The sub-thread up yonder about using 15-amp switches on 20-amp circuits
got me to thinking. Actually, have been wondering about this for a long
time, so here goes.

The thing is, we have circuits where the wiring and devices on that
circuit are designed to safely carry a certain amount of current, for
example a 15-amp circuit using devices rated for that amount and wired
with #14 wire. All well and good.

But our discussions here inevitably leave out what I'm calling the "last
inch". By that I mean such things as the wires that feed a lighting
fixture, attached inside the wall or ceiling box, or the wires
connecting a dimmer switch. These wires are always a *lot* smaller than
the cable used to wire the circuit; often they're around 16 or 18 gauge.

This would seem to violate the integrity of the circuit, because now you
have weak point. In the worst case, a short circuit at the device, you'd
have a lot of current going through these smaller wires, until the
breaker trips. Isn't there a greater chance of fire in that case?

So how does the NEC reconcile this apparent violation of the integrity
of the circuit? How do folks like us who install and work on such wiring
rationalize it? Is it simply a matter of practicality? (It's obviously
not practical to use 14-gauge wire all the way up to every device on a
lighting circuit.)


--
Found--the gene that causes belief in genetic determinism

"David Nebenzahl" wrote in message
s.com...
The sub-thread up yonder about using 15-amp switches on 20-amp circuits
got me to thinking. Actually, have been wondering about this for a long
time, so here goes.

The thing is, we have circuits where the wiring and devices on that
circuit are designed to safely carry a certain amount of current, for
example a 15-amp circuit using devices rated for that amount and wired
with #14 wire. All well and good.

But our discussions here inevitably leave out what I'm calling the "last
inch". By that I mean such things as the wires that feed a lighting
fixture, attached inside the wall or ceiling box, or the wires
connecting a dimmer switch. These wires are always a *lot* smaller than
the cable used to wire the circuit; often they're around 16 or 18 gauge.

This would seem to violate the integrity of the circuit, because now you
have weak point. In the worst case, a short circuit at the device, you'd
have a lot of current going through these smaller wires, until the
breaker trips. Isn't there a greater chance of fire in that case?

So how does the NEC reconcile this apparent violation of the integrity
of the circuit? How do folks like us who install and work on such wiring
rationalize it? Is it simply a matter of practicality? (It's obviously
not practical to use 14-gauge wire all the way up to every device on a
lighting circuit.)


--
Found--the gene that causes belief in genetic determinism

Lets say you have a light fixture with a 100 watt bulb. Inside the fixture
they use the smaller wires and something shorts out. Those wires are
encased in the fixture and the box. The short circuit lowers the resistance
and the amps shoot up and the breaker pops. The excess heat from the short
is contained and dissipated in the metal fixture long before it causes
mischief.

Now consider a different proposition. The circuit has a penny in the fuse
box and the circuit is overloaded. The wires inside the walls heat up and
catch the house on fire.

See the difference?

--

__
Roger Shoaf

Important factors in selecting a mate:
1] Depth of gene pool
2] Position on the food chain.

"David Nebenzahl" wrote in message
s.com...
The sub-thread up yonder about using 15-amp switches on 20-amp circuits
got me to thinking. Actually, have been wondering about this for a long
time, so here goes.

The thing is, we have circuits where the wiring and devices on that
circuit are designed to safely carry a certain amount of current, for
example a 15-amp circuit using devices rated for that amount and wired
with #14 wire. All well and good.

But our discussions here inevitably leave out what I'm calling the "last
inch". By that I mean such things as the wires that feed a lighting
fixture, attached inside the wall or ceiling box, or the wires connecting
a dimmer switch. These wires are always a *lot* smaller than the cable
used to wire the circuit; often they're around 16 or 18 gauge.

This would seem to violate the integrity of the circuit, because now you
have weak point. In the worst case, a short circuit at the device, you'd
have a lot of current going through these smaller wires, until the breaker
trips. Isn't there a greater chance of fire in that case?

So how does the NEC reconcile this apparent violation of the integrity of
the circuit? How do folks like us who install and work on such wiring
rationalize it? Is it simply a matter of practicality? (It's obviously not
practical to use 14-gauge wire all the way up to every device on a
lighting circuit.)


