I need to calculate the apparent power, active power and reactive power and I only got this diagram ( in german ):

enter image description here

So the equation for active power is $$P = \int_0^T u(t)i(t) \, \mathrm{d} x$$

Edit: Adjusted my solution to the answers:

$$ i(t) = kx+d = -{2t \over T} +1$$ $$ u(t) = 0.5 \rightarrow t=[0,T/2] $$ $$ u(t) = -0.5 \rightarrow t=[T/2,T] $$

$$P = {\int_0^{T/2}( {-t \over T} +0.5 \,) \mathrm{d} t + \int_{T/2}^T ({t \over T} -0.5) \, \mathrm{d} t \over T} = \frac{1}{4}W$$

Complex power: $$S = RMS(i(t)) * RMS (u(t))$$ $$RMS(i(t)) = \frac{peak}{\sqrt{3}} = 0,577$$ $$S= 0,289$$ $$Q = \sqrt{S^2-P^2}= 0,144$$ $$cos(\alpha) = \frac{P}{S}= 0,866$$

  • \$\begingroup\$ Shouldn't P be 0.5 W? Because 0.25W is only for one half. \$\endgroup\$
    – madmax
    Feb 26, 2012 at 19:05
  • 2
    \$\begingroup\$ No. You are confusing energy and power. Power is energy per time. The total energy delivered to the load grows constantly, but the power averages the same over any whole number of half cycles. That is because the time increases along with the total energy. For example, a 60 W light bulb is still 60 W whether you run it for 1 minute or 2 minutes. \$\endgroup\$ Feb 27, 2012 at 14:05
  • \$\begingroup\$ Seems plausible. \$\endgroup\$
    – madmax
    Feb 27, 2012 at 21:06

3 Answers 3


You are making this way too complicated.

From inspection you can see that during the first half of the period shown, you have a steady voltage of 500 mV. The power is then just the average of the current times this voltage, which is 250 mW. From inspection again you can see that the second half of the period is the same as the first half with the signs of the voltage and current flipped. This obviously yields the same power again, 250 mW.

The instantaneous power is a triangle wave with peaks at 0 and 500 mW, and average of 250 mW (unless I'm misunderstanding what that diagram is showing).


I forgot to mention about calculating reative power.

One way to get that is to derive the power factor. The power factor is usually described as the cosine of the phase angle between the current and voltage assuming both are sines. However, it also has a more general definition that is more appropriate in this case. You can think of the power factor as the ratio of true power to the product of RMS current and voltage.

In this case, the RMS voltage is obvious, which is 500 mV. From inspection you can see that the current is symmetric and repeating, so you only have to solve for the RMS current of a ramp from 1 to 0. From symmetry we can see that this must be the same as a ramp from 0 to 1, which will make the equation a little easier.

In other words, find the RMS current of I(t) = t from 0 to 1. To do that, first square the function, which is then t^2. The average of that from 0 to 1 is 1/3, and then the square root of that is 0.577. So the RMS voltage is 500 mV, the RMS current is 577 mA, and the product of the two is 289 mW. From above the real power is only 250 mW, so the power factor is 250mW/289mW = 0.866. The reactive power is

sqrt(289mW^2 - 250mW^2) = 144 mW

Again, there is no need to make this complicated.

  • \$\begingroup\$ I like more the explanation with apparent power than the one with power factor. \$\endgroup\$
    – clabacchio
    Feb 28, 2012 at 13:54


  • complex power \$S=V\cdot I\$
  • apparent power \$|S|=|V\cdot I|=V_\text{RMS}\cdot I_\text{RMS}\$
  • real power \$P=\mathbb{R}\{S\}\$
  • reactive power \$Q=\mathbb{I}\{S\}\$

The current-voltage product is positive for the entire period, so power flows towards the load at all times; reactive power is zero.

Don't try to define the voltage function U(t) over the entire period. Instead break up the integral into two parts: \$\int_0^{T/2}U_1(t)I(t)dt+\int_{T/2}^TU_2(t)I(t)dt\$, where U1=0.5V and U2=-0.5V.

