208v and load calculation.

[danTt]

it seems to be the week of 208v questions, so to keep the trend going, heres one that I've been wondering about for a while but never needed to solve. If I have a vl3500, drawing ~10 amps @ 208, does it draw 10 amps on each leg? or 10 total, 5 on each?

 

[STEVETERRY]

it seems to be the week of 208v questions, so to keep the trend going, heres one that I've been wondering about for a while but never needed to solve. If I have a vl3500, drawing ~10 amps @ 208, does it draw 10 amps on each leg? or 10 total, 5 on each?

A single 208V single phase unit drawing 10A, draws that 10A on each of its two line connections. However, if one 10A unit is connected to each of the pairs of line conductors (L1-L2, L2-L3, L3-L1) the line current for each of L1, L2, L3 is 10A x 1.73. This is because there is no neutral in a 208V delta connection--all loads are connected line-to-line.

So why is the line current not additive, ie 20A per line? Because of the 120 degree phase offset of L1, L2, and L3.


ST

 

[TJCornish]

Kirchhoff's circuit laws - Wikipedia, the free encyclopedia

Kirchoff's law says, among other things, that the current at any point in a simple loop must be the same, which is why in a single-phase circuit, whether that's one hot and a neutral, or two hots, the current must be the same on both legs - i.e. 10A in and 10A out. As Steve mentioned, a 3-phase system is a little different, since there are multiple circuit paths.

 

[Chris15]

Neither Kirchoff's Voltage Law nor Kirchoff's Current Law actually directly say that...
But effectively that is true.
Both KVL & KCL however will ALWAYS hold true in any circuit, three phase, single phase, one source, 1000000 sources etc.

 

[TJCornish]

Neither Kirchoff's Voltage Law nor Kirchoff's Current Law actually directly say that...
But effectively that is true.
Both KVL & KCL however will ALWAYS hold true in any circuit, three phase, single phase, one source, 1000000 sources etc.

I figured the proof was a little verbose for the scope of this thread.

 

[DCATTechie]

A single 208V single phase unit drawing 10A, draws that 10A on each of its two line connections. However, if one 10A unit is connected to each of the pairs of line conductors (L1-L2, L2-L3, L3-L1) the line current for each of L1, L2, L3 is 10A x 1.73. This is because there is no neutral in a 208V delta connection--all loads are connected line-to-line.

So why is the line current not additive, ie 20A per line? Because of the 120 degree phase offset of L1, L2, and L3.


ST

Can you explain where you got the 1.73 from? Is that the phase angle factor?

If I have a 10amp load powered at 208v connected to XY on a branch circuit for example of a 3-phase power distribution unit, what current draw would leg X and leg Y see?

 

[derekleffew]

Can you explain where you got the 1.73 from? Is that the phase angle factor? ...

The below is discussing voltage rather than current, but we've heard that the two are sometimes related.
From http://www.controlbooth.com/forums/lighting-electrics/27159-120-208-240v-questions.html :

In a logical world, one would think that two 120V hots would always add up to 240V [120+120=240], right? Wrong. If the phases are 120° apart, as they are on a three phase system, some cancellation is going to occur. The phases are 120° apart. Multiply 240 x the sine of 120° [0.866] and get 207.84; let's call it 208V. The other way to approach this is to multiply 120 by the sq.rt.(3) [1.732] to get, wait for it, 207.84. In your case, if the hot to neutral voltage is 122V rather than 120V [it varies, see here], you'll get ~211V hot to hot.
The same relationship holds for other voltages, which is why you may see equipment/connectors listed as 277/480V or 347/600V.

See also https://www.controlbooth.com/threads/amperes-current-draw-for-a-moving-light.8966/ .