--
Found--the gene that causes belief in genetic determinism

Typical dimmer is rated a 600 watts. The wiring attached to it, is of
substantial size to handle its rating. If you get a 1000, 1500, or 2000 watt
dimmer, you will find that it has larger conductors attached to it, to
handle it's respective load. A switching device is not necessarily designed
to carry the entire load of a circuit, unless it's going to carry the entire
load of a circuit.

RBM wrote:
"David Nebenzahl" wrote in message
s.com...
The sub-thread up yonder about using 15-amp switches on 20-amp
circuits got me to thinking. Actually, have been wondering about
this for a long time, so here goes.

The thing is, we have circuits where the wiring and devices on that
circuit are designed to safely carry a certain amount of current,
for example a 15-amp circuit using devices rated for that amount
and wired with #14 wire. All well and good.

But our discussions here inevitably leave out what I'm calling the
"last inch". By that I mean such things as the wires that feed a
lighting fixture, attached inside the wall or ceiling box, or the
wires connecting a dimmer switch. These wires are always a *lot*
smaller than the cable used to wire the circuit; often they're
around 16 or 18 gauge. This would seem to violate the integrity of the
circuit, because
now you have weak point. In the worst case, a short circuit at the
device, you'd have a lot of current going through these smaller
wires, until the breaker trips. Isn't there a greater chance of
fire in that case? So how does the NEC reconcile this apparent violation
of the
integrity of the circuit? How do folks like us who install and work
on such wiring rationalize it? Is it simply a matter of
practicality? (It's obviously not practical to use 14-gauge wire
all the way up to every device on a lighting circuit.)


--
Found--the gene that causes belief in genetic determinism

Typical dimmer is rated a 600 watts. The wiring attached to it, is of
substantial size to handle its rating. If you get a 1000, 1500, or
2000 watt dimmer, you will find that it has larger conductors
attached to it, to handle it's respective load. A switching device is
not necessarily designed to carry the entire load of a circuit,
unless it's going to carry the entire load of a circuit.

Also, just as a long wire needs to be a thicker gauge, compared to a normal
length of wire, to carry a fixed amount of amperages, a very short length of
wire can be rated to carry a larger amperage at a smaller gauge than
normally used. This is the rational used on appliance cords, and the
internal wiring in appliances. I have seen a formula somewhere that will
determine the exact gauge needed for a given length at a specific amperage.

On 8/29/2009 2:38 PM Roger Shoaf spake thus:

"David Nebenzahl" wrote in message
s.com...

But our discussions here inevitably leave out what I'm calling the "last
inch". By that I mean such things as the wires that feed a lighting
fixture, attached inside the wall or ceiling box, or the wires
connecting a dimmer switch. These wires are always a *lot* smaller than
the cable used to wire the circuit; often they're around 16 or 18 gauge.

This would seem to violate the integrity of the circuit, because now you
have weak point. In the worst case, a short circuit at the device, you'd
have a lot of current going through these smaller wires, until the
breaker trips. Isn't there a greater chance of fire in that case?

So how does the NEC reconcile this apparent violation of the integrity
of the circuit? How do folks like us who install and work on such wiring
rationalize it? Is it simply a matter of practicality? (It's obviously
not practical to use 14-gauge wire all the way up to every device on a
lighting circuit.)

Lets say you have a light fixture with a 100 watt bulb. Inside the fixture
they use the smaller wires and something shorts out. Those wires are
encased in the fixture and the box. The short circuit lowers the resistance
and the amps shoot up and the breaker pops. The excess heat from the short
is contained and dissipated in the metal fixture long before it causes
mischief.

Now consider a different proposition. The circuit has a penny in the fuse
box and the circuit is overloaded. The wires inside the walls heat up and
catch the house on fire.

That makes sense; the idea is to confine any potential fires within
boxes, where they presumably won't burn the damn house down.

Which is why I use metal boxes instead of plastic ones, and pay
attention to properly clamping cables going into the box (rather than
just sticking the cable through a hole in the box).


--
Found--the gene that causes belief in genetic determinism

On 8/29/2009 2:49 PM EXT spake thus:

Also, just as a long wire needs to be a thicker gauge, compared to a normal
length of wire, to carry a fixed amount of amperages,

Saying "a fixed amount of *amps*" would do.

a very short length of wire can be rated to carry a larger amperage
at a smaller gauge than normally used.

That's not true. Conductors are rated at a certain current regardless of
their length.

This is the rational used on appliance cords, and the internal wiring
in appliances. I have seen a formula somewhere that will determine
the exact gauge needed for a given length at a specific amperage.

