# Where do we use 1.73 when calculating 3 phase power

#### Thesparky

##### Member
I joined this forum to respond to this question, after joining I found that it is a closed thread which is too bad since it never really got answered... however there was a lot of engineering discussion about it.
I hope the original participants see this and join in.

#### charlie b

##### Moderator
Staff member
Is it possible for you to post a link to the thread to which you refer? That would help us to avoid duplicating comments.

The short answer to your question is that the factor "the square root of 3" (approximately 1.732) is going to find itself, one way or another, into any calculation that involves three phase systems. You will, no doubt, have seen some of the formulas. Here are two of the most common examples:

• Apparent power (VA) = 1.732 x Volts x Amps.
• Real Power (W) = 1.732 x Volts x Amps x Power Factor.

Is there something specific you are looking for?

#### junkhound

##### Senior Member
And as a side question for the student, as to why or how, one may ask, did sqrt 3 become the 'go to' number rather than tan(60)?

After all, if not a perfectly symmetric 3 phase system, the tan(phase angle) number will provide more accurate number for non symmetric systems.

#### Thesparky

##### Member

I found it while Googling my question of why is it the square root of 3 (1.73) used in the power calculations for 3 phase systems. I still haven't found that answer, just that it is used.

#### Carultch

##### Senior Member
I joined this forum to respond to this question, after joining I found that it is a closed thread which is too bad since it never really got answered... however there was a lot of engineering discussion about it.
I hope the original participants see this and join in.

Draw out the equilateral triangle, and the "Y" within it, representing the phase voltages on the complex plane. Put the origin at the center of the "Y". The projection of each leg of the "Y" onto the real axis represents each phase's voltage to neutral in real time. The setup will rotate as time moves on, similar to Smart\$'s icon which illustrates this concept very well.

The ratio between the leg length of the "Y", and the leg length of the equilateral triangle, will be the square root of 3, which is approximately 1.732. This is the ratio between line-neutral voltage and line-line voltage, in a three phase system. You can derive this with the law of sines, and you will see sin(120)/sin(30) appear in the equation. That ratio is sqrt(3).

If you use phase-to-neutral voltage, calculating power (or more generally, apparent power) is equal to 3*Voltage*current, or in symbols P=3*Vln*I. There are three phases, each carrying the equivalent power of a single phase circuit, at the same current at the phase to neutral voltage.

Now switch to phase-to-phase voltage. Start with P=3*Vln*I. Multiply by 1 in a fancy way, P=3*Vln*I*(sqrt(3)/sqrt(3)). Replace Vln*sqrt(3), with Vll, for the line-to-line voltage. Reduce 3/sqrt(3), and you get P=sqrt(3)*Vll*I.

#### Johnhall30

##### Member
VA = Volts (L-L) x 1.73 x Current (3-phase only)

FLA of each leg in a 3-phase circuit = VA / (Volts (L-L)) * 1.73)

#### Thesparky

##### Member

After reading the entire original thread I did not see the answer so here's my input. I've been doing industrial electrical and instrumentation work for 38 years, starting out as an apprentice electrician and working up through the ranks to superintendent. I now work in the corporate office as a Senior Lead estimating and providing project support. There are many times I have to use formulas to calculate things like cable size, ckt ampacity, and transformer/generator sizing to name a few. These are the formulas I use on a regular basis (3 phase only).

Amps when Voltage and HP are known
:

(Horsepower * 746) / (Volts * Efficiency (.85) * Power Factor (.80) * 1.73)

If Efficiency and/or Power Factor are not known I use .85 and .80 which are typical values.

Amps when Voltage and Kilowatts are known:

(Kilowatts * 1000) / (Volts * Power Factor (PF) * 1.73)

Kilowatts when Voltage and Amps are known:

(Volts * Amps * PF * 1.73) / 1000

Kilovolt-Amperes (KVA) when Voltage and Amps are known:

(Volts * Amps * 1.73) / 1000

Horsepower when Voltage and Amps are known:

(Volts * Amps * Efficiency % * PF * 1.73) / 746

(It takes a horse one hour to lift 746 pounds)

The phase voltage multiplied by 1.73 gives us the line voltage: 277V (As measured from phase to neutral or gnd) * 1.73 = 479.21 (or 480V nominal) which is the most common 3 phase circuit in industrial applications. 480V 3ph is used to power motors and tank heaters while the 277V single phase of the 3 phase system is commonly used for lighting.
I hope this helps to answer the original question. I'd greatly appreciate it if somebody can answer my original question which came up when I was updating my spreadsheet with these formulas and wondered... Why is it the square root of three (1.732)???

I'm sure I learned this back in college but can't remember, not that I'm getting old or have killed too many brain cells... #### Thesparky

##### Member
Thank you Carultch!!!

Thank you Carultch!!!

