ElectricalAC CircuitsmediumFE ExamCoursework

Current in a Series R-L Circuit

A series circuit has $R = 30\ \Omega$ and inductive reactance $X_L = 40\ \Omega$ , driven by a $120\ \text{V}$ rms sinusoidal source.

Find the magnitude of the rms current.

Given

R	30 ohm	Resistance
XL	40 ohm	Inductive reactance
V	120 V	Source voltage (rms)

Hint 1

Resistance and reactance are 90° apart — how do such quantities combine?

Hint 2

|Z| = √(R² + X²), not R + X.

Hint 3

√(30² + 40²) = 50 Ω, then apply Ohm's law.

Worked solution — try the problem first

Impedance magnitude. Resistance and reactance are orthogonal in the complex plane, so they combine as a Pythagorean sum, never arithmetically:

$|Z| = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \Omega$

Current:

$|I| = \frac{|V|}{|Z|} = \frac{120}{50} = 2.4\ \text{A}$

$\boxed{I = 2.4\ \text{A rms}}$

Phase angle:

$\theta = \arctan\frac{X_L}{R} = \arctan\frac{40}{30} = 53.1^\circ$

The current lags the voltage by 53.1° — inductive behaviour. In phasor form, $\mathbf{I} = 2.4\angle-53.1°$ A.

Why not 70 Ω? Adding $R$ and $X_L$ directly would give $120/70 = 1.71$ A. That treats two quantities 90° apart as collinear, which is the single most common error in AC circuit analysis.

Concepts:AC impedancePhasorsOhm's lawReactance

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