CR · Sinusoidal steady state

Phasors: Rotating Arrows

In AC, both voltage and current are sinusoids that never stop swinging, so even adding them is awkward. Turn a same-frequency sinusoid into a single rotating arrow, and the messy trigonometric addition becomes the simple vector sum of arrows laid tip to tail. Learn that one change of view.

At a right angle, 3 and 4 make 5

Take an arrow of length 3 (A1, at the 0° reference) and lay an arrow of length 4 (A2) tip to tail, turned by a phase φ. Combine the two into one sinusoid and its amplitude is R, the length of the summed arrow. Slide φ until the summed amplitude is exactly 5 — that is precisely where the two arrows meet at a right angle (90°).

Phase φ of the second arrowφ = 30°

Amplitude R of the summed sinusoid

R = √(A1²+A2²+2A1A2cosφ) = 6.77

Far from 5

One sinusoid, one arrow

A sinusoid A cos(ωt+φ) is the shadow of an arrow of length A that turns steadily from a starting angle φ. As the arrow turns through ωt, its horizontal shadow is the sinusoid’s value at that instant. So a never-resting sinusoid is written, ignoring the swing, with just two numbers: the length A and the phase φ. This is the phasor A∠φ. At the same frequency every arrow turns at the same rate, so only their angles relative to one another remain fixed.

Same frequency adds as arrows

Add two sinusoids of the same frequency and the result is again a sinusoid of that frequency. Only its amplitude and phase are new, and both are given exactly by the vector sum of the two phasor arrows laid tip to tail. The summed arrow’s horizontal part is A1 + A2 cosφ and its vertical part is A2 sinφ, so the amplitude is the hypotenuse R = √((A1+A2cosφ)² + (A2sinφ)²) = √(A1² + A2² + 2A1A2 cosφ). With no trig identity to memorize, a single picture finishes the addition.

At a right angle, the root of the sum of squares

When two arrows meet at 90°, cosφ is zero, the middle term vanishes, and only R = √(A1² + A2²) remains — the hypotenuse of a right triangle, the very relation in which 3 and 4 make 5. This right-angle addition hides throughout AC. In impedance, resistance R and reactance X add at 90° to give |Z| = √(R²+X²); in power, real power P and reactive power Q add at 90° to give apparent power S = √(P²+Q²). All are the same one arrow picture.

ObserveA cos(ωt+φ) → A∠φ

A sinusoid is a rotating arrow.

ChooseRx = ?

Horizontal part of the sum.

Fill inR = ?

The summed amplitude is the hypotenuse.

On your ownφ=90° → R = ?

At a right angle, root of the sum of squares.

Back to the first screen

As you slid the phase φ, the summed arrow’s length ran from 7 down to 1, and it was exactly 5 at φ = 90°, where the two arrows met at a right angle. Lengths 3 and 4 at a right angle give a hypotenuse of 5 — that familiar right triangle was the addition of sinusoids itself. The single change of view, seeing a swinging sinusoid as a rotating arrow, turns trig wrestling into laying arrows tip to tail. Impedance and the power triangle alike come out of this one picture.

A phasor writes a same-frequency sinusoid A cos(ωt+φ) as a rotating arrow A∠φ of length A and angle φ. Adding same-frequency sinusoids becomes the vector sum of phasor arrows laid tip to tail, so the summed amplitude is R = √(A1² + A2² + 2A1A2 cosφ). When the two arrows are at 90° the middle term drops, leaving R = √(A1² + A2²) — impedance √(R²+X²) and apparent power √(P²+Q²) are both this right-angle addition.