I was going to bring up the aspect of cords too, as our 20-amp circuits
have cords plugged into them that are rated at far less than that,
creating another potential source of fire.


--
Found--the gene that causes belief in genetic determinism

In article m,
kens says...
The sub-thread up yonder about using 15-amp switches on 20-amp circuits
got me to thinking. Actually, have been wondering about this for a long
time, so here goes.

The thing is, we have circuits where the wiring and devices on that
circuit are designed to safely carry a certain amount of current, for
example a 15-amp circuit using devices rated for that amount and wired
with #14 wire. All well and good.

But our discussions here inevitably leave out what I'm calling the "last
inch". By that I mean such things as the wires that feed a lighting
fixture, attached inside the wall or ceiling box, or the wires
connecting a dimmer switch. These wires are always a *lot* smaller than
the cable used to wire the circuit; often they're around 16 or 18 gauge.

This would seem to violate the integrity of the circuit, because now you
have weak point. In the worst case, a short circuit at the device, you'd
have a lot of current going through these smaller wires, until the
breaker trips. Isn't there a greater chance of fire in that case?

So how does the NEC reconcile this apparent violation of the integrity
of the circuit?


They specifically allow the smaller gauge wire within a fixture, and
reconcile it by permitting the fixture manufacturer to act as the
"Authority Having Jurisdiction" and testing their own fixtures. The
fixtures must remain an "end device", and cannot have anything wired in
series with them. The short length of smaller wire is proper for the
limited current draw.

I believe the code used to allow 14 gauge "drops" in 20 amp circuits,
which were runs from switches to lights with the similar logic that a
light would never draw more than 15 amps.

--
Dennis

David Nebenzahl wrote:
....
That's not true. Conductors are rated at a certain current regardless of
their length.
....
Not exactly so--look at the voltage drop tables; at a given voltage the
drop becomes excessive at a minimum conductor size and a larger
conductor is required.

All these points have been considered--in essence, the answer is that
the individual appliance/light/whatever has conductors sized
specifically for the load.

A 100W bulb, for example, on the 15A 14ga circuit doesn't need 14ga
because it draws only 1A (in round numbers)--the circuit wiring is
required to be larger to account for the loading of all devices in
simultaneous usage on the circuit.

Also, again, NEC specifically covers the wiring not the end devices;
they're under other guidelines such as UL, etc., ...

In the end, there's no increased risk in common usage as long as you
don't do something that is in obvious contravention to intended use--put
a 300W bulb in a 25W rated fixture or 3 1000W hair driers on a 25-ft
16ga light-duty extension cord, say. Sure, one _can_ do stupid, it's
presumed the Darwin rule will take care of that...

--

How practial is it, to only plug in 14 gage cords into 14
gage branch circuits? Not very. What's worse, is trying to
find a 100 watt outlet, to power your table lamp.

--
Christopher A. Young
Learn more about Jesus
www.lds.org
..


"David Nebenzahl" wrote in message
s.com...

I was going to bring up the aspect of cords too, as our
20-amp circuits
have cords plugged into them that are rated at far less than
that,
creating another potential source of fire.

David Nebenzahl wrote:
On 8/29/2009 2:38 PM Roger Shoaf spake thus:

"David Nebenzahl" wrote in message
s.com...

But our discussions here inevitably leave out what I'm calling the "last
inch". By that I mean such things as the wires that feed a lighting
fixture, attached inside the wall or ceiling box, or the wires
connecting a dimmer switch. These wires are always a *lot* smaller than
the cable used to wire the circuit; often they're around 16 or 18 gauge.

This would seem to violate the integrity of the circuit, because now you
have weak point. In the worst case, a short circuit at the device, you'd
have a lot of current going through these smaller wires, until the
breaker trips. Isn't there a greater chance of fire in that case?

So how does the NEC reconcile this apparent violation of the integrity
of the circuit? How do folks like us who install and work on such wiring
rationalize it? Is it simply a matter of practicality? (It's obviously
not practical to use 14-gauge wire all the way up to every device on a
lighting circuit.)

Lets say you have a light fixture with a 100 watt bulb. Inside the
fixture
they use the smaller wires and something shorts out. Those wires are
encased in the fixture and the box. The short circuit lowers the
resistance
and the amps shoot up and the breaker pops. The excess heat from the
short
is contained and dissipated in the metal fixture long before it causes
mischief.

