# A confusing power factor question

Discussion in 'Homework Help' started by GreenMan357, May 27, 2015.

1. ### GreenMan357 Thread Starter New Member

Apr 3, 2015
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I know how to do the work I just don't understand why we should use -i to calculate the power and reactive factors. It would be great if someone could help me out! My thinking is along the lines of "Because the power is negative the and the component is following the sign convention the voltage or the current are negative (not really negative since this is AC. I don't really know how to describe it). Anyway because of that assumption we could either multiply the current by -1 or the voltage by negative 1 and here they chose to use the current to multiply by -1."

2. ### WBahn Moderator

Mar 31, 2012
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It's probably because, as determined by Example 10.1, the box is actually acting like a source and not a load. I'd have to crank through the problem to be sure, though.

3. ### GreenMan357 Thread Starter New Member

Apr 3, 2015
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That doesn't seem like a very convincing answer. Are you implying that power and reactive factors can only be computed for loads?

4. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
No. Quite the opposite. Here you have a load that is acting like a source. That tends to confuse people. Just like dealing with a source that is acting like a load. If you are diligent about adhering to the definitions and trusting the math, it all works out nicely. But many people are formula monkeys and don't understand the limitations on the formulas they throw around.

5. ### GreenMan357 Thread Starter New Member

Apr 3, 2015
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I see what you are saying. So does that mean that when a load is acting like a source we have to change the sign of either the voltage or the current to make it behave like a load then make the power factor calculation? If so why is it so? I don't like being a formula monkey which is why I am here on this forum trying to understand the concept. I just think the way it was explained in the book didn't sufficiently provide all the nesscessary information and that hint made everything a lot more confusing because it doesn't explain why this need be the case.

6. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
I understand where you are coming from -- and keep in mind that I am speculating because I haven't worked through the problem. Also, I don't know how your text develops and presents things like Eqs. 10.10 and 10.11 using in Example 10.1 and whatever equations it is expecting that you will use in Example 10.2.

I prefer to work much closer to first principles -- or at least foundational principles for the topic at hand.

So you might try looking at those principles, namely what is power factor and, more importantly in this case, what is reactive factor and what is the definition of "magnetizing vars" and what does it mean to "deliver" and "absorb" them. My guess is that the latter is described in your text in the context of a source and, separately, in the context of a load and that that is why they felt the hint was useful.

7. ### GreenMan357 Thread Starter New Member

Apr 3, 2015
8
0
Here is what I think is justification for what I just said. The power for a circuit operating in the sinusoidal steady state is given by:
$image=http://latex.codecogs.com/gif.latex?p%20%3D%20P%20+%20Pcos%282wt%29-Qsin%282wt%29&hash=73b4f324bdc876d9655cacce177146cc$
where real power is
$image=http://latex.codecogs.com/gif.latex?P%20%3D%20%5Cfrac%7BV_mI_m%7D%7B2%7Dcos%28%5Ctheta%20_v-%5Ctheta_i%29&hash=56f71207d7761a69893e97a62f8236a6$
and reactive power is
$image=http://latex.codecogs.com/gif.latex?Q%20%3D%20%5Cfrac%7BV_mI_m%7D%7B2%7Dsin%28%5Ctheta%20_v-%5Ctheta_i%29&hash=900c6601afd22cd5766766cda7d28662$

for a purely resistive circuit
$image=http://latex.codecogs.com/gif.latex?%5Ctheta%20_v%3D%5Ctheta_i&hash=0528e471d6cfb0dde0c5c86b4d26cac8$

and the equation reduces to

$image=http://latex.codecogs.com/gif.latex?p%3D%20P%20+%20Pcos%282wt%29&hash=f5e858a8a0360f2bfbeae0366f0a08fc$

notice how this can only be positive since a resistor cannot provide power thus P has to be positive because from the equation above the P is the line about which the sinusoidal function oscillates and it is also the amplitude of that sinusoidal function.

Now for a purely capacitive circuit
$image=http://latex.codecogs.com/gif.latex?p%3D%20Qsin%282wt%29&hash=4701eb6bd20f86dcf08be9a0bb376ea9$

and for a purely inductive circuit
$image=http://latex.codecogs.com/gif.latex?p%3D%20-Qsin%282wt%29&hash=c567a3c39ccdd300efd335a9fb30f956$

as can be seen from the equations above the real/average power do not include real power. Thus in our circuit in the example the real power is negative which cannot be the case so we have to multiply either the voltage or the current by -1 to make it not so. since the negative part of the power is coming from the current which is
$image=http://latex.codecogs.com/gif.latex?i%20%3D%204cos%28wt%20-%20105%29&hash=4ba5e25b3efbd3b02b20ab4ab4269795$
we choose to multiply that by negative 1. I am not entirely certain what would happen if we multiply the voltage by -1 but i expect to get the same result and I will verify that. Would this be a good justification for why we need to use -i to calculate the power factor and reactive factor.

Edit: forgot to mention the power factor is
$image=http://latex.codecogs.com/gif.latex?cos%28%5Ctheta_v-%5Ctheta_i%29&hash=3b95f2ceea3e31a179ab253fbea22e2f$

and the reactive factor is
$image=http://latex.codecogs.com/gif.latex?sin%28%5Ctheta_v-%5Ctheta_i%29&hash=8e992c94d328096499fe16f52c03339f$

Last edited: May 27, 2015
8. ### GreenMan357 Thread Starter New Member

Apr 3, 2015
8
0
Yes I understand that sorry I should have included some context to go along with the question!

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Another approach is to firstly deduce what the equivalent circuit of the box would comprise.
This is a fairly simple exercise with many possible solutions. For instance a Thevenin Equivalent might comprise
An AC Source:
$\text{E_{th}=35.355\angle{+75^o} Volts(rms)}$
A Series impedance (purely inductive):
$\text{Z_{th}=j21.65 Ohms}$

Then one asks -"With respect to the Thevenin Equivalent source, what are the real and reactive powers generated by the Thevenin source and regarded as "flowing" towards the external AC supply connected to the box?"

The answers will differ (in sign only) from those one would give to the alternative question - "With respect to the external terminal voltage supply connected to the box, what are the real and reactive powers delivered to (or flowing into) the box by the AC supply voltage source?"

Another perspective is to consider the situation where one connected a "directional" Watt meter / Var meter between the external AC supply and the box input terminals. Clearly the result would differ (only in sign) according to the connection of the meters with respect to the anticipated or notional direction of power flow between the AC system and the box.

Last edited: May 29, 2015
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