Why can phasors be used simplify getting instantaneous power from instantaneous curent and voltage?

Question

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?

Answer 1

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:

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

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:

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

Answer 2

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.

Answer 3

The power calculated with phasors uses RMS variables and results in the average power. Thus, the calculation of P1 should be P1 = V/sqrt(2)*I/sqrt(2), which will give the correct power 6 W (which is the average power).

Answer 4

Nobody has pointed out a simple mistake in the power formula.

$$ P=Re(\tilde{S})=Re(\dot{V}I^*)=Re(6\angle30\cdot2\angle{-30})=12 $$

Where \$I^*\$ is complex conjugate of \$\dot{I}\$.

Answer 5

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