Makes perfect sense!

#### Thesparky

##### Member

Is it possible for you to post a link to the thread to which you refer? That would help us to avoid duplicating comments.

The short answer to your question is that the factor "the square root of 3" (approximately 1.732) is going to find itself, one way or another, into any calculation that involves three phase systems. You will, no doubt, have seen some of the formulas. Here are two of the most common examples:

• Apparent power (VA) = 1.732 x Volts x Amps.
• Real Power (W) = 1.732 x Volts x Amps x Power Factor.

Is there something specific you are looking for?

Here's the link, I posted it once but don't see it.

#### gar

##### Senior Member
190619-2143 EDT

Thesparky:

The constant 1.732 is based on phasor analysis (vectors), and some assumptions.

Assumptions:
3 phase supply.
All waveforms are pure sine waves, no harmonics. Thus, no waveform distortion.
The 3 phase system is perfectly balanced.
A phasor diagram of interest is an equilateral triangle. Thus. three 60 deg angles.
This can be redrawn as two 30, 60, 90 triangles.
Such a triangle has relations 1^2 + (1/2)^2 = 3/4. The sq-root of 3/4 is the other side or (sq-root 3)/2.
This is not the phasor we want unless you are interested in the voltage from a high leg delta.
But now we have another triangle to be split into two right triangles to get the line to neutral phasor.
See if you can continue to do the analysis to get line to neutral.
Keep in mind that the sine of 30 is exactly 1/2, and cos sin of 30 is exactly (sq-root 3)/2.

All of this depends upon sine waves and perfect balance.

Check for any mistakes I might have made.

.

#### Besoeker3

##### Senior Member
It’s the ratio of VLL to VLN

#### oldsparky52

##### Senior Member
In case the OP is looking for a less technical answer, I submit this (please note, this is very basic and ignores power factor or other such things).

Let's assume you have a balanced 3-phase load of 3600 watts at 208V and a single phase load of 3600 watts at 240V. How much amperage is on each conductor.

Well, the single phase is easy, W=EI, so
3600 = 240*I
3600/240 = I
15 = I.

So, now how do you figure out the 3-phase since you have 208V in 3 different configurations (AB,BC,CA). Because of the relationships of the 3-phase voltages explained above, you have to do something to make the calculations correct. That "something" is using 1.732 as a multiplier to the voltage, then use the same formula you would for the single phase calculation above. The new formula for 3-phase is W=(E*1.732)*I.

So 3600=(208*1.732)*I
3600 = 360*I
10 = I

You would work in reverse if you know the amperage and need to know the watts.

W = (208*1.732)*10
W = (360)*10
W = 3600

Just as an aside, in single phase, if you add the line to neutral voltages for each phase you come up with 240 (120+120=240).
In 3-phase, if you add the line to neutral voltages for each phase you come up with 360 (120+120+120) or ... 208*1.732

#### junkhound

##### Senior Member
For the OP, here is a picture and basic equations, might be easier to understand than just verbiage.
Note that sqrt 3 is valid (as already noted by multiple responders) ONLY for a balanced system
V = line to line voltage for unity phase voltage.

1st column is Pythagoras relationships of right triangle, 2nd column is trig.
Tan 60 = sqrt 3 = 1.73....

Exercise for the student: displace phases by a few degrees and amplitude by a few percent and calculate positive, negative, and zero sequence components. (good luck) hmy:
(note; positive, negative, and zero are math tools only, 'solution' not programmed into most scopemeters so not easy to 'measure' - only ones I've seen with the capability are the windows based 4 channel Tek bench scopes) #### gar

##### Senior Member
190620-0925 EDT

The invention of the concept of phasors to do AC calculations on steady state AC sine waves was by Steinmetz of General Electric late 1800s. See https://en.wikipedia.org/wiki/Phasor

You must be dealing with sine waves and linear loads.

.

#### RichB

##### Senior Member
And as a side question for the student, as to why or how, one may ask, did sqrt 3 become the 'go to' number rather than tan(60)?

After all, if not a perfectly symmetric 3 phase system, the tan(phase angle) number will provide more accurate number for non symmetric systems.

Who cares==They are the same out to at least 20 digits and 3 digit accuracy sent a man to the moon:roll: #### junkhound

##### Senior Member
Who cares==They are the same out to at least 20 digits and 3 digit accuracy sent a man to the moon:roll: Well, For a perfectly balanced system, they are the SAME BY DEFINITION of tangent. My comment was for a non-balanced system.

#### synchro

##### Senior Member
While this has probably been beat to death, I think I might be able to provide a simple perspective.

Putting the OP’s original question into its most basic form:
How do we calculate the power delivered to a load from a given power source by using physically accessible and measurable properties of that power source?

