactances combine algebraically when both are reactive; confusing magnitude with real part.

Final Answer:
Use the vector magnitude: Z = sqrt(R^2 + XL^2)