"For each of the values calculate P and Q" (average)

V = 250 \$\cos (\omega t +45)\$
I = 4 \$\cos (\omega t -30)\$

The way I would try solving this would be by converting to phasors, taking the product and multiplying by a half. I will then be left with a phasor with phase shift of \$15\$ deg. But turns out the final phase shift is \$75\$ deg. I assumed this may be as a result of taking the current max at \$t = 0\$. So that way we just shift both the current and voltage by \$30\$ deg. But then the next question...

V = 18 \$\cos (\omega t -30)\$
I = 5 \$\cos (\omega t - 75)\$

If I now shift by \$75\$ deg, then \$-30 + 75 = 45\$ deg and not \$105\$ deg as stated in the memo. What is happening?

  • \$\begingroup\$ Just curious...was it a typo or what else? \$\endgroup\$ – Lorenzo Donati May 10 '15 at 12:27
  • \$\begingroup\$ The question was taken directly from Pearson's Electric Circuits 10th ed. It is one of four questions, each with different amplitudes and phase shifts for current and voltage. The other ones seem to result in the correct answer by what you described above, so I would assume it to be a mistake. I will confirm tomorrow. \$\endgroup\$ – Chris-Al May 10 '15 at 12:58
  • \$\begingroup\$ Just another quick question; the multiplication by 0.5 is because we are finding the average, right? \$\endgroup\$ – Chris-Al May 10 '15 at 13:00
  • \$\begingroup\$ Well, in a sense, but probably not what you probably imagine (not arithmetic average). If you used phasors with RMS amplitude (i.e. scaled by 1/sqrt(2)) that 0.5 factor won't be there. The fact that the complex power has that expression depends on the definition of average power in the time domain and trigonometric identities. If you multiply two sinusoidal signals and take the average on a period (using integrals) you end up with P=0.5VIcos(phi) if V and I are peak (amplitude) values or VIcos(phi) if they are RMS values. The derivation is not difficult but not feasible in a comment. \$\endgroup\$ – Lorenzo Donati May 10 '15 at 13:10
  • \$\begingroup\$ To confirm, the second question was a mistake in the book. \$\endgroup\$ – Chris-Al May 12 '15 at 16:55

Using phasor (Steinmetz') notation with peak values (not RMS):

\$ V = 250 \angle{45°}\$ \$ I = 4 \angle{-30°}\$

Complex power is \$ S=\dfrac{1}{2}\; V\cdot I^{*}\$ where the asterisk means "complex conjugate", an operation which inverts the sign of the phase. Therefore:

\$ S = \dfrac{1}{2}\; (250 \angle{45°}) \cdot (4 \angle{-30°})^{*} = \dfrac{1}{2}\; (250 \angle{45°}) \cdot (4 \angle{30°}) = 500\angle{75°}\$

Analogously for the second example:

\$ S = \dfrac{1}{2}\; (18 \angle{-30°}) \cdot (5 \angle{-75°})^{*} = \dfrac{1}{2}\; (18 \angle{-30°}) \cdot (5\angle{75°}) = 45\angle{45°}\$

So effectively it seems there is an incoherence somewhere. Maybe a typo? In the second case if V had a phase of +30° the results would match.


Just to incorporate and expand a bit of theory I explained in a comment to the OP.

Let's consider a linear load driven by a sinusoidal current \$i(t)\$ and having the voltage \$v(t)\$ across its terminals; because of the linearity of the load the voltage is sinusoidal too:

\begin{align*} v(t) &= V_m \; \cos(\omega t + \phi_V) && \Leftrightarrow & V &= V_m \, \angle \phi_V = V_m \, e^{j\phi_V} \\ i(t) &= I_m \; \cos(\omega t + \phi_I) && \Leftrightarrow & I &= I_m \, \angle \phi_I = I_m \, e^{j \phi_I} \\ \end{align*}

Assuming the directions of i and v are associated, the instantaneous power absorbed by the load is:

\begin{align*} p(t) &= v(t) \cdot i(t) = V_m \; \cos(\omega t + \phi_V) \cdot I_m \; \cos(\omega t + \phi_I) = \\[1em] &= \dfrac{1}{2} \, V_m \, I_m\; \left[{ \cos(2\omega t + \phi_V + \phi_I) + \cos(\phi_V - \phi_I) }\right] \end{align*}

where the trig formula \$ \quad \cos(a)\cos(b) = \dfrac{1}{2}[cos(a+b) + cos(a-b)] \quad\$ was used.

The average power \$P\$, since \$p(t)\$ is periodic, can be computed averaging on a single period, thus:

\begin{align*} P &= \dfrac{1}{T} \int_0^T {p(t)}\, dt = \dfrac{1}{T} \int_0^T \dfrac{1}{2} \, V_m \, I_m\; \left[{ \cos(2\omega t + \phi_V + \phi_I) + \cos(\phi_V - \phi_I) }\right]\, dt = \\[1em] &= \dfrac{1}{T} \int_0^T \dfrac{V_m \, I_m}{2} \cos(2\omega t + \phi_V + \phi_I) \, dt + \dfrac{1}{T} \int_0^T \dfrac{V_m \, I_m}{2} \cos(\phi_V - \phi_I) \, dt \end{align*}

The first integral is zero, because it is the integral of a sinusoidal function over its period, while the second integral is the average of a constant, so it is that constant. Therefore the average power becomes:

\begin{align*} P = \dfrac{V_m \, I_m}{2} \cos(\phi_V - \phi_I) = \dfrac{V_m \, I_m}{2} \cos \phi \end{align*}

where \$ \phi = \phi_V - \phi_I \$ is the phase difference between the voltage and the current.

Without entering in further details, the average power is also called active power, whereas the reactive power is defined as:

\begin{align*} Q = \dfrac{V_m \, I_m}{2} \sin \phi \end{align*}

On the other hand, if we define the complex power S as :

\begin{align*} S = \dfrac{1}{2} \, V \cdot I^{*} \end{align*}

we obtain:

\begin{align*} S &= \dfrac{1}{2} \, V_m \, e^{j \phi_V} \cdot (I_m \, e^{j \phi_I})^{*} = \dfrac{1}{2} \, V_m \, e^{j \phi_V} \cdot I_m \, e^{-j \phi_I} = \dfrac{V_m \, I_m}{2} \, e^{j (\phi_V - \phi_I)} = \dfrac{V_m \, I_m}{2} \, e^{j \phi} = \\[1em] &= \dfrac{V_m \, I_m}{2} \cos \phi + j \, \dfrac{V_m \, I_m}{2} \sin \phi = P + j \, Q \end{align*}

Hence it's easy to see why P and Q are respectively the real and the imaginary part of S.


Background: Average Power of an AC load is V.I.Cos(angle)

For the first example: -

  • V = 250 V(peak) and
  • I = 4 A(peak)

The RMS of each are 176.78 V and 2.8284 A. The power factor is the cosine of the angle between them. The angle between them is 75 degrees.

Power (real) = V.I.Cos(angle) = 500 x 0.2588 i.e. real power is 129.41 watts

For the 2nd example, V. I. Cos(45deg) = 45 x 0.7071 = 31.82 watts.


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