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Voltage Equation

08/01/2011 4:20 AM

why line voltage = 1.732 phase voltage?

can any one tell me how to derive this equation

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#1

Re: voltage equation

08/01/2011 4:44 AM

Most people with electrical training can do this.

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#2

Re: voltage equation

08/01/2011 5:58 AM

Try having a look on Google!

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#3

Re: Voltage Equation

08/01/2011 8:54 AM

Read B L Theraja thoroughly.

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#4
In reply to #3

Re: Voltage Equation

08/01/2011 12:18 PM

I don't know about the current version of BLT, however in our time it was strictly no no for electrical engineering students (there were at least on those editions innumerable errors, and one of our pastimes would be to find the number of them in a given time)

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#7
In reply to #4

Re: Voltage Equation

08/01/2011 8:47 PM

I very much agree with you that for advanced leraning BLT a strict NO-NO. But, at least for the question in point (i.e.) voltage equation, BLT has made no mistake.

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#5

Re: Voltage Equation

08/01/2011 12:24 PM

Line voltage is stated as "phase to phase". Phase voltage is stated as "phase to ground".
In a three phase system, each phase is 120 degrees out of phase with respect to the other two, one leading, and the other trailing.
Draw the vector diagram for this and you get three triangles inside a larger triangle, the outer sides being phase to phase and the inner sides being phase to ground. The outer triangle is equilateral, with angles of 60 degrees, and the inner triangles are isosceles with angles at the outer triangle's vertices of 30 degrees.
Look at one of the inner triangles and bisect it with a vector from ground perpendicular to the vector for phase to phase. You see a right triangle. Now you can do trigonometry...
The base is one half the phase to phase voltage. Lets call that X. In trig, cosine(theta) = X (one half phase to phase) over hypotenuse (phase to ground). Cosine 30 is 0.866, so phase to ground is one half phase to phase over 0.866, or phase to phase over 1.732.
A typical US distribution system has a three phase power at 13.2kv, phase to phase. We make that, simply, 13.2. If you measure phase to ground while the system is in relatively good balance, which it is most of the time, you get 7.62kv. We call that 7620. This is in the ratio of 1.732.

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#6

Re: Voltage Equation

08/01/2011 12:42 PM
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#8

Re: Voltage Equation

08/02/2011 2:47 AM

1.732 may seem arbitrary, but √3 is the pure geometry of an equilateral triangle plus the segments from each vertex to the centroid.

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#9

Re: Voltage Equation

08/02/2011 5:32 PM

In the forum "nino" has given the answer for the relation between the Line Voltage and Phase voltage by equilateral triangle with geometrically presentation as

Distance between any two Vertices of equilateral Triangle = √3x distance between any vertex to Centriod (Neutral).

I try to show the same based on the power balance.

For simplicity assume the STAR connected load.

3-Phase power balance power load = 3x Voltage (phase to neutral, Vn) x Phase Current (Ip) given as

3-Phase Power = 3x Vn x Ip…………………………………………………………………………………………..1

Same load power can be given by Line to line voltage(VL) and current in line(IL) as

Three phase Power = √3x VL x IL ……………………………………………………………………………………2

Since Equation 1 & 2 are same then:

3-Phase Power = √3x VL x IL = 3x VnxIp

√3x VL x IL = 3x Vn x Ip; or VL x IL = √3x Vn x Ip …………………………………………………….3

Since Star connection Load then IL = Ip

Equation-3 becomes as VL = √3x Vn

Or Line to Line Voltage, VL = √3x Phase to Neutral Voltage

This also be applicable for Delta connected load.

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Anonymous Poster (1); electricalexpert65 (2); Mushtaq Hussainh (1); nino (1); PWSlack (1); TonyS (1); Tornado (1); wareagle (1)

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