12-2-2 NM cable only has one grounding conductor?

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Yes, that's true, but you're talking about resistances measured in tenths of ohms, in series with resistances measured in hundreds of ohms. As a practical matter, except on *very* long runs, the resistance of the conductors is not really a concern.

I'm *not* ignoring the neutral. But you're missing the fact that the voltage drop between the source and the load is almost completely independent of the resistance *elsewhere* in the circuit. The only effect that the resistance of the neutral has on the circuit is a *minuscule* difference in the current that flows through it.

The effect on voltage drop is still so small as to be utterly negligible.

Do you know just *how* low the resistance of copper conductors actually is?

I don't think I ever said it wasn't "real". But it certainly isn't important.
-- Regards, Doug Miller (alphageek at milmac dot com)
Nobody ever left footprints in the sands of time by sitting on his butt. And who wants to leave buttprints in the sands of time?
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Very good, I agree with this. Voltage drop due to the conductors only matters for long runs. But it is still a reasonable question whether sharing a neutral between two circuits will reduce that voltage drop, even if it only matters for long runs.

True, the voltage drop between the source and the load only depends on the resistance of the hot. But the "available voltage" at the load depends on the voltage drop of both the hot and the neutral.
Consider an example: exactly 120V available at the panel, a single circuit with (long) wiring with a resistance of 0.1 ohms each for the hot and the neutral, and an electric heater with a resistance of 10 ohms (nominally a 1200W heater). If the heater were wired directly to the panel, the total resistance is 10 ohms and the heater would draw 12 amps. With the (long) wiring, the total resistance is 10.2 ohms, and the heater draws 11.76 amps (120/10.2). The voltage drop across the hot is 1.176 volts, the drop across the heater is 117.6 volts, and the drop across the neutral is 1.176 volts.
Now suppose we have two 10 ohm heaters on an Edison circuit (shared neutral). This represents the best case of totally balanced loads, and the current on the neutral will be 0. So this is equivalent to having the two heaters in series across a 240V supply. Each hot conductor has the same resistance of 0.1 ohms. The total resistance is 20.2 ohms, and the current is 11.88 amps (240/20.2). The voltage drop across the first hot is 1.188 volts, the drop across each heater is 118.8 volts, and the drop across the second hot is again 1.188 volts.
In the single circuit case, the "voltage drop" the single heater saw was 2.4 volts (120 - 117.6). In the shared neutral case, the voltage drop was only 1.2 volts (120 - 118.8). This illustrates the best case scenario of using a shared neutral: with perfectly balanced loads, the voltage drop due to the wiring with be half as much.
Cheers, Wayne
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Yes, *if* you have perfectly balanced, purely resistive loads.
But the difference has no practical significance in any case.
-- Regards, Doug Miller (alphageek at milmac dot com)
Nobody ever left footprints in the sands of time by sitting on his butt. And who wants to leave buttprints in the sands of time?
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wrote:

Ouch. Good catch!
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wrote:

It's perfectly fine, as long as both of these conditions are met: a) the two hot conductors are on *opposite* legs of the service, and b) both hot conductors are disconnected by the same device. Both requirements are easily met by using a standard double-pole (240V) breaker.

That'll definitely be easier in the long run, and less confusing to the next homeowner.
-- Regards, Doug Miller (alphageek at milmac dot com)
Nobody ever left footprints in the sands of time by sitting on his butt. And who wants to leave buttprints in the sands of time?
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I believe (b) above is not always required by the NEC, although it seems like a good idea for a variety of reasons. 210.4(B) requires this when the circuit supplies "more than one device or equipment on the same yoke". And 210.4(C) requires this if the circuit serves any 240V loads.
Cheers, Wayne
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