3 phase 208V on two elements

[avrepair]

One element 12.8 ohms is connected between L1 and L2 and the other identical element is connected between L2 and L3, and they are on the same 3 pole breaker. What would the current be on each line?

 

[kbsparky]

Is your Amprobe broke?

What do you think the current measurements should be?

 

[avrepair]

In theory only, have not done it.

In theory only, have not done it.

I think each element would draw 16.3 amps but I am not clear on how the currents would divide at common point. Would L2 carry 32.6 amps or would it divide because of the vector differences.

 

[jumper]

Look up Ohm's Law.

 

[gar]

100313-0732 EST

avrepair:

In theory yes for 16.25, but maybe about about 10 to 20% less because of the temperature coefficient of resistance change.

Yes you need to use vector arithmetic.

.

 

[hardworkingstiff]

I think each element would draw 16.3 amps but I am not clear on how the currents would divide at common point. Would L2 carry 32.6 amps or would it divide because of the vector differences.

Great question.

100313-0732 EST

avrepair:

In theory yes for 16.25, but maybe about about 10 to 20% less because of the temperature coefficient of resistance change.

Yes you need to use vector arithmetic.

.

What's the answer (is it 28.23?), and can you show how to get it? Thanks.

 

[jumper]

Look up Ohm's Law.

Ignore this post, my darn computer went into offline and I did not see post #3 when I replied.

 

[gar]

100313-0942 EST

There are two equal length vectors at an angle of 120 deg.

Draw the two vectors. You need to calculate the third side of the triangle. To do this you create a right triangle.

One side of the right triangle is X + X sin 30 deg = 3*X/2. The other side is X cos 30 deg = (sq-root 3) * X / 2 .
Now calculate the hypotenuse of the right triangle = sq-root ( square each of the above two terms and sum the squares )
sq-root of (9*X^2/4 + 3* X^2/4) = sq-root of (12*X^2/4) = sq-root of (3*X^2) = sq-root of 3 * X = 1.732 * X .

Thus, the result in the original post is 1.732 * 16.25 = 28.145 .

To remember 1732 associate it with the birth year of George Washington.

.

 

[augie47]

avrepair, assuming you are still addressing your water heater project, don't forget 422.13 requires that load to be treated as continuous so the 28.145 becomes a calculated load of 35.18 preventing you from using your #10 with the 30 amp required {240.4(D)} breaker.

 

[kwired]

avrepair, assuming you are still addressing your water heater project, don't forget 422.13 requires that load to be treated as continuous so the 28.145 becomes a calculated load of 35.18 preventing you from using your #10 with the 30 amp required {240.4(D)} breaker.

If that is the case his decision is 2 25 amp 2 pole circuits or 1 40 amp 3 pole circuit. Each with its own set of circumstances.

 

[K8MHZ]

To remember 1732 associate it with the birth year of George Washington.

Or, if you are like me, use 1.732 to remember when Ol' Georgie was born. (I hated history.....)