# complex power circuit question

Discussion in 'Homework Help' started by Ramiel, Apr 14, 2019.

1. ### Ramiel Thread Starter Member

Feb 19, 2018
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0
Hello everyone,
There this one idea I am confused about.
There is this circuit for example and I am asked to find the complex power of the source.
I get the idea of finding I1, I2, add them and find I by kcl.
I can find Vs by adding (I(0.2 + j0.4) + 240<0)
Then here is my question, to find the power of the source S = VI*.
Why in the solution they directly multiply VI*? Shouldnt they multiply I with a minus sign to make the current enter from the positive terminal
of the voltage source to make it conform with the passive sign convention.

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2. ### Papabravo Expert

Feb 24, 2006
11,995
2,574
The real part never gets a minus sign. If you look at the different ways of computing power according to ohms law they involve Voltage and Current squared. The implication is that power can never be negative. Using the complex conjugate keeps the sign correct when computing power because squaring j introduces a factor of -1 to the imaginary part.

Twinkle, twinkle little star
Power's equal I squared R

3. ### Ramiel Thread Starter Member

Feb 19, 2018
59
0
Then how can I know which direction do I take the current when measuring the complex power? Is it going in the positive terminal or not?
I remember that power can be negative which shows that power is releases and positive to show that power is absorbed.

4. ### Papabravo Expert

Feb 24, 2006
11,995
2,574
If you like, power consumed by devices such as motors, heaters, and the like is considered positive. It is just the time rate of change of energy. Energy is always positive, but it can increase or decrease, and thus power can be negative.

5. ### Ramiel Thread Starter Member

Feb 19, 2018
59
0
I
I am sorry but I cant see how this answered my question in the post

6. ### Papabravo Expert

Feb 24, 2006
11,995
2,574
You take it in the direction to compute a positive power from a load or other passive device. Loads consume positive power and sources produce negative power. You can do it the other way around if you like. It really doesn't matter as long as you apply the rules consistently.

https://en.wikipedia.org/wiki/AC_power

In particular it explains why, by convention, we use the conjugate of current to compute complex power, with units of VA (Volt-Amperes).

Last edited: Apr 14, 2019
7. ### WBahn Moderator

Mar 31, 2012
23,881
7,388
Two problems here. Power very much can be negative. If positive power is power being dissipated by a device, then negative power is power being delivered by that device. If you take the time to define your voltage and current polarities properly and then develop your equations consistently with respect to those definitions, the math will take care of the rest.

Second, the complex conjugate is not used so that the sign is correct due to squaring the j. It is there to compensate for the use of phasor quantities, which were developed to model sinusoidal voltages and currents in linear circuits, to compute a nonlinear quantity like power. When we use phasors, we get to use all of our DC relationships for voltage and current just as they were. But that's not the case for power. However, it's close; we just need to make that one tweak.

You may have to try a different browser to get the equations to render properly.

8. ### WBahn Moderator

Mar 31, 2012
23,881
7,388
You get to choose. You just then have to be consistent.

Imagine I have a circuit with six elements in it and they are all in black boxes so I don't know which ones are loads and which ones are sources. In that case I would probably define the current at each device so that it is entering the positive terminal. Know I know that if I get a positive power that the device in that box is acting as a load and absorbing power while if it is negative it is acting as a source and delivering power.

But if I happen to know which boxes contain loads and which contain sources, I might choose (and this is a very common choice to make) to define the currents on my sources as leaving the positive terminal because I then expect the power associated with all of my devices to be positive, thus getting rid of a lot of pesky minus signs that humans are really good at messing up. I just need to remember that positive power for a load means it is absorbing power while positive power for a source means it is delivering power.

9. ### MrAl AAC Fanatic!

Jun 17, 2014
5,756
1,221
I have a question for you.

In your schematic, is that 240v 0 degrees across the 10kw and 12kw loads or is that the mains voltage?
I ask because that's a typical mains voltage yet it is across the two loads. That's entirely possible, but thought i would ask.

Feb 19, 2018
59
0

11. ### Ramiel Thread Starter Member

Feb 19, 2018
59
0
It is acroos the 10kw and 12 kw loads

12. ### MrAl AAC Fanatic!

Jun 17, 2014
5,756
1,221
Hi,

Oh then that makes it even simpler because you dont have to calculation that from the mains voltage.