# How to calculate power from this diagram? (Another one)

I asked a similar question before, but now with another diagram.
I have a solution but I am not sure if it is right.

First half: $$i_1(t)=\frac{t}{T}A$$ $$u_1(t)=1V$$

Second half:
$$i_2(t)=\frac{t}{T}A$$ $$u_2(t)=-1V$$

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

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You definitely should learn to write your units in your equations. They're the first indication that what you're doing makes sense. –  stevenvh Feb 26 '12 at 17:35
-1/4 (what?) = 1/4 (what?) ???? –  stevenvh Feb 26 '12 at 17:36
Your current should be defined as $i(t)=\frac{t}{T} [A]$. At the end of a period it reaches 1 A not 2 A. –  Count Zero Feb 26 '12 at 20:36
Well, I wrote it for the first and second half of the period. So its t in the time of T/2 and that is 2t/T. When I use it like that i get the correct value. When I use it like you say, I get -1/4. –  madmax Feb 26 '12 at 21:14
Of course not! The function isn't defined by the domain, and it's the same as before. –  clabacchio Feb 26 '12 at 21:21
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This is easiest done in two parts, for the first and second halves of the repeating waveform. This time the average power will obviously be negative since the power magnitude in the second half clearly is larger than in the first half, and the first half is obviously positive and the second negative. This means the load is supplying net power.

The total power magnitude is the RMS voltage times the RMS current. The RMS voltage is obviously 1 V. The current is a ramp from 0 to 1, or I(t) = t. That squared is t^2, the average of that is 1/3, and the square root of that is 0.577. The power magnitude is therefore 1 V * 577 mA = 577 mW.

The real power is the integral of the instantaneous voltage and current over one repeating cycle. From inspection, this is 250 mW in the first half and -750 mW in the second half, for a total of -500 mW.

The power factor is -500 mW / 577 mW = -0.866. Often the absolute value of the power factor is quoted with a statement as to whether the reactive load is inductive or capacitive. You can take the arc cosine of the power factor to get the phase angle of -30 deg in this case, but that has little meaning when the voltage and current are so non-sinusoidal.

The total complex power is the vector sum of the real and reactive power. Since we know the real power and the power magnitude, we can calculate the reactive power, which is -289 mW. The sign only indicates whether it is capacitive or inductive (whether it is a +90 deg or -90 deg).

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Thank you for your answer. It would be a bit easier to understand if you would use some formular coding and not just text. –  madmax Feb 26 '12 at 18:06
Can u tell me whats wrong with my integral? I get -0.75W for the real power. –  madmax Feb 26 '12 at 18:34
@madmax it's true, but this time you should be able to solve the exercise with the former question, so please take the answer as is and try to do some work :) And your integral is wrong because you've taken it from the other question, and now i(t) is different –  clabacchio Feb 26 '12 at 18:45
Here it looks logic that you add up the two halves for the real power. But why don't you add it up at my other question? I don't get that. –  madmax Feb 26 '12 at 18:49
@madmax: I don't remember the details of your other question, but I sortof remember that it was easier to solve by inspection. Here the sign of the power changes, whereas in your other question it was always positive. –  Olin Lathrop Feb 26 '12 at 19:16

You wrote:

first half: $$i_1(t)=\frac{2t}{T}$$ $$u_1(t)=1$$
second half:
$$i_2(t)=\frac{2t}{T}+0.5$$ $$u_2(t)=-1$$

But $$i(t) = \frac{t}{T}$$ in the whole period, so you get the wrong power.

And the power becomes (normalizing T to 1): $$P = {\int_0^{T/2} \left( {t \over T} \, \right) \, dt + \int_{T/2}^T \left({-t \over T} \right) \, \,dt \over T} = { \left[ \frac{t^2}{2} \right] _0^\frac{1}{2}}-{ \left[ \frac{t^2}{2} \right] _\frac{1}{2}^1}= \frac{1}{8}-0-\frac{1}{2}+\frac{1}{8}= - \frac{1}{4}W$$

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Corrected that. –  madmax Feb 26 '12 at 19:03
I am sorry, it's wrong: 0.125+0.125-0.5 = -.025 ! and not - 0.5 –  madmax Feb 28 '12 at 13:36
@madmax you're right, it's 0.25. But now don't ask about power anymore ;) –  clabacchio Feb 28 '12 at 13:51
I am sorry, 6 days till exam, but I don't feel very confident. ;) –  madmax Feb 28 '12 at 14:09