Transformer Open/Short Circuit test

Discussion in 'Homework Help' started by Max Kreeger, Sep 5, 2015.

1. Max Kreeger Thread Starter Member

Oct 1, 2013
71
0
Hi guys,
The problem isn't with the question itself its more to do with the fact about how they got Z and R. Just need some clarification.
In every other AC power question. I have always worked out impedance with whatever parameters I had be it P/I^2 or V/I. I'm confused as to what the difference here is? Now I'm being thrown off and don't know what I'm calculating in other questions :S

2. Max Kreeger Thread Starter Member

Oct 1, 2013
71
0
Both of those formulas work out 'R' as can be seen here..so why are they getting two different numbers?

3. WBahn Moderator

Mar 31, 2012
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4,789
Do you understand the difference between real power and complex power?

How much real power is dissipated in the inductance?

4. WBahn Moderator

Mar 31, 2012
17,737
4,789
Really? You still need this formula wheel?

5. GopherT AAC Fanatic!

Nov 23, 2012
6,029
3,790
Did he NEED it or was he using it to clarify his question? If trying to clarify his question, it didn't work because he only got insults instead of someone pointing him in the right direction.

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6. WBahn Moderator

Mar 31, 2012
17,737
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Ah... so you are saying perhaps he thinks that the people he is asking for help regarding complex impedance haven't grasped middle school algebra. That's just so much better.

7. Max Kreeger Thread Starter Member

Oct 1, 2013
71
0
I was using to clarify...
Don't worry I figured it out..Z = V/I ohms law for AC circuits
Z equals R when there is no reactive component.

I've been at it for hours, brain clearly stopped. Apologies for wasting your time.

8. WBahn Moderator

Mar 31, 2012
17,737
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I don't think you yet quite understand because there IS a reactive component.

The voltage is, indeed, applied across the series combination of the resistor and the inductor and so the relationship between V and I is

V = IZ = I(R+jX)

You can't ignore the reactance.

But P is the real power (as opposed to the apparent or the complex power) and the only place that that appears is in the resistor since the inductor absorbs zero real power. So now you are not talking about the voltage of the source, you are talking only about the voltage that appears across the resistor -- which is NOT the same voltage (since you have a voltage divider). So, yes, you could use Ohm's Law instead of the power law, but you have to first determine the voltage across the resistor, which you don't know.

If you just blindly throw the applied voltage and the current at the formula P=VI then you do not get real power, you get apparent power because you are not taking the phase angle into account. Consider that V = 35 V and I = 41.667 A. If you multiply these together you get 1460 VA (volt-amps, as opposed to watts), which is a FAR cry from 300 W.

Last edited: Sep 5, 2015
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9. Max Kreeger Thread Starter Member

Oct 1, 2013
71
0
Yes, although what I stated above was generalized and not in relation to the transformer problem. Thank you

10. WBahn Moderator

Mar 31, 2012
17,737
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Then perhaps the problem is that you haven't quite grasped a fundamental fact about Ohm's Law and the power formula. Ohm's Law relates a resistance to the voltage across THAT resistance and the current through THAT resistance. It's a common misconception, so don't fret too much; just learn to AWAYS keep it in mind. Otherwise you will find yourself frequently throwing the nearest V and I into Ohm's Law or the handiest P and V at the power law formula without considering whether those formulas actually apply to those quantities.