Double Pole Circuit Breakers

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I understand that double pole circuit breakers feed 240v (two120's) to water heaters, stoves, ect...
But, Im confused about the amperage marking on the middle tab; I see that most say 30A. Does that mean the each pole from that circuit breaker can handle up to 30A each pole? 30 + 30 = 60A total? or is the amps on each hot leg split into two; 15 amps each pole; 15 + 15 = 30A? I never worked with 240V circuits before, obviosly. I just need to install a small baseboard heater in my shop for the upcoming winters; it calls for a 20A 240v breaker.
Another quick question, but off topic: Why don't wire manufactures insulate the bare ground wire inside romex cables?
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Each pole of a breaker, single, double or triple, provides the amperage marked on the breaker. Double pole 20 amp gives you 20 amps on each legs or 20 amps @240 volts I too, would like the ground wires to be insulated for my own safety

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this)@optonline.net> wrote:

How would insulating the ground wire improve safety?
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Doug Miller (alphageek at milmac dot com)
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If they were insulated, I wouldn't have to be so careful while flailing them around inside panel boxes
volts

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On Mon, 12 Jun 2006 07:09:40 -0400, "RBM" <rbm2(remove

imho:
Ground Wires 'do not' carry current except under ground fault conditions. In that case, the breaker/fuse will break the circuit asap(by design).
So it normally carries 0 voltage, 0 current.
hth,
tom

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volts

You are correct that it normally carries nothing, but why would a breaker open on a ground fault? (unless it is a gfci...)
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A ground fault is simply a leak to the grounding conductor - it could be a few milliamps, or many amps.
Breakers trip on big ones. GFCIs trip on small ones.
--
Chris Lewis, Una confibula non set est
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Chris Lewis wrote:

A ground fault is a leak to ground. It may or may not pass through the grounding conductor.

Breakers trip on excessive current regardless of the path that it takes. GFCIs trip on an unbalance between the two wires that compose the active circuit.
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--John
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If a hot comes in contact with a metal casing, that is grounded, it is called a ground fault. Following the NEC all grounded equipment conductors (the ground wire) have to have as lose impedance to electricy back to the source.
So using ohms law E=IR
E is electrical potencial, or Voltage I is current R is resistance.
Solve for I (current) results in I = E/R
substitute numbers for lowest norm voltage. I = 120 / ~0
*Note: Used 0 since a small house has almost 0 ohms back to the panel on the ground wires.
So calc Current, and you have an almost infinite amount of current, a short, and the breaker will open on this ground.
A normal breaker should open on any ground fault over thier set points. Meaning a 15 amp breaker should open on any ground faults over 15 amps, on over current protection. Ofcourse an almost infinite current, should trip the breaker on short circuit protection.
hth, (please ignore spelling errors)
tom
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snipped-for-privacy@yahoo.com ( snipped-for-privacy@yahoo.com) said...

Overcurrent protection is placed on the hot side of a circuit, so with 120V circuits, there is only a single pole breaker for the single hot.
On a 240V circuit, you get 120 volts (relative to the neutral) per hot (from separate legs). The current that goes through one returns through the other. Overcurrent protection is needed on BOTH hots and must be rated the same because it is the SAME current that passes through both.
Since the voltage is doubled, even though the current is the same, the POWER that can be provided by the circuit is doubled.
A neutral conductor is only needed on a 240V circuit if there are loads that require only 120V on the circuit. In those cases, the neutral will carry the DIFFERENCE between the current in each hot. (e.g.: if one hot had a current of 10A and the other had 9A, the neutral would be carrying the 1A difference).

Why bother? There is really no safety issue as it only serves to bond metal chassis and boxes to ground. It will carry a current in fault situations, but no greater than the overcurrent protection on the circuit.
However, in the case of those ORANGE outlets, the ground pin on the receptacle is ISOLATED from the bare ground that the box is bonded by. A separate INSULATED ground conductor is needed to bond the ground pin to the grounding bus in the panel. Often the red conductor of a 3-wire cable is used for this purpose.
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Calvin Henry-Cotnam
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snipped-for-privacy@yahoo.com wrote:

Depends on how it's wired, actually. A double-pole breaker can provide one 240V circuit, or two 120V circuits. In your example, a 30A double-pole breaker will provide one 30A circuit at 240V, or two 30A circuits at 120V.

Because it would be pointless to do so. The grounding wire is intended to insure that the metal frames of equipment (e.g. a washing machine) and exposed metal components of the premises wiring (e.g. conduits or receptacle boxes) are grounded and cannot become live. In other words -- it's connected to things that are not insulated anyway, and thus no purpose would be served by insulating it.
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Doug Miller (alphageek at milmac dot com)
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On Mon, 12 Jun 2006 11:46:52 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

That's one thing people often do not understand. If you have a 100A service and only use 120V items in the house, you can actually draw 200A. But if you have all 240V appliances, you can only use 100A. Of course most homes have both, so that's where the mathematics comes into play.
Mark
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snipped-for-privacy@UNLISTED.com wrote:

The total power is the same either way or in combination.
100A x 240V = 24,000W or 200A x 120V = 24,000W or 50A x 240V = 12,000W + 100A x 120V = 12,000W 24,000W total etc...
Pete C.
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wrote:

If you happen to have a perfectly balanced system, which is most unlikely...
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Toller wrote:

It makes no difference how balanced or unbalanced you have the system, you will not be pulling more than 24,000W at least not for more than a few min before the main breaker trips.
It could well be less than that number if you max out one leg first. You need to have the system balanced to reach the 24,000W load, but the total amount of power available from the service is 24,000W no matter what.
Pete C.
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wrote:

Limited to 12,000W per side.

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On Mon, 12 Jun 2006 08:34:35 -0500, snipped-for-privacy@UNLISTED.com wrote:

No way.
Your total load calc may be 200 A at 120 V, but you are still pulling 100 amps. You've got two 120 volt loads in series, that's all. You are STILl pulling 100 amps at 240 volts. Since the individual loads are in series/parallel, they share half teh voltage, so the total load calc comes out to 200 amps, but only 100 amps are flowing through the service. Really.
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Which is exactly the same electrical power as 200A at 120V.

100A at 240V -- and, as noted above, if only 120V loads are in use, the total current draw is (up to) 200A at 120V.
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Doug Miller (alphageek at milmac dot com)
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On Fri, 23 Jun 2006 13:22:41 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

The power consumption is equivalent to 200A at 120V, but only 100A are flowing. Kirchoff's Law will not be violated!
Currents in series do not add; =voltages= add. It's still 100A at 240V.
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On Fri, 23 Jun 2006 08:11:40 -0700, ~^Johnny^~

Better yet, Look at it this way: If I hang ten 12V lamps in series across a 120V line, and each lamp is 120 watts, am I pulling a total of 100 amps? I am not! I am drawing ten amps! Each lamp gets 12V at 10A. 100A doesn't ever flow, anywhere in the circuit. But I'm still consuming 1200W: 120V at 10A, _NOT_ 12V at 100A. If the lamps were in parallel, they would require 100A at 12V. But they're not. That's a load calculation only.
An Edison circuit is just a 240 volt circuit with a grounded center-tap. In theory, if loads are balanced, the neutral drops out of the circuit, as the individial loads become a series-parallel voltage divider network. Each load receives 120V at its rated =power=.
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