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I calculated the question and I did not get the same answer as in the book. Could someone please explain why my method is not correct?

Question: enter image description here

My solution: My Solution

Book solution: enter image description here

The problem is from the book "PPI FE Electrical and Computer Review Manual" by Michael R. Lindeburg PE.

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  • \$\begingroup\$ I also wanted to add that the books solution is confusing to me which is why I am not able to figure out what I did wrong. I am still trying to familiarize my self with different ways of expressing Reactive, Real, and apparent power \$\endgroup\$ Commented Dec 22, 2023 at 19:42
  • \$\begingroup\$ Is there a way to do it the same method I did but fix it to match the final solution? I wasn't sure how to go about doing that. Any insight will be greatly appreciated. I am trying to understand the concepts of intelligibly using apparent, real, and reactive power equations instead of just getting solutions. The book solution is confusing. \$\endgroup\$ Commented Dec 22, 2023 at 19:48
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    \$\begingroup\$ DuaTheEngineer - Hi, Thanks for giving the book's title for the copied information. Please see the site rule for referencing. It's not clear from your question if that book is published online or not. If it is published (legitimately) online, then please add a link to it. Or if it is offline, the rule requires that you "include the source to the best of your ability (title, author, page number, etc)". Therefore whichever one applies for that source, please add more information since just a book's title isn't enough. Thanks. \$\endgroup\$
    – SamGibson
    Commented Dec 22, 2023 at 20:39

2 Answers 2

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Could some one please explain why my method is not correct?

The mistakes I see are that you have calculated total apparent power and not individual reactive powers. You have not recognized that the reactive power of a leading power factor load cancels with (subtracts from) the reactive power of a lagging power factor load. In other words, adding the apparent powers gets you nowhere close to solving the problem.

When you divided actual power (12 kW) with power factor (0.7) you actually arrived at apparent power (volts x amps): -

enter image description here

Image from here.

Power Factor is \$\cos(\theta)\$ and this equals real divided by apparent and not real divided by reactive power. So, you need to calculate the phase angle using \$\arccos(PF)\$ then use \$\tan\$ to calculate the ratio of reactive to real power.

Then, remember that it is the difference between the leading and lagging reactive powers that is important.

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  • \$\begingroup\$ Well the reason I calculated Apparent power is to find it to use Reactive power with the equation Qtotal = Sqrt (S^2-P^2). \$\endgroup\$ Commented Dec 22, 2023 at 19:56
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    \$\begingroup\$ @DuaTheEngineer you have to compute them individually then subtract. If we are done here, please take note of this: What should I do when someone answers my question. If you are still confused about something then leave a comment to request further clarification. \$\endgroup\$
    – Andy aka
    Commented Dec 22, 2023 at 19:59
  • \$\begingroup\$ I am still not fully understanding the steps, so arc cost (0.8)= 36.86 and arccos(0.7) = 45.57 . Real power is p=12kWcos (36.86), 18cos(36.86 ), reactive is q= 12kwsin(45.57) and 18kwsin(45.57) \$\endgroup\$ Commented Dec 22, 2023 at 20:08
  • \$\begingroup\$ @DuaTheEngineer what two individual reactive powers did you get? Please use @andyaka or I don't always get a notification when you leave a comment. \$\endgroup\$
    – Andy aka
    Commented Dec 22, 2023 at 20:57
  • \$\begingroup\$ And one Q (lagging) is up +, while the other Q (leading) is down - (- angle). As Andy says, the capacitive load supplies reactive power to the lagging load. \$\endgroup\$ Commented Dec 22, 2023 at 21:20
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You made a big conceptual error in adding the two apparent powers together as scalars rather than as complex quantities. If you calculate your S values as complex values, then you can add them together to get the total S. The total reactive power is the imaginary part of the total S. The key is to think of S as a complex quantity with real and imaginary components. It is similar to a vector with vertical and horizontal components, for example. Just like you can't add the magnitude of two vectors together to get the magnitude of their sum, you can't just add apparent power magnitudes together.

$$S_i = P_i + j Q_i$$ $$ S_T = \sum{S_i} = \sum{P_i} + j \sum{Q_i} = P_T + j Q_T$$ $$ \left|S_T\right| = \sqrt{P_T^2 + Q_T^2} \ne \sum{\left|S_i\right|}$$

The only time that you can add apparent power magnitudes together is when all the apparent power values have the same power factor. In that situation, it's like adding vector magnitudes when their directions are all aligned.

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