Now consider a different proposition. The circuit has a penny in the
fuse
box and the circuit is overloaded. The wires inside the walls heat up
and
catch the house on fire.

That makes sense; the idea is to confine any potential fires within
boxes, where they presumably won't burn the damn house down.

Which is why I use metal boxes instead of plastic ones, and pay
attention to properly clamping cables going into the box (rather than
just sticking the cable through a hole in the box).


Hmmm,
Is there anyone who does not do that when wiring? Then it'll fail
inspection. Every thing is simple math and cool head. Actually life is.

David Nebenzahl wrote:
specific amperage.

I was going to bring up the aspect of cords too, as our 20-amp circuits
have cords plugged into them that are rated at far less than that,
creating another potential source of fire.


Hmm,
If only that cord has a load carrying 20 whole amps.
I hope you still remember Ohm's law in hi school physics class.
I left hi school in 1960.

On Sat, 29 Aug 2009 15:07:40 -0700, David Nebenzahl
wrote:

On 8/29/2009 2:49 PM EXT spake thus:

Also, just as a long wire needs to be a thicker gauge, compared to a normal
length of wire, to carry a fixed amount of amperages,

Saying "a fixed amount of *amps*" would do.

a very short length of wire can be rated to carry a larger amperage
at a smaller gauge than normally used.

That's not true. Conductors are rated at a certain current regardless of
their length.

No, the ampacity of cables varies with length, becuase the resistance
per foot causes more voltage drop on a long cord than on a short one.

This is the rational used on appliance cords, and the internal wiring
in appliances. I have seen a formula somewhere that will determine
the exact gauge needed for a given length at a specific amperage.

I was going to bring up the aspect of cords too, as our 20-amp circuits
have cords plugged into them that are rated at far less than that,
creating another potential source of fire.

On Sat, 29 Aug 2009 21:20:27 -0400, wrote:
On Sat, 29 Aug 2009 15:07:40 -0700, David Nebenzahl
wrote:

On 8/29/2009 2:49 PM EXT spake thus:

Also, just as a long wire needs to be a thicker gauge, compared to a normal
length of wire, to carry a fixed amount of amperages,

Saying "a fixed amount of *amps*" would do.

a very short length of wire can be rated to carry a larger amperage
at a smaller gauge than normally used.

That's not true. Conductors are rated at a certain current regardless of
their length.

No, the ampacity of cables varies with length, becuase the resistance
per foot causes more voltage drop on a long cord than on a short one.

Wrong. The current through a cable is constant no matter what the
length as is the maximum current capacity of a wire.

Resistance and voltage drop are a separate matter.

"AZ Nomad" wrote in message
...
On Sat, 29 Aug 2009 21:20:27 -0400,
wrote:
On Sat, 29 Aug 2009 15:07:40 -0700, David Nebenzahl
wrote:

On 8/29/2009 2:49 PM EXT spake thus:

Also, just as a long wire needs to be a thicker gauge, compared to a
normal
length of wire, to carry a fixed amount of amperages,

Saying "a fixed amount of *amps*" would do.

a very short length of wire can be rated to carry a larger amperage
at a smaller gauge than normally used.

That's not true. Conductors are rated at a certain current regardless of
their length.

No, the ampacity of cables varies with length, becuase the resistance
per foot causes more voltage drop on a long cord than on a short one.

Wrong. The current through a cable is constant no matter what the
length as is the maximum current capacity of a wire.

Resistance and voltage drop are a separate matter.

Seems to me there are two pertinent factors.

One is voltage drop which is an IR drop and really only affects the
operation of the device at the far end.

Second is the power dissipation in the wire that is the I(squared)R value.
That dissipation is spread linearly along the conductor and gives the rise
in temperature that is relevant to ignition.

Charlie

In article , dpb wrote:
David Nebenzahl wrote:
....
That's not true. Conductors are rated at a certain current regardless of
their length.
....
Not exactly so--look at the voltage drop tables; at a given voltage the
drop becomes excessive at a minimum conductor size and a larger
conductor is required.

Yes, exactly so. Voltage and current are not the same. A long conductor *does*
cause voltage to drop, but it does *not* affect current.

In article , wrote:
On Sat, 29 Aug 2009 15:07:40 -0700, David Nebenzahl
wrote:

On 8/29/2009 2:49 PM EXT spake thus:

Also, just as a long wire needs to be a thicker gauge, compared to a normal
length of wire, to carry a fixed amount of amperages,

Saying "a fixed amount of *amps*" would do.

a very short length of wire can be rated to carry a larger amperage
at a smaller gauge than normally used.

