# PE exam question 49

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#### yaoyaodes

##### Member
A balanced 3 phase load with a power factor of 0.8 lagging is connected as show.

I am so confused with this question and solution.

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#### DuranDuran

##### "It wasn't me."
You can do a search for any NCEES problem and find it here. Just type "NCEES #____" in the search bar. Also remember problem #49 used to be numbered #509 in the older NCEES practice exam books.

#### akyip

##### Member
I'll try to explain this as best as I can with what I remember Zach Stone's Electrical PE Review course taught about 2-wattmeter method.

In a 2-wattmeter method, the two wattmeters are ALWAYS connected such that:
• They share the same common negative terminal connection point.
Here, the common negative terminal point is the B phase.
• They have different positive terminal connection points.
Wattmeter W1's positive terminal is the C phase, and Wattmeter W2's negative terminal is the A phase.
For each wattmeter's connection going from the positive terminal to the negative terminal:
• The wattmeter that is connected in the positive or ABC sequence (AB, BC, or CA) will have its real power formula as: P = VLL * IL * cos(theta + 30).
In this problem, Wattmeter 2 is connected from A to B, which is the positive or ABC sequence.
So: P W2 = VLL * IL * cos(theta + 30)
• The wattmeter that is connected in the negative or CBA sequence (CB, BA, or AC) will have its real power formulas as: P = VLL * IL * cos(theta - 30)
In this problem, Wattmeter 1 is connected from C to B, which is the negative or CBA sequence.
So: P W1 = VLL * IL * cos(theta - 30)
The 2 wattmeters will have the same real power readings when theta = 0, which corresponds to pf = 1. The reason is that the cosine function is symmetrical about the origin theta = 0 (it is an even or evenly symmetrical function about theta = 0). So cos(-x) = cos(x).

When theta = 0:
• cos(0 - 30) = cos(0 + 30)
• cos(-30) = cos(+30) = 0.866
• Power factor: pf = cos(0) = 1
I hope that this explanation helps. I also included a few illustrations to try to reinforce my explanation.

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Last edited:

#### Zach Stone P.E.

##### Learn how to Pass the Power PE Exam at electricalp
I'll try to explain this as best as I can with what I remember Zach Stone's Electrical PE Review course taught about 2-wattmeter method.

In a 2-wattmeter method, the two wattmeters are ALWAYS connected such that:
• They share the same common negative terminal connection point.
Here, the common negative terminal point is the B phase.
• They have different positive terminal connection points.
Wattmeter W1's positive terminal is the C phase, and Wattmeter W2's negative terminal is the A phase.
For each wattmeter's connection going from the positive terminal to the negative terminal:
• The wattmeter that is connected in the positive or ABC sequence (AB, BC, or CA) will have its real power formula as: P = VLL * IL * cos(theta + 30).
In this problem, Wattmeter 2 is connected from A to B, which is the positive or ABC sequence.
So: P W2 = VLL * IL * cos(theta + 30)
• The wattmeter that is connected in the negative or CBA sequence (CB, BA, or AC) will have its real power formulas as: P = VLL * IL * cos(theta - 30)
In this problem, Wattmeter 1 is connected from C to B, which is the negative or CBA sequence.
So: P W1 = VLL * IL * cos(theta - 30)
The 2 wattmeters will have the same real power readings when theta = 0, which corresponds to pf = 1. The reason is that the cosine function is symmetrical about the origin theta = 0 (it is an even or evenly symmetrical function about theta = 0). So cos(-x) = cos(x).

When theta = 0:
• cos(0 - 30) = cos(0 + 30)
• cos(-30) = cos(+30) = 0.866
• Power factor: pf = cos(0) = 1
I hope that this explanation helps. I also included a few illustrations to try to reinforce my explanation.
Thanks for the mention @akyip!

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