3 phase voltages | on ElectriciansForums

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H

hughesy

hi when you measure between phase and phase you get 415 volts and between pase and neutral you get 240 why is it 240 and not say half of 415 .
 
You are opening a WHOLE can of worms now:D

Lenny or someone else may be along later to explain all in great detail.
cheers i was hoping they woul be, not after a major discussion although its very welcome just need a basic anser i know the phases are lagged at 120 deggrees and that each phase is at a differant level than each other phase and a few other basic things just not this one.:confused:
 
Interesting question. I know it's to do with the fact AC is delivered as a sine wave and so the voltage specified is sort of an average, and that the voltage is expressed in terms of RMS to show this, but that's as far as I can go and that may not be properly correct, so I'm also looking forward to the more detailed explanation. I'd like to know what the actual voltage level is when considering peak to peak. You have to consider that during one cycle of AC voltage, the voltage is climbing from relative 0V to whatever the peak voltage is, and then back to 0V before starting a negative cycle from 0V to the peak. Ug, you've made me try to use my brain, and now look what's happened. Danger Will Robinson.
 
Hi,

Let's see if I can get this right -

The voltage between two phases is 400 v - this is the potential difference

between the two lines at any moment in time

eg. If L1 = 180 v then L2 will be (minus) - 220 v giving

a PD of 400 v

This is because they are 120 degrees out of phase with each other.

To see it properly you would have to draw the two sine waves.

The formula that gives this result (for a star connection) is:

line voltage = sq rt of 3 x phase voltage

so

line voltage = 1.732 x 230 = 400 v (approx)

I think that's right:)

Cheers

Wayne
 
Hi,

Let's see if I can get this right -

The voltage between two phases is 400 v - this is the potential difference

between the two lines at any moment in time

eg. If L1 = 180 v then L2 will be (minus) - 220 v giving

a PD of 400 v

This is because they are 120 degrees out of phase with each other.

To see it properly you would have to draw the two sine waves.

The formula that gives this result (for a star connection) is:

line voltage = sq rt of 3 x phase voltage

so

line voltage = 1.732 x 230 = 400 v (approx)

I think that's right:)

Cheers

Wayne
very good answer but not quite what i was looking for although very helpfull and a good learning point .thanks mate
 
Its very difficult to explain in words so I have
attatched a sketch of how 240v line voltage's relation to 415v phase voltage is derived from.

Not very good at sketching
I hope this helps


Ray
 

Attachments

  • Phase Exp005.pdf
    14 KB · Views: 88
Good answers wayne and raylewis ! so how does split phase work ? 460-480v lol only joking ! southern electric use this alot in remote areas like farms etc saves expence sending 3 phases across fields etc !
 
Hey.

Beat me to it Ray.:D:D

I'd managed to draw one on the computer but couldn't then get it on to the forum, tried everything kepyt saying "file type not valid" anyone any ideas why??

Cheers?
 
Its very difficult to explain in words so I have
attatched a sketch of how 240v line voltage's relation to 415v phase voltage is derived from.

Not very good at sketching
I hope this helps


Ray

great sketch this was usefull i ave also found it whilst going through my old college work .

Good answer Wayne.:D

Hughesy mate, what are you looking for then??

have put the drawing to it now and checked some old college work and understand it just needed some time to refresh my memory.cheers.:)

Hi,

Let's see if I can get this right -

The voltage between two phases is 400 v - this is the potential difference

between the two lines at any moment in time

eg. If L1 = 180 v then L2 will be (minus) - 220 v giving

a PD of 400 v

This is because they are 120 degrees out of phase with each other.

To see it properly you would have to draw the two sine waves.

The formula that gives this result (for a star connection) is:

line voltage = sq rt of 3 x phase voltage

so

line voltage = 1.732 x 230 = 400 v (approx)

I think that's right:)

Cheers

Wayne

Hi mate i get it now after some revision and thinking hope i wasnt offensive when i said it wasnt the anser i was looking for, thank you for your help
 
Last edited by a moderator:
Interesting question. I know it's to do with the fact AC is delivered as a sine wave and so the voltage specified is sort of an average, and that the voltage is expressed in terms of RMS to show this, but that's as far as I can go and that may not be properly correct, so I'm also looking forward to the more detailed explanation. I'd like to know what the actual voltage level is when considering peak to peak. You have to consider that during one cycle of AC voltage, the voltage is climbing from relative 0V to whatever the peak voltage is, and then back to 0V before starting a negative cycle from 0V to the peak. Ug, you've made me try to use my brain, and now look what's happened. Danger Will Robinson.

just in case you were wondering your voltage peak level is calculated by multiplying your rms by sq root of 2 so for 230volt your peak is about 325volt:)
 

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