I am stuck on working out power consumption when both the current and voltage is represented as phasors.

\$I(motor) = 4.5+j8.2 amps\$
\$V(motor) = 9.4+j5.5 amps\$

The answer I have been given to this question is 43.70 watts, however I do not know how to get to this answer. I've tried
\$Power = Voltage\times Current\$
Which game me -2.8+j101.83


1 Answer 1


The way to calculate power with two V and I rms phasors is

\$S = V \cdot I^{*}\$

S is apparent power, which is composed of the following:

\$S = P + jQ\$

Q is reactive power, and P is active power, which is what you are looking for.

So, let's apply this to your case:

\$S = (4.35 + j8.2) \cdot (9.4 - j5.5) = 87.4 + j52.33 \$

Now, this doesn't match your expected result, but there is one missing piece of information, you didn't mention if the phasors are rms phasors or peak phasors. It seems like they are peak phasors, so then we can transform through this equation (which only applies for sinusoidal signals):

\$V_{rms} = \frac{V_{peak}}{\sqrt{2}}\$

And then this gives

\$S = \frac{ (4.35 + j8.2) }{\sqrt{2}} \cdot \frac{(9.4 - j5.5)}{\sqrt{2}} = 43.7 + j26.16 \$

And, the real part of that result is the real power P, so

\$ P = 43.7 \$

  • \$\begingroup\$ Thanks for your answer. It seems like uni didn't actually teach me about rms and peak phasors as the question does not mention anything about them. However, one part I do no properly understand is how you got \$87.4+j52.33\$. If possible, can you please go into more detail? Thanks \$\endgroup\$
    – Yuxie
    Apr 10, 2016 at 8:11
  • \$\begingroup\$ That is the result of multiplying the two complex phasors V and I conjugate, and it doesn't really mean anything, since the formula only applies to rms phasors. \$\endgroup\$
    – payala
    Apr 10, 2016 at 8:26

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