When I learned phasors I learned we passed from sinusoidal to phasor this way:

$$\sqrt{2} V\cos(\omega t+\phi) \rightarrow \sqrt{2} Ve^{j\phi}$$

However I'm now studying for a book that does this:

$$\sqrt{2} V\sin(\omega t+\phi) \rightarrow Ve^{j\phi}$$

Is this some kind of notation that I am not familiar with? Can I accept this as correct? I«m facing this notation studying three-phase systems, while the first notation was used in the context of circuit analysis.

Thank you!

  • \$\begingroup\$ What formula do your two sources give for calculating the power delivered by a (phasor) voltage applied to a load? \$\endgroup\$
    – The Photon
    Feb 25, 2019 at 4:36
  • 2
    \$\begingroup\$ Is there a mathematical relationship between sines and cosines? \$\endgroup\$
    – Solar Mike
    Feb 25, 2019 at 6:06
  • \$\begingroup\$ Most modern textbooks I saw use a cosine-phasor. A sine-phasor is also present, but not that popular. It is actually just about selecting a Real axis on the phasor space. \$\endgroup\$ Feb 25, 2019 at 14:50

2 Answers 2


I also saw the first definition in my education:

$$v(t) = \mathcal{R}\left\{\underline{V}\cdot e^{j\omega t}\right\} = \mathcal{R}\left\{(V\cdot e^{j\phi})\cdot e^{j\omega t}\right\} = V\cdot\cos(\omega t+\phi)$$

But as long as you are consistent with your formulation, both will lead to the same results when analyzing problems. The phasor amplitude will now represent the RMS value of the sinusoidal signal, and the phase will be relative to a sine wave (90 degrees shifted compared to the cosine).

[EDIT] If you're also interested in how exactly they're linked:

Both definitions are still strongly linked. Using the second definition is basically identical to

$$\begin{align} v(t) &= \mathcal{I}\left\{\underline{V}'\cdot \sqrt{2} e^{j\omega t}\right\}\\ &= \mathcal{I}\left\{(V'\cdot e^{j\phi'})\cdot \sqrt{2} e^{j\omega t}\right\}\\ &=\sqrt{2}V'\cdot \sin(\omega t + \phi')\\ &= (\sqrt{2}V')\cdot \cos\left(\omega t + \left(\phi' - \frac{\pi}{2}\right)\right) \end{align}$$

And you can find that

$$\begin{align} V &= \sqrt{2}V'\\ \phi &= \phi' - \frac{\pi}{2}\\ \underline{V} &= Ve^{j\phi} = (\sqrt{2}V')e^{j(\phi'-\frac{\pi}{2})} = -j\sqrt{2}\cdot \underline{V'} \end{align}$$

  • \$\begingroup\$ What is being proved here? This doesn't proof the second equation PO posted. \$\endgroup\$
    – Huisman
    Feb 25, 2019 at 8:37
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    \$\begingroup\$ @Huisman I'm not sure what there is to prove about a definition. Both are valid definitions of a phasor in the sense that they will lead to the same relative results, but they are not equivalent (ie. you can't compare phasors of definition 1 with that of definition 2). \$\endgroup\$
    – Sven B
    Feb 25, 2019 at 8:50
  • \$\begingroup\$ I wouldn't use two different definitions of phasors and telling you can't compare them. Just use \$sin(\omega t) = cos(-\omega t +\pi/2)\$ and you can compare them. \$\endgroup\$
    – Huisman
    Feb 25, 2019 at 9:26
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    \$\begingroup\$ Well yes, but I meant to not compare them directly as I outlined in my answer (the resulting phasors are still linked by a factor \$-j\sqrt{2}\$). But OP's question was if it could be accepted as correct, to which the answer is yes, but they are not identical definitions - so not just a different notation. \$\endgroup\$
    – Sven B
    Feb 25, 2019 at 13:23
  • \$\begingroup\$ Thank you for the detailed explanation! I got confused by my book but it's just a different definition. \$\endgroup\$ Feb 25, 2019 at 19:09

It is common practice to give phasor magnitude in RMS. Good example is your 120V wall outlet. Peak of that sine wave us around 169V. It would be nice if they would state that fact somewhere in the text to avoid confusing students unnecessarily.


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