I'm looking at a 50A range circuit, and just started wondering: what is the "logic" behind 6/3 SER cable having such a smaller gauge ground wire?

Is it that it really only needs to handle a single pole (120V,25A) in the case of an open-neutral ground fault? Any sort of 240V dead short would carry its own current (tripping breaker) right?

Thanks

 

During a short circuit, hot to ground, the current on both wires is much higher than the maximum allowed by the breaker. It is the breaker's job to see that excessive current and then trip.

 

OK, that much is clear. I guess my question has more to do with partial shorts to ground. This is more of a "theory" question than anything practical, so apologies if I'm asking nonsense here.

On a typical 120V 20A 12ga circuit, imagine a device develops a low-resistance (say 8 ohm) short to ground, energizing the chasis. That'd be about 15A flowing to ground (if my ohm's law if correct). Not enough to trip the breaker (unless it were GFCI) but the 12ga ground wire could handle the current until problem detected.

So, I wondered, is it possible for a device on a 240V 50A 6ga circuit to develop a partial short to ground that isn't enough to trip the breaker, yet is more than the smaller ground wire is rated to handle? Wouldn't that pose a fire hazard on the over-current ground wire?

My only though was that since the 240 is really just 2 opposite pole 120's, that you'd never see the sum of their current on the ground no matter how they shorted. At worst you'd see the difference of their currents. At least, that's the only "logic" I could come up with for a 6ga cable having only a 12ga ground. Am I missing something here?

 

It's because a "partial shorts to ground" is an exceptionally rare phenomenon.

 

Lets just say for example:

You have billions or electrons flowing thru a wire
now add resistance, a lose connection or a "short" that cant trip a circuit breaker.
The billions of electrons moving about the speed of light hit the restricted area, the electrons are going to create friction.
Generating heat at the point of resistance.

The filament in a light bulb gets red hot due to the resistance.

Its the resistance that will answer your question.
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Say a 40 amp ground "short".
using a 12 ga ground wire, 50 feet long. 120 volts at 40 amps.

Ohms law:
The resistance of the 12ga wire is about 0.097 ohms at 50 feet.
0.097 ohms x 40 amps = 3.88 volts
The voltage drop across the ground wire (50 ft) will be about 3.88 volts with that 40 amp "short" to ground.
3.88 volts x 40 amps = 155 watts.
That 12 ga wire will be dissipating 155 watts the full length of 50 feet, with 40 amps flowing thru it.

The resistance of the "short" is about 3 ohms.
120 volts / 40 amps = 3 ohms
120 volts x 40 amps = 4800 watts
3 ohm "short" to ground.

a 3-ohm "short" is going to generate more heat at the "short" area then the 12 ga wire at 0.097 ohms.

The "short" location of 3 ohms will be like a 4800 watt heater. the device area will melt down.
That 12 ga wire will be dissipating 155 watts, across the full length of the 50 feet at 40 amps.
I don't think 155 watts is going to heat the 12 ga wire very much.

If you have that big of a "short" your device will melt down before the ground wire will become a problem.
Any-any lose connections will create heat when power is flowing thru them.

Your house Circuit Breakers are more for protecting the house wiring not the device plugged in.