If instantanoeus power is p(t) and instantaneous voltage and current saying for a resistor are v(t) and i(t), then p(t)=i(t)*v(t). But if you convert v(t) and i(t) to their respective phasors and multiply them, the product is NOT the phasor of p(t).

For example, for R=3 Ohms,$$v(t)=6*cos(120\pi+30^{\circ})$$ and $$i(t)=2*cos(120\pi+30^{\circ})$$. Therefore, $$p(t)=12*cos^2(120\pi+30^{\circ})$$, or $$p(t)=6(1+cos(240\pi+60^{\circ}))$$.

The problem is if I try to multiply $$V=6\angle{30^{\circ}}$$, which is the phasor of v(t), by $$I=2\angle{30^{\circ}}$$, the phasor of i(t). The resulting product $$P=12\angle{60^{\circ}}$$ is equivalent to $$p_1(t)=12cos(120\pi+60^{\circ})$$.

p1(t) is not equal to p(t). What did I do wrong?

  • \$\begingroup\$ Please show your work between your third and fourth equations. Also, do you know how the cosine of 120\$\pi\$ is related to the cosine of 240\$\pi\$? \$\endgroup\$ – Elliot Alderson Mar 6 at 13:04
  • \$\begingroup\$ I used the trig identity cos2x=2cos^2 x-1 \$\endgroup\$ – most venerable sir Mar 7 at 1:47

Multiplying current and voltage in phasors form cannot represent the instantaneous power in an AC circuit. Specifically for a resistive load the power factor is unity (\$\phi=0\$) and the instantaneous power is:

$$ p(t) = \frac{1}{2}V_mI_m(1+ \cos 2\omega t) $$

Resistive load AC Power

Notice that resulting wave appears shifted vertically, not having negative values. In other words, only delivered power (positive). In other hand, if voltage and current are \$ V_{rms}\angle \phi_v \$ and \$ I_{rms}\angle \phi_i \$, the AVERAGE POWER \$P\$ is:

$$ P = V_{rms}I_{rms}\cos (\phi_v-\phi_i) $$ or $$ P = V_{rms}I_{rms}\cos (\phi) $$

For a resistive load:

$$ P = V_{rms}I_{rms} $$

  • \$\begingroup\$ Just multiplying two phasors would be equivalent to multiplying two complex numbers so that the module of the phasor result will be the multiplication of the modules and the resulting angle is the sum of the angles. If nothing else is said (or without context), one could associate it with a sine-wave with positive and negative values; which does not correspond to the case in question - instantaneous power. \$\endgroup\$ – Dirceu Rodrigues Jr Mar 6 at 14:21

What did I do wrong?

In the first round of equations you have dealt with things properly i.e. you have used properly defined phasors and, as such, you get a true representation of power as a steady level of 6 watts superimposed with a cosine wave that takes the peak power to 12 watts and the minimum power to 0 watts at a rate of twice the frequency of the individual voltage and current waveforms. Average power is 6 watts.

In your second round you have lost the "phasor" part and are just multiplying two vectors that correspond with the peaks of voltage and current waveforms hence, you get a peak power of 12 watts. You have gone on to try and reconcile (short cut) this by saying p1 is equivalent to your final equation and, that is incorrect. That's where you went wrong.


Power is a scalar quantity, therefore it cannot be expressed in vector (or phasor) form.

  • \$\begingroup\$ Complex power can be expressed in phasor form. Average (real) power and apparent power are scalars, but complex power is a vector. You need to be precise with the words you use. \$\endgroup\$ – Elliot Alderson Mar 6 at 13:40

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