  • \$\begingroup\$ Well I also told my professor that the reactive power is zero. But he told me that it is wrong and that I can't say that the two signals are in phase because it is only possible with sinus signals. I am a bit confused now ... \$\endgroup\$
    – madmax
    Feb 21, 2012 at 20:51
  • \$\begingroup\$ Just a tip: usually time-dependent variables are represented with lowercase, to distinguish them from rms and-or average... \$\endgroup\$
    – clabacchio
    Feb 21, 2012 at 21:03
  • \$\begingroup\$ At least what you said was correct as far as it went, but how about actually answering the question of what the real and reactive powers are? \$\endgroup\$ Feb 23, 2012 at 13:30
  • \$\begingroup\$ Olin has demonstrated that reactive power is different than 0, I would suggest to remove that statement :) \$\endgroup\$
    – clabacchio
    Mar 5, 2012 at 9:18
  • \$\begingroup\$ @clabacchio, I hadn't heard of his definition of reactive power. Feel free to vote mine down. \$\endgroup\$
    – tyblu
    Mar 5, 2012 at 15:29

First: I would represent I as function of t instead of x:

$$ i (t) = -{2t \over T} +1 $$


$$ u(t) = 0.5 \rightarrow t=[0,T/2] $$ $$ u(t) = -0.5 \rightarrow t=[T/2,T] $$

Than you can assume that their (base) frequency is equal, because they have the same period T. They appear to have also the same phase, because are both positive in the first semi-period and negative in the second.

So the power will be:

$$ p(t) = 0.5* \left( \frac{-2t}{T} \right) +0.5 = \left( \frac{-t}{T} \right) +0.5 \rightarrow t=[0,T/2] $$ $$ p(t) = -0.5* \left( \frac{-2t}{T} \right) +0.5 = \left( \frac{t}{T} \right) -0.5 \rightarrow t=[T/2,T] $$

Now you apply the integral you showed to calculate the average:

$$ P = {\int_0^{T/2} \left( {-t \over T} +0.5 \, \right) \mathrm{d} t + \int_{T/2}^T \left( {t \over T} -0.5 \right) \, \mathrm{d} t \over T} = \frac{1}{4}W $$

(I may be wrong but if you do it graphically you should obtain this)

For the reactive power see Olin's answer :)

  • \$\begingroup\$ Thanks for your answer but my solution would be 0. That cant be. Also I dont know how you got -2/T. -2 is the pitch but why 1/T? \$\endgroup\$
    – madmax
    Feb 21, 2012 at 19:48
  • 2
    \$\begingroup\$ -2/T because i(t) drops of 2 (A?) in the time T; try to substitute the values and you should get it right; then, cannot be 0 because the signals are both positive or negative at the same time \$\endgroup\$
    – clabacchio
    Feb 21, 2012 at 20:03
  • \$\begingroup\$ Ok thanks. But I have to correct you. The reactive power is not zero. I met with my professor today and he told me that you cant decide if the signals are in phase or not in this case. He also said that the reactive power is not zero but he wasn't in the mood to explain it to me ... \$\endgroup\$
    – madmax
    Feb 21, 2012 at 20:41
  • \$\begingroup\$ No, I didn't want to say that they have no phase. It's just that the current is not the same type of signal as the voltage. (square wave) And that was my professor's argument to say that I can't decide that it is in phase although it seems as it is. \$\endgroup\$
    – madmax
    Feb 21, 2012 at 21:51
  • 1
    \$\begingroup\$ @clabacchio: No, reactive power is not zero. Having the signs always match doesn't guarantee a power factor of 1, which is what is necessary for reactive power to be 0. You can't really talk about phase here since the signals are not sinusoids, but that doesn't mean there can't be reactive power. It should be obvious from inspection that there is reactive power since the voltage and current aren't proportional to each other. Remember that power factor of 1, reactive power of 0, and load being purely resistive all say the same thing. Clearly the load is not resistive. \$\endgroup\$ Feb 23, 2012 at 13:23

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