#### ebeowulf17

Joined Aug 12, 2014
3,282
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?
Other than the dubious inverse cosine number, the rest of your process looks like exactly what I did. I'm not entirely sure I'm right either, but my method is the same as yours, for whatever that's worth.

#### VulcanCCIT

Joined Sep 14, 2019
19
a lot of the examples in books and on the internet talk about the power factor being leading or lagging but the test question does not indicate leading or lagging. when lagging for example the power factor would be - so the vector math they are adding positive and negative angles giving you a result. Not sure if they are assuming this is a resistive load? so angle is in phase? but in phase with what? with the 12 amp load the phase is 0. the .8pf is angle 36 and some change... so no angle addition or subtraction... all of this is confusing me lol. I had a circuit analysis class in college and still have the book, but it was so long ago about 38 years ago... difficult to re-learn the math. So far it seems no one has come up with an exact match to the test answer but very close!!!

#### vu2nan

Joined Sep 11, 2014
188
Hi Vulcan,

Here goes!

Lamp load current = I₁ = 12 A

AM Transmitter current = I₂ = 20 A

Resultant phase current = Iph

'X' component of phase current = Xph

'Y' component of phase current = Yph

Xph = I₁ Cosφ₁ + I₂ Cosφ₂ = (12 * 1 + 20 * 0.8) = 28

Yph = I₁ Sinφ₁ + I₂ Sinφ₂ = (12 * 0 + 20 * 0.6) = 12

Iph² = Xph² + Yph² = 28² + 12² = 928

Iph = √ 928 = 30.46

Line current = Iph * 1.73 = 52.7

Regards,

Nandu.

Last edited:
• ebeowulf17

#### VulcanCCIT

Joined Sep 14, 2019
19
Hi Vulcan,

Here goes!

Lamp load current = I₁ = 12 A

AM Transmitter current = I₂ = 20 A

Resultant phase current = Iph

'X' component of phase current = Xph

'Y' component of phase current = Yph

Xph = I₁ Cosφ₁ + I₂ Cosφ₂ = (12 * 1 + 20 * 0.8) = 28

Yph = I₁ Sinφ₁ + I₂ Sinφ₂ = (12 * 0 + 20 * 0.6) = 12

Iph² = Xph² + Yph² = 28² + 12² = 928

Iph = √ 928 = 30.46

Line current = Iph * 1.73 = 52.7

Regards,

Nandu.
Omg so easy!!!!! I am going to study this and practice it with other values thank you!!!!!!!!!! This forum has been amazing!!!!

• cmartinez and ebeowulf17

#### ebeowulf17

Joined Aug 12, 2014
3,282
a lot of the examples in books and on the internet talk about the power factor being leading or lagging but the test question does not indicate leading or lagging. when lagging for example the power factor would be - so the vector math they are adding positive and negative angles giving you a result. Not sure if they are assuming this is a resistive load? so angle is in phase? but in phase with what? with the 12 amp load the phase is 0. the .8pf is angle 36 and some change... so no angle addition or subtraction... all of this is confusing me lol. I had a circuit analysis class in college and still have the book, but it was so long ago about 38 years ago... difficult to re-learn the math. So far it seems no one has come up with an exact match to the test answer but very close!!!
As I said before, it doesn't matter if it's leading or lagging to solve this question. Here's a link to a better tutorial on vector addition:

Here's a good image from that website, helping visualize a vector addition: And here you'll see that if the angle between the first and second lines goes in the opposite direction (positive vs. negative angle, leading vs. lagging, capacitive vs. inductive, etc.) the resulting vector still has the same magnitude. Since the question is only looking for the resulting RMS current, not any phase information or instantaneous current levels, you don't need to know whether it's leading or lagging. You can assume either one you want and calculate accordingly - the result is the same either way.

#### ebeowulf17

Joined Aug 12, 2014
3,282
Hi Vulcan,

Here goes!

Lamp load current = I₁ = 12 A

AM Transmitter current = I₂ = 20 A

Resultant phase current = Iph

'X' component of phase current = Xph

'Y' component of phase current = Yph

Xph = I₁ Cosφ₁ + I₂ Cosφ₂ = (12 * 1 + 20 * 0.8) = 28

Yph = I₁ Sinφ₁ + I₂ Sinφ₂ = (12 * 0 + 20 * 0.6) = 12

Iph² = Xph² + Yph² = 28² + 12² = 928

Iph = √ 928 = 30.46

Line current = Iph * 1.73 = 52.7

Regards,

Nandu.
Thanks for posting this - I see now where rounding got me!

Since I created formulas in a spreadsheet and let it crunch the numbers, there was no rounding happening at intermediate steps in the calculations (or at least the rounding was happening at a much higher level of precision, to the point of being negligible.) If you round to two decimal places at each step, as you appear to have done, then the final value clearly rounds quite neatly to 52.7. This rounding scenario explains that last 1/10th of an amp difference between my result and the test question's official answer. Cheers!

#### VulcanCCIT

Joined Sep 14, 2019
19
Hi Vulcan,

Here goes!

