The question below is something that I certainly used to know the answer to, but the answer has apparently fallen out of my brain.

I think it belongs in this section because it is education related. (I wasn't sure if it belonged here or in Physics)

It relates to the amperage load when connecting two heating elements on a 3-Phase delta supply.

The only way I can describe my question is by the examples below. I'll start with a Wye system to hopefully make my question clear. (Yeah, sure....)

For simplicity, the amp draw on the sample loads will be 20 each, even though the voltage is different on each example. There are no real parts involved, just hypothetical.

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WYE SYSTEM WE KNOW THE ANSWER-

Picturing 2 purely resistance loads (for example heating elements) of 20 amps per item on a 120/208 volt Wye system as:

Element One is wired across L-1 & L-2

Element Two is wired across L-2 & L-3

Reading the amperage between L-1 & the element, we will have 20 amps.

Reading the amperage between L-3 & the element, we will have 20 amps.

Reading the amperage between L-2 & where both elements are connected, we will have 40 amps.

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BUT ON A DELTA SYSTEM = ?

Picturing 2 purely resistance loads (for example heating elements) of 20 amps per item on a 125/240 volt Delta system, where

L-1 & L-3 to ground are each 125 volts

L-2 to ground is 208 volts

L-1 to L-2 reads 240 volts

L-1 to L-3 reads 240 volts

L-2 to L-3 reads 240 volts

Element One is across L-1 and L-2

Element Two is across L-3 and L-2

Reading the amperage between L-1 and the element, we will read 20 amps.

Reading the amperage between L-3 and the element, we will read 20 amps.

THE QUESTION-

But, reading the amperage between L-2 and where both elements are tied together, will we have 20 amps, 40 amps or

20 amps times the square root of 3?

Thank You All for putting up with this "I should remember this" question. I appreciate your help!

Paul