Phasors, Reactance and Impedance

By the end you should be able to: Represent sinusoidal quantities as phasors in trigonometric, rectangular, polar and complex forms, and analyse purely resistive, inductive and capacitive circuits using reactance and impedance.

Why phasors

In an AC circuit the element relations are $v = L\,di/dt$ and $i = C\,dv/dt$ , so applying KVL to an RLC circuit produces a differential equation. Solving one for every circuit would be intolerable.

The escape: if every quantity in the circuit is a sinusoid of the same frequency, then a sinusoid is fully described by just two numbers — its magnitude and its phase. Represent it by a rotating vector, a phasor, and differentiation becomes multiplication. The differential equations collapse into algebra with complex numbers.

The condition matters. Phasors apply only to steady-state sinusoidal circuits at a single frequency. They say nothing about transients, and nothing about a circuit driven by two different frequencies at once.

The four representations

A voltage $v = 100\sin(\omega t + 30°)$ can be written:

Form	Expression
Trigonometric	$100\sin(\omega t + 30°)$
Polar	$70.7\angle 30°$ (RMS)
Rectangular	$61.2 + j35.4$
Complex exponential	$70.7\,e^{j30°}$

Phasors are conventionally RMS, not peak — so the 100 V peak becomes 70.7 V. Mixing the two is a common and expensive error.

Convert between polar $R\angle\theta$ and rectangular $a + jb$ by

a = R\cos\theta, \quad b = R\sin\theta; \qquad R = \sqrt{a^2+b^2}, \quad \theta = \tan^{-1}\frac{b}{a}

Use rectangular form for addition and subtraction, polar for multiplication and division. Adding two polar quantities directly, by adding magnitudes, is wrong unless they happen to be in phase.

Purely resistive circuit

v = V_m\sin\omega t \Rightarrow i = \frac{V_m}{R}\sin\omega t

Current and voltage are in phase, $\phi = 0$ . Power is $P = VI$ , all of it dissipated.

Purely inductive circuit

With $v = V_m \sin\omega t$ and $v = L\,di/dt$ , integrating gives

i = \frac{V_m}{\omega L}\sin(\omega t - 90°)

The current lags the voltage by 90°. The opposition is the inductive reactance:

X_L = \omega L = 2\pi f L \qquad (\Omega)

Note $X_L \propto f$ : an inductor blocks high frequencies and passes DC ( $X_L = 0$ at $f = 0$ , consistent with "short circuit to DC").

Purely capacitive circuit

i = V_m\omega C\sin(\omega t + 90°)

The current leads the voltage by 90°. The capacitive reactance:

X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \qquad (\Omega)

Here $X_C \propto 1/f$ : a capacitor passes high frequencies and blocks DC ( $X_C \to \infty$ at $f = 0$ ).

The mnemonic: CIVIL. In a Capacitor, I comes before V; V comes before I in an L. Reliable, and it settles the sign question in a second under exam pressure.

Neither reactance dissipates power. Over a cycle, energy flows into the field for a quarter cycle and back out the next. The average is zero. That is the physical origin of reactive power in the next topic.

Impedance

The general opposition, combining both:

Z = R + j(X_L - X_C) \qquad (\Omega)

|Z| = \sqrt{R^2 + (X_L-X_C)^2}, \qquad \phi = \tan^{-1}\frac{X_L - X_C}{R}

Ohm's law in phasor form is then simply

\bar{V} = \bar{I}\bar{Z}

The sign of $\phi$ classifies the circuit:

$X_L > X_C$ : $\phi$ positive, current lags — inductive.
$X_C > X_L$ : $\phi$ negative, current leads — capacitive.
$X_L = X_C$ : $\phi = 0$ , purely resistive — resonance.