# Voltage drop equation confusion



## EngrinSF (Jan 19, 2018)

Hi,

The equation for voltage drop is I*L/1000[ R*cos(theta) + X*sin(theta)]. But I am confused on when to use the plus or minus sign in between the resistance and reactance?  Different problems in CI show different approaches. Someone please help!


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## Omer (Jan 19, 2018)

reema_suresh said:


> Hi,
> 
> The equation for voltage drop is I*L/1000[ R*cos(theta) + X*sin(theta)]. But I am confused on when to use the plus or minus sign in between the resistance and reactance?  Different problems in CI show different approaches. Someone please help!


Use the plus sign when the power factor is lagging (the most common case), and the minus sign when the power factor is leading.


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## EngrinSF (Jan 19, 2018)

I thought it's the other way round? plus for leading and minus for lagging?


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## Zach Stone P.E. (Jan 19, 2018)

reema_suresh said:


> Hi,
> 
> The equation for voltage drop is I*L/1000[ R*cos(theta) + X*sin(theta)]. But I am confused on when to use the plus or minus sign in between the resistance and reactance?  Different problems in CI show different approaches. Someone please help!


Hi Reema

The key to remember is that voltage drop is just the voltage across the line impedance connecting the source to the load (unless you are calculating the percent voltage drop instead of the voltage).

For example, the equation you wrote:

I*L/1000[ R*cos(theta) + X*sin(theta)]

Is really just Ohm's law solving for the voltage across the conductor impedance depending on the total line current:

V = (I)•(Z)

Where:

Z = (L)•(Conductor Impedance in ohms)

To answer the specific question you asked:

"I am confused on when to use the plus or minus sign in between the resistance and reactance?"

This depends on the type of load. Sine the load in question in the conductor connecting the source to the load, the sign will almost always be positive since conductors tend to be more inductive than they are capacitive.

This means that the sign between the resistive and reactive term will be a plus sign in the form of:

 Z per 1,000 ft = (R per 1,000ft + jX per 1,000ft)

Where R is the AC resistance of the conductor

And X is the reactance of the conductor.


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## EngrinSF (Jan 19, 2018)

There is a question on the CI tests question 48: 







Why is there a negative sign in the formula? and XL is inductive!


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## rg1 (Jan 20, 2018)

reema_suresh said:


> There is a question on the CI tests question 48:
> 
> View attachment 10721
> 
> ...


The Theta is power factor of the load. It does not have anything to do with Z of Line feeder. So be careful. The formula is for inductive line and inductive load. Theta is positive for lagging load and negative for leading load. In the question given load is capacitive, (leading) so take theta negative. Try to derive the formula yourself by drawing phasor diagram. You will find a good discussion on the subject including phasor diagrams by members including me here in the forum. If you do not get it, let me know , I will draw again for benefit of all.


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## EngrinSF (Jan 21, 2018)

Thank you for the reply! I understood the phasor diagrams but the one question that remains is , how do you know if the load is capacitive or inductive in the mentioned problem? I do not see any info on type of load.


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## rg1 (Jan 22, 2018)

reema_suresh said:


> Thank you for the reply! I understood the phasor diagrams but the one question that remains is , how do you know if the load is capacitive or inductive in the mentioned problem? I do not see any info on type of load.


Hint

 I= 200&lt;25

Can you get something from angle 25 degrees here. What if there is negative sign here?


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## EngrinSF (Jan 22, 2018)

Yes! Positive for leading load and negative for lagging load!

Thank you!


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## rg1 (Jan 22, 2018)

reema_suresh said:


> Yes! Positive for leading load and negative for lagging load!
> 
> Thank you!


Great you got it right!


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## bobbilly (Jan 23, 2018)

reema_suresh said:


> There is a question on the CI tests question 48:
> 
> View attachment 10721
> 
> ...


What edition book do you have, 2nd for me? I just came across this problem today and their answers are different but the same exact problem. Looks my problem is wrong.

https://imgur.com/a/xkcJS

In terms of flipping the signs this is what I saw: https://www.wolframalpha.com/input/?i=200[(.062%2F1000*175*cos(25))+-+(.041%2F1000*175*sin(25))]+%3D%3D+200[(.062%2F1000*175*cos(-25))+%2B+(.041%2F1000*175*sin(-25))]


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## EngrinSF (Jan 23, 2018)

@bobbilly I have the 2011 edition! I do not know what edition it is. I think your maybe right cause they corrected it?


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## bobbilly (Jan 23, 2018)

rg1 said:


> The Theta is power factor of the load. It does not have anything to do with Z of Line feeder. So be careful. The formula is for inductive line and inductive load. Theta is positive for lagging load and negative for leading load. In the question given load is capacitive, (leading) so take theta negative. Try to derive the formula yourself by drawing phasor diagram. You will find a good discussion on the subject including phasor diagrams by members including me here in the forum. If you do not get it, let me know , I will draw again for benefit of all.


Could you draw out the phasor diagram? I think I am just confusing myself here. You are saying the opposite of what Justin said above. 



EngrinSF said:


> @bobbilly I have the 2011 edition! I do not know what edition it is. I think your maybe right cause they corrected it?


2nd edition is 2015, so I guess they corrected it to be positive.


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## rg1 (Jan 26, 2018)

bobbilly said:


> Could you draw out the phasor diagram? I think I am just confusing myself here. You are saying the opposite of what Justin said above.
> 
> 2nd edition is 2015, so I guess they corrected it to be positive.


I have drawn a phasor diagram to explain VD for lagging load and Inductive line. For leading load inductive line or any combination of load and line you can draw yourself.


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## ellen3720 (Sep 3, 2018)

I'm having trouble determining when to use the complex voltage drop equation including cos and sin, as used above.

For example, the CI problem above uses Rcos and Xsin. However, similar CI voltage drop problem 4-19 doesn't. It just multiplies the values from NEC table 9 by the current.

Is there a rule of thumb for when to use the complex equation?

Problem 4-19 attached for reference.


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## Szar (Sep 4, 2018)

ellen3720 said:


> I'm having trouble determining when to use the complex voltage drop equation including cos and sin, as used above.
> 
> For example, the CI problem above uses Rcos and Xsin. However, similar CI voltage drop problem 4-19 doesn't. It just multiplies the values from NEC table 9 by the current.
> 
> ...


If the question asks for the NEC, you use the NEC.  If the question doesn't ask for the NEC, you use Ohms Law (with impedance).


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