Introduction to Phasors

AC Circuits · 12 min read

Every AC circuit problem eventually turns into an equation full of $\sin(\omega t + \phi)$ terms that have to be added, differentiated, and compared. Doing that with trig identities gets painful fast. Phasors are the workaround: replace every sinusoid with a rotating vector on the complex plane, and the trig collapses into ordinary vector arithmetic.

A phasor is a rotating arrow

Any sinusoid $v(t) = A \sin(\omega t + \phi)$ can be written as the imaginary component of a complex exponential:

v(t) = \operatorname{Im}\bigl[ A\, e^{j(\omega t + \phi)} \bigr]

The complex number $A e^{j(\omega t + \phi)}$ is a vector of length $A$ that spins around the complex-plane origin at angular speed $\omega$ . Its shadow on the imaginary axis is the sine. We call this rotating vector a phasor.

Left: a rotating arrow on the complex plane with length

A

and instantaneous angle

\omega t

. Right: its vertical projection on a horizontal time axis is the sinusoid

A \sin(\omega t)

The crucial simplification: every sinusoid in the circuit is rotating at the same $\omega$ . The rotation is therefore a common factor that can be stripped off, leaving only the static magnitude and phase:

\boxed{\;\mathbf{V} = A \angle \phi\;}

This frozen-in-time vector is what “the phasor of v(t)” usually means. Two sinusoids at the same frequency can now be compared by comparing their two phasors — just two static arrows on the complex plane.

Two notations for the same arrow

A phasor is a complex number. Complex numbers have two standard notations, and AC analysis uses both freely:

Polar form: $V = |V| \angle \theta$ (length and angle). Ideal for multiplication, division, and reading a phasor off a diagram.
Rectangular form: $V = a + jb$ (real and imaginary components). Ideal for addition and subtraction — components add term-by-term.

The conversion is a right-triangle identity:

a = |V| \cos\theta \quad b = |V| \sin\theta \quad |V| = \sqrt{a^2 + b^2} \quad \theta = \arctan(b/a)

The same phasor carries two equivalent addresses on the complex plane:

|V| \angle \theta

(polar — length + angle) or

a + jb

(rectangular — real + imaginary). Pick whichever makes the current problem easier.

A practical rule of thumb: add in rectangular, multiply in polar. Need to add two voltages? Break both into $a + jb$ , add the real parts, add the imaginary parts, convert back to polar if you want the answer as magnitude and angle. Need to multiply (e.g. $I = V / Z$ )? Polar is trivial (divide magnitudes, subtract angles); rectangular is ugly.

Adding phasors

Two sinusoids at the same frequency add up to a third sinusoid at the same frequency. With phasors this is pure vector addition — either tip-to-tail or via the parallelogram law.

Adding two phasors: place

V_2

tip-to-tail on the end of

V_1

. The resultant

V = V_1 + V_2

runs from the origin to the final tip. (A parallelogram construction — dashed — gives the same answer.)

In rectangular form:

V_1 + V_2 = (a_1 + a_2) + j(b_1 + b_2)

In polar form: convert, add, convert back.

The jω operator — calculus made geometric

Here is the payoff that makes phasors worth the setup. Differentiating a sinusoid shifts its phase by +90° and multiplies its amplitude by $\omega$ :

\frac{d}{dt}\bigl[A \sin(\omega t + \phi)\bigr] = A\omega \sin(\omega t + \phi + 90°)

On the phasor plane, shifting by +90° is exactly the same as multiplying by $j$ . So in the phasor domain:

\frac{d}{dt} \;\longleftrightarrow\; j\omega

The time-derivative operator, which made capacitor and inductor equations messy in the time domain, becomes a plain multiplication in the phasor domain.

Multiplying a phasor by

j

rotates it 90° counter-clockwise; multiplying by

j\omega

rotates and scales by

\omega

. That's the whole geometric content of the derivative operator in the phasor domain —

d/dt \Leftrightarrow j\omega

All three element laws collapse onto phasor impedances thanks to this trick:

Resistor: $V = R I$ — no phase shift.
Inductor: $V = L\,dI/dt \Rightarrow V = j\omega L \cdot I$ — impedance $Z_L = j\omega L$ , purely imaginary and positive. (Voltage leads current by 90°, matching ELI.)
Capacitor: $I = C\,dV/dt \Rightarrow V = \frac{1}{j\omega C} \cdot I$ — impedance $Z_C = 1/(j\omega C) = -j/(\omega C)$ , purely imaginary and negative. (Current leads voltage by 90°, matching ICE.)

Together these three give you a universal impedance that generalises Ohm's law to AC: $V = Z I$ , where Z is a complex number containing everything you need about both amplitude scaling and phase shift. That's the key that unlocks Series RLC, Frequency Response, and Resonance — the next three topics.

A phasor is a rotating arrow on the complex plane; in steady state we freeze out the rotation and write it as

V = |V| \angle \theta

(polar) or

a + jb

(rectangular). Add in rectangular, multiply in polar, and use

d/dt \leftrightarrow j\omega

to turn calculus into algebra. Every AC analysis technique from here on is built on this scaffold.

Common applications

Power-factor correction. Phasor addition of load currents (real + reactive) is how you size a compensating capacitor to bring a factory's current back into phase with its voltage.
Three-phase systems. Three sinusoids 120° apart become three phasors at 0°, 120°, 240°. Their vector sum is zero — which is why you don't need a neutral return wire in a balanced three-phase load.
Filter design. Transfer-function magnitude and phase plots (Bode) are really $|V_{out} / V_{in}|$ and the angle between them — phasor thinking all the way down.
Radio / signal processing. I / Q demodulation is phasor algebra running in silicon.

Try the rotating phasor live in the Simulate stage or lock in the reflex with the Quiz.