Why use 1.732 for phase voltage in electrical thechnology?

Answer 1

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.

Addition: It's also just the square root of 3.

Comment">Comment

The three 'hot' conductors that supply a three-phase load are called 'line conductors', which is why the voltage between any pair is called a 'line voltage'. There is simply no such thing as a 'phase-to-phase voltage' or a 'phase-to-ground' voltage!

In a three-phase, four-wire, system, a 'phase voltage' is measured between any line conductor and the neutral conductor. In this case, the line voltage is 1.732 times the phase voltage.

In a three-phase, three-wire, system, a 'phase voltage' is measured between any pair of line conductors (there is no neutral conductor), because it is numerically-equal to the line voltage.

Answer 2

A good example is the 208 v 3-phase system which has 208 v between live wires and 120 v from each live to the neutral.

The voltages in the live wires do not all peak at the same time. They are timed to peak at one-third of a cycle apart, so they are spaced by 120 degrees in phase.

So in this system the 120 v lines never get a chance to add up to 240 v, because when each one peaks the others are off-peak. The most they ever sum to is 208 v.

This system has the advantage that it provides efficient power transmission. If you happened to draw 20 amps from live to neutral on all three phases, the current in the neutral would be zero, and that means it produces no power loss. Therefore you only get the power loss in three supply wires.

If the same power was supplied by three separate single-phase systems, there would be a power loss in six wires, obviously.

Another Answer

There is no such thing as a 'phase-to-phase' voltage; the correct term is 'line-to-line voltage' or, simply, 'line voltage'. In a three-phase, four-wire, star (wye) connected load, 'line-to-neutral' voltages are called 'phase voltages'.

A line voltage is the phasor sum of two phase voltages. This works out to 1.732 times the value of one of the phase voltages.

The Europeans use star-connected loads, in which the nominal line voltage is 400 V, resulting in a nominal phase voltage of 230 V.

Answer 3

Because it is the phasor-sum (vectorial sum) of two phase voltages. Assuming that the phase voltages are identified as VAN, VBN, and VCN, each displaced from each other by 120 degrees, then the line voltage (e.g.) VAB is the vector sum of phase voltages VAN plus VNB (which is VBN, reversed). In other words, you are vectorially-adding two identical values that are 60-degrees apart, and the resulting value is always 1.732 (or the square-root of 3) larger than either. Try it yourself, draw two lines 30 degrees apart, vectorially add them, and measure the resulting line -it will be 1.732 times longer than either of the original lines.

Answer 4

First of all, let's get the terminology correct. In a three-phase, four-wire, system, there are two categories of voltage: line voltages (measured between any pair of line conductors) and phase voltages (measured between any line conductor and the neutral conductor). Providing the voltages are balanced, then a line voltage is 1.732 (i.e. the square-root of 3) larger than a phase voltage.

The reason for this is that a line voltage is the phasor (or vector) sum of two phase voltages. For example, is the lines are labelled A-B-C and the neutral is labelled N, then the line voltage VAB is equal to the vector sum of phase voltages VANand VNB, which are displaced from each other by 60 degrees. Application of basic trigonometry will reveal that VAB will be equal to 1.732 times the value of VAN or VNB. Alternatively, you could draw the phasors to scale and measure the relationships.