You are correct.

Each current element of a form 2 meter is designed to produce 1/2 the flux proportionate to the current going through it. The left hand half of the element being wound opposite to the right hand element. And, as you mentioned, the potential coil is connected across each leg (240 volts).
If a 120 v load is pulling "A" amount of current (and for simplicity assuming Watts = Volts x Amps and ignore PF), then 240 x 1/2 A = Watts.

Which of course, is the same as 120 x A = Watts.

For a 240 volt load, the "A" current is returning in the other leg of the meter current coil in the opposite direction. But since the coil is wound in reverse, the flux fields "coming and going" are cumulative.

So 240 volts x (1/2 A + 1/2 A) = Watts.

Which of course, is the same as 240 x A = Watts.

So take an unbalanced load....100 amps on one leg and 10 on the other; and assume a 4 volt drop in the service wire on the heavier loaded leg.

Actual watts,
(116 volts x 100 amps)+(120 volts x 10 amps)=12800 watts or 12.8 Kw.

The meter sees,
236 volts x (1/2 of 100 amps + 1/2 of 10 amps) = 12980 or 12.98 Kw

Or a 1.4% overcharge (if I did my math right). It pays to be balanced for more than voltage drop. Of course, I'm just grabbing arbitrary values at random here.

Hey RODALCO...how about checking my math. Did I blow it? [Linked Image]



[This message has been edited by WFO (edited 12-27-2005).]