In two different books I'm reading the authors start the discussion of instantaneous power delivered to a load in a steady state AC-circuit by assuming
\$i(t) = I_0 \sin(\omega t + 0)\$ and \$u(t) = U_0 \sin(\omega t + \alpha)\$\

Then they derive or simply state the formula \$p(t) = i(t)u(t) = \dfrac{U_0I_0}{2}\left(\cos(\alpha)(1-\cos(2\omega t)) + \sin(\alpha)\sin(2\omega t) \right) \quad (1.) \$

I thought it was strange to assume the phase \$\beta\$ of \$i(t)\$ to be zero, wouldn't we want to be able to calculate the instantaneous power for arbritary \$i(t)\$?
When i set \$i(t) = I_0 \sin(\omega t + \beta)\$ and \$u(t) = U_0 \sin(\omega t + \alpha)\$ I reached the formula
\$ p(t) = \dfrac{U_0I_0}{2}(-cos(2\omega t + \alpha + \beta) + \cos(\alpha - \beta) \quad (2.) \$
When \$\beta = 0\$ \$(2.)\$ is equal to \$(1.)\$ but with other values for \$\beta\$ this is not necessarily so.\

So I wonder, what is the point with deriving a formula for instantaneous power only for the case when the phase of the current is \$0\$?

EDIT in response to the two first answers

About my derivation being wrong, I can't see why it would be wrong. wolframalpha seems to agree that is is correct (I've used \$K\$ instead of \$I_0\$). Here is how I derived my formula:

\$I_0\sin(\omega t + \beta) \cdot U_0 \sin(\omega t + \alpha) \$ =
\$\dfrac{I_0 U_0}{2}(2\sin(\omega t + \beta) \sin(\omega t + \alpha)) = \$
\$\dfrac{I_0 U_0}{2}(2(-\cos(\omega t + \alpha + \omega t + \beta) + \cos(\omega t + \alpha) \cos(\omega t + \beta) ) ) = \$
\$\dfrac{I_0 U_0}{2}(-\cos(\omega t + \alpha + \omega t + \beta) + 2\cos(\omega t + \alpha) \cos(\omega t + \beta) -\cos(\omega t + \alpha + \omega t + \beta) ) = \$
\$\dfrac{I_0 U_0}{2}(-\cos(2\omega t + \alpha + \beta) + \cos(\omega t + \alpha) \cos(\omega t + \beta) +\sin(\omega t + \alpha) \sin(\omega t + \beta) ) = \$
\$\dfrac{I_0 U_0}{2}(-\cos(2\omega t + \alpha + \beta) + \cos(\omega t + \alpha - (\omega t + \beta) ) ) = \$
\$\dfrac{I_0 U_0}{2}(-\cos(2\omega t + \alpha + \beta) + \cos(\omega t + \alpha - (\omega t + \beta) ) ) = \$ \$\dfrac{I_0 U_0}{2}(-\cos(2\omega t + \alpha + \beta) + \cos(\alpha - \beta ) ) = (2.) \$

Regarding that only the difference in phase matters, I can't see how this can be. If we only imagine two sine curves being shifted back and forth along the time axis it seems obvious to me that the product of their values at some time \$t\$ will change, and hence the power will change when we shift them even if we shift both the curves by the same phase.

For example if \$\alpha = \pi/4\$ and \$\beta = 0\$ and the amplitudes are \$1\$ then at \$t=0\$ we have \$\sin(\omega t + \beta)\sin(\omega t + \alpha) = \sin(0 + 0)*\sin(0 + \pi/4) = 0\$

but if \$\beta = \pi/4\$ so that \$\alpha = \pi/2\$ we have at \$t=0\$ that
\$\sin(\omega t + \beta)\sin(\omega t + \alpha) = \sin(0 + \pi/4)*\sin(0 + \pi/2) = \dfrac{1}{\sqrt{2}}1 \neq 0\$

  • \$\begingroup\$ What I meant by stating that only difference in I and U phases matters is that the result will be the same except for a shift of phase (which should be obvious). Do you understand what phase means and what it means to chose a phase origin that makes \$\beta=0\$? Phase is proportional to time and the principles of physics don't change if you chose a different origin in time. So when you shift time for your input variables the result will stay the same (except for being also shift in time). \$\endgroup\$
    – Curd
    Sep 27, 2019 at 12:18
  • \$\begingroup\$ @Curd Ooooh, "Except for a shift in phase" made all the difference for my understanding, actually I didn't think of that at all! It makes perfect sense though. I was to focused on what specific formula to use if confronted with a problem when the phase of the current was not 0. I'll accept your answer but I encourage you to add the bold part of your comment to your answer :) \$\endgroup\$ Sep 27, 2019 at 12:35
  • \$\begingroup\$ Since sin and cos are infinite, the positioning of a vertical axis where one of α or β is set to 0 (waveform starts on y-axis) is a basic convention to simplify the math. \$\endgroup\$ Sep 27, 2019 at 18:41

2 Answers 2


In general the origin of phase measurment is arbitrary and doesn't matter as long as the origin is constant over time.
I.e. if the phase of input varables is shift (by chosing a different phase origin) the result will stay the same except for also getting shift in phase.
What matters is only the difference between phases of current and voltage.

So in order to treat all possible cases you can define the origin of phase measurent just so that one of the phases is 0 (\$\beta\$) and only vary the other one (\$\alpha\$).

This kind of narrowing down of all possible cases (here: any \$\beta\$) to a special case (here: \$\beta=0\$) is done to simplify the situation without weakening the problem. It is quite a common thing in mathematical and physical proofs or derivation of formulas and it is typically accompanied by a standard remark like "without loss of generality" (or in German "ohne Beschränkung der Allgemeinheit").

  • \$\begingroup\$ I don't see how this can be. I edited my post if you'd care to read. \$\endgroup\$ Sep 27, 2019 at 10:35
  • \$\begingroup\$ The authors just decided to define the origin of phase to be 0 for current (they can do this, because phase origin is basically arbitrary). If you want to compare your general formula with theirs you have to shift it exactly by \$\beta\$ to match their (arbirtrary but convenient) phase measurement convention. You should get the same result then. \$\endgroup\$
    – Curd
    Sep 27, 2019 at 11:09

The instantaneous power is the product of instantaneous voltage and current. These two can be out of phase. Current can be lagging or leading the voltage, but you can also turn that voltage is leading or lagging the current, it doesn't make any difference in the math. Now one of them is used as reference and the other has a phase shift with respect to reference (the other). Why would you need two phase differences? Which is now reference, universal time? Just keep it simple.

Your derivation of formula seems wrong to me, if you do the math carefully, then you should get the same result.


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$$sin(\omega t+\alpha)\cdot sin(\omega t +\beta)=\dfrac{1}{2}\cdot ( cos(\alpha - \beta) - cos(2\omega t + \alpha + \beta))$$

So, your calc is good and their is wrong.


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