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I learned that, when superimposing two waves on top of each other to calculate the resulting wave's amplitude, it's helpful to use phasors. From what I gathered, phasors are vectors originating at the origin and ending at any position along the unit circle. Of interest is a phasor's projection onto the x-axis, because that projection gives the height of the wave (represented by the phasor) at any position.

I read here (go to "Phasor Addition using Rectangular Form") that when superimposing waves in phasor setups, the waves' phasors are added as normal vectors. That is, each phasor's horizontal components is added to the other, and each vertical component is added to the other. If, however, wave (phasor) height is given by the projection onto the x axis, why not just add the $x$ components?

Borrowing the example from the linked site, if we need to add a wave 30 units in magnitude to a different wave 20 units in magnitude and 60 degrees out of phase, why isn't the calculation simply $30 + 20cos(60)$?

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  • $\begingroup$ 1. It is not a requirement for a phasor to "end ... along the unit circle". In your example you can see the two initial phasors don't have unit length, so they don't obey this (non-)rule. $\endgroup$
    – The Photon
    Commented Aug 28, 2015 at 16:56

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If all you want to know is the voltage at one instant, you could indeed simply add the x-components.

If you want to know the amplitude of the resulting sinusoid, you need to do complete vector addition (both x and y coordinates) to get the full amplitude (43.6 V. in your example). This will tell you not just what the voltage is at one instant in time, but how it will vary as the sum signal evolves through a full period.

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