why phase-ground voltage = phase-neutral + ground-neutral Voltage ?

 

The voltage drop due to current through the neutral is what generates the ground-neutral voltage. Ground should always be ZERO volts. Neutral may be slightly above ZERO at the load due to the IR drop in the wire.

 

In an unbalanced 3 phase system, there will be a difference between neutral and ground. Same for a single phase system.

 

why phase-ground voltage = phase-neutral + ground-neutral Voltage ?

Shouldn't that be like this?

phase-ground = phase-neutral + neutral-ground

Just write it mathematically:

(p - g) = (p - n) + (n - g)

Then remove the parenthesis...

p - g = p - n + n - g

...and re-arrange the terms...

p - g = p - g + n - n

...and remove the unnecessary "+ n - n" at the end...

p - g = p - g

...and then it is rather obvious why they are the same.

 

why phase-ground voltage = phase-neutral + ground-neutral Voltage ?

Shouldn't that be like this?

phase-ground = phase-neutral + neutral-ground

Just write it mathematically:

(p - g) = (p - n) + (n - g)

Then remove the parenthesis...

p - g = p - n + n- g

...and re-arrange the terms...

p - g = p - g + n - n

You missed a spot in math. After removing parenthesis, you arrive back at p-g = p-g

Treating them as mathematical variables means they are the same thing, but they aren't.

Current flows through the neutral wire, it does NOT normally flow through ground wire.

 

Treating them as mathematical variables means they are the same thing, but they aren't.

Sure they are. Measuring the relative voltage between two points is finding the difference in the absolute voltages of those two points, even if we can't easily know what those absolute voltages are. The phase has an absolute voltage, the neutral has an absolute voltage, and the difference between those two voltages is the phase-neutral voltage, which is p - n, which is the difference between two numbers, p and n.

Current flows through the neutral wire, it does NOT normally flow through ground wire.

Why is that relevant to a formula about voltages?

 

If zero current was flowing, and there are no faults, they would be the same. If current is flowing, you get a different answer.

In a normal operating circuit, use a house for example, current flows through the neutral and hot conductors equally, all current in hot should equal all current in the neutral. NO current flows in ground wire if there isn't a fault.

Copper has resistance, so any current through it creates a voltage drop. Since there is a voltage drop in the neutral wire and not in the hot, the two are not at the same potential when current is flowing.

 

if there is unbalanced operation due to (unbalanced generation,unbalanced loads,L-L faults and other causes of Vn ~= 0, there would be a difference between Vn & Vg
Note: V(ground) may not be zero ,All addition processes are Vectorial

 

Note: V(ground) may not be zero

If we are talking about power distribution, such as to a receptacle, then ground should always be zero volts, assuming no faults, since it is there only as a safety, not a normal current path. That is the job of the neutral line. True, both wires may tie to the same buss bar in the distribution box but that same bar should, by code, be tied to a driven ground rod, therefore zero volts.The original question, I assumed, (yes, we know what that means),was that measurements were taken at the load point.