# Trouble with complex power

Discussion in 'Homework Help' started by db21, Feb 22, 2016.

1. ### db21 Thread Starter New Member

Feb 22, 2016
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Hey, I'm seriously stuck on my homework regarding complex power. I'm at a loss for what to do. These are the two problems:

1. Calculate the complex power for a resistor, R=10Ohms that has a current of i(t)=14,1*cos(314t-45°) running through.
2. Calculate the complex power for a condensator, C= uF which has a voltage of v(t) = 14,1*cos(1000t+45°).

I know that the complex power can be found by multiplying the RMS voltage and the complex conjugate of the current i, but I'm struggling. Any help is greatly appreciated. Thanks a lot!

2. ### Papabravo Expert

Feb 24, 2006
10,144
1,790
You have mentioned one way to calculate power, but using Ohm's Law and a bit of algebra you can find two additional expressions. One of the expressions involves just current and impedance, and the other involves just voltage and impedance.

3. ### db21 Thread Starter New Member

Feb 22, 2016
10
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Thank you for your response . I'm not entirely sure what you mean. Are you suggesting just I^2*R?. Thanks a bunch.

4. ### Papabravo Expert

Feb 24, 2006
10,144
1,790
Two out of three is not bad. The third should be a piece of cake.

5. ### db21 Thread Starter New Member

Feb 22, 2016
10
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I'm puzzled. i(t)=(14,1*cos(314t-45°) )^2 * 10 Ohms?

Edit: The rules for complex power are different, no?

6. ### db21 Thread Starter New Member

Feb 22, 2016
10
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Could I find v(t) from R and I and then use those two to get P? Is that what you were suggesting?

7. ### db21 Thread Starter New Member

Feb 22, 2016
10
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My best bet atm is convert the current to a phasor and multiply that with the resistor. Convert the result back to "normal" and multiply with the current?

8. ### WBahn Moderator

Mar 31, 2012
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4,796
How can that be i(t)?

9. ### WBahn Moderator

Mar 31, 2012
17,747
4,796
If you are given the current in a resistor can you find the voltage across it?

If you are given the voltage across a capacitor can you find the current through it?

If you know the voltage across a device and the current through a device, can you find the complex power in the device?

10. ### db21 Thread Starter New Member

Feb 22, 2016
10
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So first up is finding the voltage across it and then using that to find the complex power in the device

11. ### WBahn Moderator

Mar 31, 2012
17,747
4,796
Finding the voltage across and the current through -- you need both. That's the most fundamental way and will always work. There are shortcuts that will work in specific instances, including these, because of what you know, for instance, about the reactive power in a resistor and the real power in a capacitor.

12. ### db21 Thread Starter New Member

Feb 22, 2016
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So you're telling me there is a short cut?

13. ### WBahn Moderator

Mar 31, 2012
17,747
4,796
Yes -- but shortcuts are best taken only when you understand the fundamentals well enough to see the shortcut on your own. So focus on the fundamentals first.

14. ### db21 Thread Starter New Member

Feb 22, 2016
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Of course. Couldn't I just do S=I^2*Z. So for this example it would be (14,1A)^2*10Ohms, or is that completely wrong ?

15. ### WBahn Moderator

Mar 31, 2012
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4,796
Yes, you can do that.

Do you know where that S=I²·Z formula comes from? Can you derive it? Or is it just a formula on a sheet that you grab and use.

I ask because, if it's the latter, then there's a good chance that you will get the next one wrong.

16. ### db21 Thread Starter New Member

Feb 22, 2016
10
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As I understand it as and I've read in my book it is based on the following:

"In a linear circuit the sum of the complex powers produced by all the sources is equal to the sum of the complex powers delivered to all of the passive loads."

You are right about the next one tho . I can't figure out where to start.

17. ### WBahn Moderator

Mar 31, 2012
17,747
4,796
While this is true, it has no bearing on the formulas you are trying to work with.

So it's time to take a step back and get the fundamentals under your belt. The following might help: