#### VulcanCCIT

Joined Sep 14, 2019
19
I am studying for an Broadcast Engineering test and I have a sample test. I believe my problem is in how they worded the question. Perhaps I have over analyzed it... the question is:

"A three-phase delta connected 100V line is attached to three groups of lamps, each requiring 12A at unity power factor, and an AM transmitter with a power supply draw of 20A in each phase at 80% power factor. Find the current for each load carried by each line wire."

I know that IL = IP in a Delta... I have been using Pl= Il X El X 1.73 / 100 = Amps per line. So they say one load is 20A per phase.... I think it is the "three groups of lamps, each requiring 12A at Unity Power Factor" that is the confusing part. is it 3 sets of lamps with 12A per lamp? 12A per group? is it 1 set of lamps per phase? 3 sets per phase? I and another engineer (who is much more seasoned than I) have run the number in all scenarios, and we cant get that same answer. It must be something simple that we are both missing. The 1.73 is the Square root of 3. Any help would be amazing lol. out of the 50 question test this is the only one that I have gotten wrong. Thank you all!!

#### Hymie

Joined Mar 30, 2018
834
The question clearly states that there are 3 groups of lamps, each requiring 12A at unity power factor. Presumably each of these groups of lamps is across a phase to phase (what else could they be?).

In parallel with each of these groups of lamps is an AM transmitter drawing 20A at 80% power factor (the question does not state whether this is the result of a capacitive or inductive load).

So the question is essentially in a balanced 3 phase (delta) circuit, the current between each phase is split between 12A at unity power factor and 20A at 80% power factor – what is the current in each line?

Since the loads are balanced, the current in each line must be the same – I’ll let you work out the rest.

#### VulcanCCIT

Joined Sep 14, 2019
19
The question clearly states that there are 3 groups of lamps, each requiring 12A at unity power factor. Presumably each of these groups of lamps is across a phase to phase (what else could they be?).

In parallel with each of these groups of lamps is an AM transmitter drawing 20A at 80% power factor (the question does not state whether this is the result of a capacitive or inductive load).

So the question is essentially in a balanced 3 phase (delta) circuit, the current between each phase is split between 12A at unity power factor and 20A at 80% power factor – what is the current in each line?

Since the loads are balanced, the current in each line must be the same – I’ll let you work out the rest.
In my original post I put that the sample test answer was 57.2 dyslexia kicked in and the sample test says the answer is 52.7. No matter how I crunch it I can’t get that answer I get 55.3... 12x1.73 = 20.76 for the lamps 20 x 1.73 = 34.6 for the transmitter 20.76+34.6= 55.3

#### Hymie

Joined Mar 30, 2018
834
Why are you multiplying the currents by 1.73 (√3)?

The currents quoted are those flowing at 1.73 x line voltage (173V~).

In each leg of the delta phases you have 12A at unity power factor and 20A at 80% power factor – now work out the total current in each phase leg (the answer is not 32A) and then each line current.

#### VulcanCCIT

Joined Sep 14, 2019
19
I still get 55.36... I don't see how you got a line voltage of 173 as in a Delta circuit Ephase = Eline. It is stated Eline is 100v. if we have the transmitter on a phase at 20a and the lamps are at 12a, at unity that would be 32a. Iline = √3XIphase= 1.73X32=55.36 at unity. The question ask for the load in in LINE. So I think the confusion on my part is the 12a is unity and the 20 is at .8pf. It states the lamps REQUIRE 12a at unity. if the load is at .8 would not current for the lamps at .8pf be 100X12X.8/100? Power/V? or 9.6A? if that was the case 20 + 9.6 = 29.6a per phase
line current then would be 1.73 X 29.6 = 51.2 but the sample test answer is 52.7....

I will upload a picture of the exact test question... maybe that will help. Also, vector math I do not think applies as it does not indicate leading or lagging for the power factor. Every example I can find using the angles still shows the current say 20z +32degrees, or -45degrees, etc.

#### VulcanCCIT

Joined Sep 14, 2019
19
Also, if the test answer is 52.7, given the above that Iline = phase X√3. then if you have Iline, then I phase is 52.7/√3 = 30.46a which given that I have no idea how that amperage is calculated given the test question.

#### Hymie

Joined Mar 30, 2018
834
I would not always believe the given answer as being correct.

Although i have not done the calculations, but the figure of 52.7A would indicate that the 20A at 80% power factor has turned into a current of 40.7A.

#### VulcanCCIT

Joined Sep 14, 2019
19
I have actually asked the test maker for the solution which I found out is an option... I have not heard back yet.

#### ebeowulf17

Joined Aug 12, 2014
3,274
I'm fascinated, and somewhat confused by this question. I tried working out the math myself and never came up with their answer. I'm very inexperienced with power factor calculations, so I'm not terribly surprised by my failure.

Finally, I created an LTspice simulation with voltage sources as the three phase power supply and current sources as "loads" forcing the stated amount of current through each lamp or AM transmitter load location. I then added very low value resistors in series with each line so that I'd have a place to measure line current. Sure enough, phase current is 30.45, slightly higher than I expected, but not too far off. Then, starting with 30.45A in each phase, multiply by sqrt(3) and you get 52.7.

Like I said, even after seeing the simulation confirm the result, I'm still struggling to understand it at multiple levels. Still, for what it's worth, I'm pretty confident that their answer is correct - LTspice hasn't given me a wrong answer yet... except when I fed it bad data! #### Attachments

• 4.5 KB Views: 0
• cmartinez

#### VulcanCCIT

Joined Sep 14, 2019
19
I'm fascinated, and somewhat confused by this question. I tried working out the math myself and never came up with their answer. I'm very inexperienced with power factor calculations, so I'm not terribly surprised by my failure.

Finally, I created an LTspice simulation with voltage sources as the three phase power supply and current sources as "loads" forcing the stated amount of current through each lamp or AM transmitter load location. I then added very low value resistors in series with each line so that I'd have a place to measure line current. Sure enough, phase current is 30.45, slightly higher than I expected, but not too far off. Then, starting with 30.45A in each phase, multiply by sqrt(3) and you get 52.7.

Like I said, even after seeing the simulation confirm the result, I'm still struggling to understand it at multiple levels. Still, for what it's worth, I'm pretty confident that their answer is correct - LTspice hasn't given me a wrong answer yet... except when I fed it bad data!
View attachment 186475
But you can’t do it by math? I have heard of lSpice but not sure what it is... glad this was hard for someone else lol

#### ebeowulf17

Joined Aug 12, 2014
3,274
But you can’t do it by math? I have heard of lSpice but not sure what it is... glad this was hard for someone else lol
Well, like I said, no electronics or electrical training of any kind - just self taught through the internet. My rudimentary understanding of power factor calculations has been good enough to predict (approximately) phase shift based on the difference between linear and vector sums of two amperage readings, and vice-versa, but that's about it.

As for delta transformers, I know what they look like on a schematic, and have a vague conceptual understanding, but don't know any of the math (I didn't know about the sqrt(3) line current to phase current relationship until you brought it up!)

As for the math in this particular scenario, based on my very limited understanding, I would've expected phase current to be 29.6 instead of 30.45A. On the one hand, I'm within 3%, which would be close enough for a lot of basic estimation work. On the other hand, I'm obviously doing it totally wrong or there wouldn't be any significant error! Anyway, once you've got the phase current, you were obviously right about the sqrt(3) line current calculation. That part checks out perfectly in LTspice.

#### VulcanCCIT

Joined Sep 14, 2019
19
Well, like I said, no electronics or electrical training of any kind - just self taught through the internet. My rudimentary understanding of power factor calculations has been good enough to predict (approximately) phase shift based on the difference between linear and vector sums of two amperage readings, and vice-versa, but that's about it.

As for delta transformers, I know what they look like on a schematic, and have a vague conceptual understanding, but don't know any of the math (I didn't know about the sqrt(3) line current to phase current relationship until you brought it up!)

As for the math in this particular scenario, based on my very limited understanding, I would've expected phase current to be 29.6 instead of 30.45A. On the one hand, I'm within 3%, which would be close enough for a lot of basic estimation work. On the other hand, I'm obviously doing it totally wrong or there wouldn't be any significant error! Anyway, once you've got the phase current, you were obviously right about the sqrt(3) line current calculation. That part checks out perfectly in LTspice.
Well I have asked the testing people for the solution which they said is an option... just waiting to hear back... thank you!!!!!!

• ebeowulf17

#### GetDeviceInfo

Joined Jun 7, 2009
1,727
Why wouldn’t you vectorily add your phase currents(in and out of phase), then calculate your line current. The question is a bit funky, seemingly asking for both phase and line currents

• ebeowulf17

#### VulcanCCIT

Joined Sep 14, 2019
19
Why wouldn’t you vectorily add your phase currents(in and out of phase), then calculate your line current. The question is a bit funky, seemingly asking for both phase and line currents
Well it does not indicate a capacitive or inductive load so how would I know the actual phase angle but I’ll look at that

#### ebeowulf17

Joined Aug 12, 2014
3,274
Well it does not indicate a capacitive or inductive load so how would I know the actual phase angle but I’ll look at that
I don't think inductive vs. capacitive will change the result in this case. Whether the phase angle you calculate is a negative or positive amount won't change the rest of the equations.

I get it now! (thanks to @GetDeviceInfo) I wasn't actually doing vector math before - I was just scaling the smaller of the two current values by the power factor, which neglects the other axis in the vector math. Now that I see it, I feel silly for not seeing my error sooner. This page (https://www.allaboutcircuits.com/textbook/alternating-current/chpt-2/complex-vector-addition/) does a good job of explaining the vector math for anyone reading this who isn't already familiar with it.

In the case of this problem, I think you can find the phase angle for the AM transmitter at 80% power factor by using the inverse cosine function (Arccos or ACOS) on the power factor (still not totally sure, but it seems to be working.) Use the resulting phase angle in your vector addition of the two currents and you should hopefully get the right answer. In my case, I'm still not totally sure I've got it, because of rounding questions. I come up with 30.463A phase current and 52.764 line current. If I round that to one decimal place, I'd call it 52.8 instead of 52.7. So I don't know if I'm following different rounding rules than they are, or if this means I'm still not doing some part of the math quite right. It's pretty close though!

• BobaMosfet and cmartinez

#### BobaMosfet

Joined Jul 1, 2009
1,117
As we know, PF = 80%.

Apparent power is VA = 100V * 20A = 2000W.
True Power being 80% of Apparent Power = 1600W

acos(0.80) = 38.659808254090090604005862335173° (phase angle)

Vector math:

1st Phasor: 12 ∠ 0°
2nd Phasor: 20 ∠ 38.659808254090090604005862335173°

I used standard trig to derive the right-triangles run/rise:

Phasor 1 (X/Y): 12.0/0.0
Phasor 2 (X/Y): 15.6174/12.4939

Final Vector: 30.3119948A ∠ 24.34165138°

Final current: (30.3119948A * 1.732) = 52.5003749936A

Obviously, there are rounding errors, but this is as close as I could get to the purported Test answer. My $0.02..... Any glaring mistakes with what I've done? #### cmartinez Joined Jan 17, 2007 6,991 As we know, PF = 80%. Apparent power is VA = 100V * 20A = 2000W. True Power being 80% of Apparent Power = 1600W acos(0.80) = 38.659808254090090604005862335173° (phase angle) Vector math: 1st Phasor: 12 ∠ 0° 2nd Phasor: 20 ∠ 38.659808254090090604005862335173° I used standard trig to derive the right-triangles run/rise: Phasor 1 (X/Y): 12.0/0.0 Phasor 2 (X/Y): 15.6174/12.4939 Final Vector: 30.3119948A ∠ 24.34165138° Final current: (30.3119948A * 1.732) = 52.5003749936A Obviously, there are rounding errors, but this is as close as I could get to the purported Test answer. My$0.02..... Any glaring mistakes with what I've done?
If what you mean by acos(0.80) is the inverse cos of 0.80, then the result should be 36.869897645844021296855612559093°

• ebeowulf17

#### VulcanCCIT

Joined Sep 14, 2019
19
As we know, PF = 80%.

Apparent power is VA = 100V * 20A = 2000W.
True Power being 80% of Apparent Power = 1600W

acos(0.80) = 38.659808254090090604005862335173° (phase angle)

Vector math:

1st Phasor: 12 ∠ 0°
2nd Phasor: 20 ∠ 38.659808254090090604005862335173°

I used standard trig to derive the right-triangles run/rise:

Phasor 1 (X/Y): 12.0/0.0
Phasor 2 (X/Y): 15.6174/12.4939

Final Vector: 30.3119948A ∠ 24.34165138°

Final current: (30.3119948A * 1.732) = 52.5003749936A

Obviously, there are rounding errors, but this is as close as I could get to the purported Test answer. My \$0.02..... Any glaring mistakes with what I've done?
For my education ... I understand vector 1 ... x=12 y=0 but can you explain or state the formula for vector 2?

#### ebeowulf17

Joined Aug 12, 2014
3,274
If what you mean by acos(0.80) is the inverse cos of 0.80, then the result should be 36.869897645844021296855612559093°
Agreed - I just double checked my spreadsheet against the rapid tables online calculator to make sure I hadn't messed up the first time. I get the same number as you.

#### ebeowulf17

Joined Aug 12, 2014
3,274
For my education ... I understand vector 1 ... x=12 y=0 but can you explain or state the formula for vector 2?
x = cos(phase angle) * amperage
y = sin(phase angle) * amperage