Can you show a diagram of this? If you rip out the N of a 1ph CT xrmer, anything that was wired to the N for half voltage will not suddenly run line-line. MWBC would be the only thing that would run wonky, where turning 'on' just one device (line-CT) does nothing, but turning both 'on' runs just fine (line1-device1-CT-device2-line2).
So, if the N is moved to that location, no xfrmr "pole" will be able to return to the CT, and the 120v ckts will not function.
Not properly anyway. Unless you have every circuit completly balaced (unlikely) and no multiwire circuits, you will have "attempted operation" and voltage fluctuations high/low voltage on either hot leg depending on loads attached. Everything is somewhat tied together in the service panel.
This only relevant on a single branch circuit not if loss of primary or service neutral or a MWBC. Every circuit that is 120V is semi interconnected via the neutral and grounding bus and via the bonded N/G point. You would see this effect if you have on the branch circuits with multiple neutrals tied together in a second point of reference such as a Jbox. Lift one neutral at the device end and you will get some variations of voltage on that neutral in relation to the EGC depending on loads on the other circuits linked in the Jbox. Seen it and with a shocking effect of 30V to 90V referenced N/G (apprentice got a shock on N/G contact.)
It doesn't magically connect as you describe but there us still a reference to ground on the system as a whole. Imbalaces from single phase 120 loads will still try to return to source through any means possible, even if it is a high impedance means through the earth (NEV). So not only would you need to lift the xfer neutral but also all references to ground by removing all N/G bonds everywhere in the system. The higher the impedance you could be limiting equipment damage as not enough getting thru to cause dammage and essentially is "off" but will get some measurable current thru.
Almost correct. Both won’t run just fine.
You are missing that all the 120V loads are connected to each other through the neutral. When the neutral is disconnected at the source the circuit becomes a voltage divider network with the voltage on the neutral dependent on the ratio of aggregate L1 to N and L2 to N resistances. The L1 to N loads have a return path through the L2 to N loads and vice versa.
Yeah, well, but put a megohn load on one side and 1 ohm on the other and one of them will get pretty dang close!
Back when test lights were used it was pretty obvious when one wattage was less than the other.No single load will see240V. But with two loads in series the voltage in the lower wattage load approaches arbitrarily close to 240V as the impedance ratio increases. If the only loads are two 100 W incandescent bulbs, neither would light if only one was turned on.
When you turn in the second, on the opposite "phase", they might both light fine if their thermal characteristics match closely enough. But if one heats faster it could well hog the voltage drop and burn out. Then both would be dark again.
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And, the two loads are 120v loads, so it's Line1--> 120vLoad-1 -->N1--> bar in panel --> N2--> 120vLoad2--> Line2
The panels not the source.
It's a MWBC, nothing more.
Same thing.
As long as the lumped resistance from L1 to N and L2 to N are the same, everything is hunky if the neutral is disconnected; if the neutral were there there would be no current on it, anyway. If the resistances are not balanced, though, the voltage to neutral from the legs is entirely dependent on the difference in the lumped load resistances L1 to N and L2 to N, and the line with the higher resistance to neutral will get the higher voltage, although V(L1 to N) and V(L2 to N) will always sum to 240V. This is EE101 stuff.
It won't, but if resistance of one is low enough it can come close to being full 240 across the other load.