Lamp load current = I₁ = 12 A

AM Transmitter current = I₂ = 20 A

Resultant phase current = Iph

'X' component of phase current = Xph

'Y' component of phase current = Yph

Xph = I₁ Cosφ₁ + I₂ Cosφ₂ = (12 * 1 + 20 * 0.8) = 28

Yph = I₁ Sinφ₁ + I₂ Sinφ₂ = (12 * 0 + 20 * 0.6) = 12

Iph² = Xph² + Yph² = 28² + 12² = 928

Iph = √ 928 = 30.46

Line current = Iph * 1.73 = 52.7

Regards,

Nandu.
One question. On Xph 12*cos(degree) ... at unity isn’t the angle 0 instead of 1?

#### VulcanCCIT

Joined Sep 14, 2019
19
Thanks for posting this - I see now where rounding got me!

Since I created formulas in a spreadsheet and let it crunch the numbers, there was no rounding happening at intermediate steps in the calculations (or at least the rounding was happening at a much higher level of precision, to the point of being negligible.) If you round to two decimal places at each step, as you appear to have done, then the final value clearly rounds quite neatly to 52.7. This rounding scenario explains that last 1/10th of an amp difference between my result and the test question's official answer. Cheers!
Thank you ebe!!! More to study it has been 39 years since my circuit analysis class and in the real world of broadcast Engineering it is more hands on... the designers did the math, we jut keep it running but basic theory should be in the brain for sure!!!!

• ebeowulf17

#### VulcanCCIT

Joined Sep 14, 2019
19
Vu2 disregard ... cos(0) is 1 lol

#### vu2nan

Joined Sep 11, 2014
188
Omg so easy!!!!! I am going to study this and practice it with other values thank you!!!!!!!!!! This forum has been amazing!!!!
Any time, Vulcan!

Regards,

Nandu.

#### vu2nan

Joined Sep 11, 2014
188
Vu2 disregard ... cos(0) is 1 lol .

#### vu2nan

Joined Sep 11, 2014
188
a lot of the examples in books and on the internet talk about the power factor being leading or lagging but the test question does not indicate leading or lagging. when lagging for example the power factor would be - so the vector math they are adding positive and negative angles giving you a result. Not sure if they are assuming this is a resistive load? so angle is in phase? but in phase with what? with the 12 amp load the phase is 0. the .8pf is angle 36 and some change... so no angle addition or subtraction... all of this is confusing me lol. I had a circuit analysis class in college and still have the book, but it was so long ago about 38 years ago... difficult to re-learn the math. So far it seems no one has come up with an exact match to the test answer but very close!!!
Hi Vulcan,

The resistive lamp load current is in phase with the voltage (along the X axis). Power factor = Cos φ₁ = Cos 0 = 1.
The reactive AM transmitter current may either lead or lag the voltage. Power factor = Cos φ₂ = 0.8. Power factor cannot be negative since Cos -φ = Cos φ.

Regards,

Nandu.

Last edited:

#### vu2nan

Joined Sep 11, 2014
188
Thanks for posting this - I see now where rounding got me!
Since I created formulas in a spreadsheet and let it crunch the numbers, there was no rounding happening at intermediate steps in the calculations (or at least the rounding was happening at a much higher level of precision, to the point of being negligible.) If you round to two decimal places at each step, as you appear to have done, then the final value clearly rounds quite neatly to 52.7. This rounding scenario explains that last 1/10th of an amp difference between my result and the test question's official answer. Cheers!
Hi ebeowulf17,

You're most welcome.

Yes, I do understand that it was through sheer luck that I got the exact result!

Regards,

Nandu.

Last edited:

#### VulcanCCIT

Joined Sep 14, 2019
19
Hi Vulcan,

The resistive lamp load current is in phase with the voltage (along the X axis). Power factor = Cos φ₁ = Cos 0 = 1.
The reactive AM transmitter current may either lead or lag the voltage. Power factor = Cos φ₂ = 0.8. Power factor cannot be negative since Cos -φ = Cos φ.

Regards,

Nandu.
Thank you!!

#### VulcanCCIT

Joined Sep 14, 2019
19
So based on this entire discussion, can you all verify this PIC as a proper drawing and analysis?

See attached

#### vu2nan

Joined Sep 11, 2014
188
View attachment 186718 Yes makes more sense I was using a vector math example that I found but it was based on other values...yours is actually the accurate way. I redrew it... how does this look?
See attached.
Hi Vulcan,

I had wrongly mentioned the Iph value as 28, instead of 30.46, in the vector diagram. I regret the error.

Here's the corrected vector diagram. Regards,

Nandu.

#### VulcanCCIT

Joined Sep 14, 2019
19
I also wanted to thank ALL of you who helped me with this. It has been about 38 or so years since I studied circuit analysis. I appreciate all of you and I hope this thread helps someone in the future. Thank you!!!    • BobaMosfet and cmartinez

#### BobaMosfet

Joined Jul 1, 2009
1,858
If what you mean by acos(0.80) is the inverse cos of 0.80, then the result should be 36.869897645844021296855612559093°
You are correct, I think I must have mis-typed it. Sorry about that.