That's not true. Conductors are rated at a certain current regardless of
their length.

No, the ampacity of cables varies with length,

Nonsense.

becuase the resistance
per foot causes more voltage drop on a long cord than on a short one.

Voltage and current are not the same.

Conductor ampacity is dependent on:
- the conductor material (Cu vs Al)
- the insulation material
- the conductor diameter
and NOT on the conductor length. If you think rated ampacity is in any way
dependent upon conductor length, then please cite the section of either the
CEC or the NEC which says so. I'm not going to hold my breath.

certinally a 18 gauge light cord on a 20 amp circuit is a fire hazard.

the plugs of all such cords should be required to have a built in
fuse.

I once had a customer using a 18 gauge ungrounded light extension cord
on a 18 amp grounded machine that tended to burn wires off inside the
unit.

I wrote it up as a safety hazard, warning them buy a AC extension

next time I went they had improved things by stapling the extension
cord to a carpeted wall wire got very hot showed signgs of
melting.....

I fixed the machine then went shopping in the mall, and bought a AC
extension cord. the customer got mad when I cut their junk cord in
pieces and gave them free the air conditioner cord. told them the new
extension cord was far cheaper than even the paperwork for the
insurance claim when they burned down south hills village.

the customer was ****ed and never called me again. frankly i didnt
care.

have you noticed christmas lights now have fuses in each plug? that
should be required for all plugs

On Aug 30, 7:50*am, (Doug Miller) wrote:
In article , dpb wrote:
David Nebenzahl wrote:
....
That's not true. Conductors are rated at a certain current regardless of
their length.
....
Not exactly so--look at the voltage drop tables; at a given voltage the
drop becomes excessive at a minimum conductor size and a larger
conductor is required.

Yes, exactly so. Voltage and current are not the same. A long conductor *does*
cause voltage to drop, but it does *not* affect current.

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total circuit.
Voltage E, Current I and Resistance R are not independent of each
other under normal conditions.

E=IR
I=E/R
R=E/I

Jimmie

On Aug 30, 11:24*am, JIMMIE wrote:
On Aug 30, 7:50*am, (Doug Miller) wrote:

In article , dpb wrote:
David Nebenzahl wrote:
....
That's not true. Conductors are rated at a certain current regardless of
their length.
....
Not exactly so--look at the voltage drop tables; at a given voltage the
drop becomes excessive at a minimum conductor size and a larger
conductor is required.

Yes, exactly so. Voltage and current are not the same. A long conductor *does*
cause voltage to drop, but it does *not* affect current.

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will *the current
in the total circuit because it adds resistance to the total circuit.
Voltage E, Current I and Resistance R are not independent of each
other under normal conditions.

E=IR
I=E/R
R=E/I

Jimmie
I WROTE WHAT!!!!!
Sorry answered two phone calls and three questions from my wife while
trying to write this. Please ignore my dribble.

Jimmie

In article , JIMMIE wrote:
On Aug 30, 11:24=A0am, JIMMIE wrote:
On Aug 30, 7:50=A0am, (Doug Miller) wrote:

In article , dpb =
wrote:
David Nebenzahl wrote:
....
That's not true. Conductors are rated at a certain current regardles=
s of
their length.
....
Not exactly so--look at the voltage drop tables; at a given voltage th=
e
drop becomes excessive at a minimum conductor size and a larger
conductor is required.

Yes, exactly so. Voltage and current are not the same. A long conductor *does*
cause voltage to drop, but it does *not* affect current.

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total circuit.
Voltage E, Current I and Resistance R are not independent of each
other under normal conditions.

E=IR
I=E/R
R=E/I

Jimmie
I WROTE WHAT!!!!!
Sorry answered two phone calls and three questions from my wife while
trying to write this. Please ignore my dribble.

I normally do anyway... g

In article , JIMMIE wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total circuit.

Adding resistance to a circuit changes voltage, not current.

On Sun, 30 Aug 2009 20:58:38 GMT, (Doug Miller)
wrote:

In article , JIMMIE wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total circuit.

Adding resistance to a circuit changes voltage, not current.

I can't believe what I am hearing.

On 8/30/2009 1:58 PM Doug Miller spake thus:

In article
,
JIMMIE wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total
circuit.

Adding resistance to a circuit changes voltage, not current.