Let the power source be one or more transformer windings, each of which has its own RMS voltage Vphase across it, and RMS current Iphase through it.
The voltage and currents on the load are Vline and Iline.
Assume a unity power factor and balanced voltages and currents where appropriate..

With 2-wire single phase, the power into the load is simply Vphase x Iphase = Vline x Iline.

In the 4-wire wye case Vphase = Vline and Iphase = Iline. The total power is simply 3 Vphase x Iphase = 3 Vline x Iline because we have three windings each contributing power, and the individual phase voltages and phase currents are both directly accessible for measurement. Therefore no scaling by sqrt (3) is necessary.

In the 3-wire wye case the power available is the same as the 4-wire above, but because the neutral point is no longer accessible Vphase is not a measurable quantity. So we have to use the line-to-line voltage Vline, but this overstates the actual voltage from the source winding by sqrt (3). So as mentioned before the power is 3 Vphase x Iphase = (3 /sqrt(3)) Vline x Iphase = sqrt (3) x Vline x Iphase = sqrt (3) x Vline x Iline.

In the 3-wire delta case the voltage Vphase is measurable because it equals Vline, but the phase current Iphase in an individual winding is not. The measurable quantity Iline overstates the current Iphase in the source winding (which is not accessible) by sqrt(3), which can be seen from a phasor diagram of the currents. So in the delta case the power is 3 Vphase x Iphase = 3 Vphase x (1/sqrt(3)) x Iline = sqrt(3) x Vphase x Iline = sqrt(3) x Vline x Iline

So the reason sqrt(3) appears is because:
In the wye connection the line-to-line voltage is a vector addition of two phase voltages resulting in a factor of sqrt(3) more voltage than that on one phase.
In the delta connection each line output current is a vector addition of two phase currents resulting in a factor of sqrt(3) more current than that on one phase.
So in each case either the measurable and delivered voltage or current is sqrt(3) higher than that on a phase (the transformer winding delivering the power). Therefore it must be divided by sqrt(3) to calculate the power delivered from the source.

Last edited:

#### domnic

##### Senior Member
1,73

1,73

After reading the entire original thread I did not see the answer so here's my input. I've been doing industrial electrical and instrumentation work for 38 years, starting out as an apprentice electrician and working up through the ranks to superintendent. I now work in the corporate office as a Senior Lead estimating and providing project support. There are many times I have to use formulas to calculate things like cable size, ckt ampacity, and transformer/generator sizing to name a few. These are the formulas I use on a regular basis (3 phase only).

Amps when Voltage and HP are known
:

(Horsepower * 746) / (Volts * Efficiency (.85) * Power Factor (.80) * 1.73)

If Efficiency and/or Power Factor are not known I use .85 and .80 which are typical values.

Amps when Voltage and Kilowatts are known:

(Kilowatts * 1000) / (Volts * Power Factor (PF) * 1.73)

Kilowatts when Voltage and Amps are known:

(Volts * Amps * PF * 1.73) / 1000

Kilovolt-Amperes (KVA) when Voltage and Amps are known:

(Volts * Amps * 1.73) / 1000

Horsepower when Voltage and Amps are known:

(Volts * Amps * Efficiency % * PF * 1.73) / 746

(It takes a horse one hour to lift 746 pounds)

The phase voltage multiplied by 1.73 gives us the line voltage: 277V (As measured from phase to neutral or gnd) * 1.73 = 479.21 (or 480V nominal) which is the most common 3 phase circuit in industrial applications. 480V 3ph is used to power motors and tank heaters while the 277V single phase of the 3 phase system is commonly used for lighting.
I hope this helps to answer the original question. I'd greatly appreciate it if somebody can answer my original question which came up when I was updating my spreadsheet with these formulas and wondered... Why is it the square root of three (1.732)???

I'm sure I learned this back in college but can't remember, not that I'm getting old or have killed too many brain cells... IT takes a horse one hour to lift 746 pounds. 1 horse power can lift 1980000 pounds in one hour.

#### Besoeker3

##### Senior Member
IT takes a horse one hour to lift 746 pounds. 1 horse power can lift 1980000 pounds in one hour.
Go with SI. Much simpler.

#### Carultch

##### Senior Member
IT takes a horse one hour to lift 746 pounds. 1 horse power can lift 1980000 pounds in one hour.

That's not even remotely close to what the unit horsepower means at all. It makes no mention of how high the horse lifts the load, and 746 is really the number for the conversion factor of HP to Watts.

What happened is, James Watt needed a performance metric to sell the steam engine, and compare it to something people already understood. He studied the performance of draft horses, and how much load they could lift by pulling a mill wheel. Based on the speed the horses walked, and the load the millwheel lifted, he determined that horses could pull with the power equivalent to lifting a 550 pound vertical load at a rate of 1 ft per second.