Changes current, too. E.g., a current-limiting resistor in series with a
LED, sized so that the LED won't draw excess current and burn itself out.


--
Found--the gene that causes belief in genetic determinism

In article , Metspitzer wrote:
On Sun, 30 Aug 2009 20:58:38 GMT, (Doug Miller)
wrote:

In article
, JIMMIE
wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total circuit.

Adding resistance to a circuit changes voltage, not current.

I can't believe what I am hearing.

Oh? What's the problem?

In article m, David Nebenzahl wrote:
On 8/30/2009 1:58 PM Doug Miller spake thus:

In article
,
JIMMIE wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total
circuit.

Adding resistance to a circuit changes voltage, not current.

Changes current, too. E.g., a current-limiting resistor in series with a
LED, sized so that the LED won't draw excess current and burn itself out.

Wrong.

If you want to limit the current passing through an LED, you put a resistor in
*parallel* with it, not series.

Total current in the circuit remains the same.

David Nebenzahl wrote:
On 8/30/2009 1:58 PM Doug Miller spake thus:

In article
,
JIMMIE wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total
circuit.

Adding resistance to a circuit changes voltage, not current.

Changes current, too. E.g., a current-limiting resistor in series with a
LED, sized so that the LED won't draw excess current and burn itself out.


Hi,
If you are not joking trying to be funny. Your basic knowledge is
LACKing. Please be quiet if you don't know about some thing.

Metspitzer wrote:
On Sun, 30 Aug 2009 20:58:38 GMT, (Doug Miller)
wrote:

In article , JIMMIE wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total circuit.
Adding resistance to a circuit changes voltage, not current.

I can't believe what I am hearing.
Hmm,
You are very funny. He was right.

On Mon, 31 Aug 2009 02:45:56 GMT, (Doug Miller)
wrote:

In article , Metspitzer wrote:
On Sun, 30 Aug 2009 20:58:38 GMT, (Doug Miller)
wrote:

In article
, JIMMIE
wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total circuit.

Adding resistance to a circuit changes voltage, not current.

I can't believe what I am hearing.

Oh? What's the problem?

If you add resistance in series to the destination load (which is what
higher wire resistance does), the *total* R seen by the source is
higher. Assuming a constant V at the source, this means I (current)
through the wire will be lower.

At the *load*, V will be lower also (often called the "IR drop" of the
wire) -- the load is now part of a "voltage divider".

So yes, adding resistance changes both voltage (away from the source)
and current.

Another way to think of it -- if the V across the final load (constant
R) is lower, the current (I) must also be lower. Circuit analysis can
be kind of fun -- you can often approach it from multiple perspectives
and get the same answer.

This assumes a constant resistance load, which a light bulb isn't
completely, but is sufficient for this purpose.

Josh

In article , Josh wrote:
On Mon, 31 Aug 2009 02:45:56 GMT, (Doug Miller)
wrote:

In article , Metspitzer
wrote:
On Sun, 30 Aug 2009 20:58:38 GMT, (Doug Miller)
wrote:

In article
, JIMMIE
wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total circuit.

Adding resistance to a circuit changes voltage, not current.

I can't believe what I am hearing.

Oh? What's the problem?

If you add resistance in series to the destination load (which is what
higher wire resistance does), the *total* R seen by the source is
higher. Assuming a constant V at the source, this means I (current)
through the wire will be lower.

No, it doesn't. It means *voltage* on the other side of the resistance will be
lower. Current is the same at all points in a series circuit.

At the *load*, V will be lower also
^^^^
You misspelled "only".

(often called the "IR drop" of the
wire) -- the load is now part of a "voltage divider".

So yes, adding resistance changes both voltage (away from the source)
and current.

No, it doesn't.

Another way to think of it -- if the V across the final load (constant
R) is lower, the current (I) must also be lower.

Wrong again. Current is the same at all points in a series circuit.

Circuit analysis can
be kind of fun -- you can often approach it from multiple perspectives
and get the same answer.

Evidently you've found multiple ways to get the same wrong answer.

Back to Circuit Analysis 101 for you, and this time pay attention when the
instructor discusses Kirchoff's Current Law.

This assumes a constant resistance load, which a light bulb isn't
completely, but is sufficient for this purpose.

You're assuming a *lot* of things; unfortunately, almost none of them are
correct.

Josh wrote:
On Mon, 31 Aug 2009 02:45:56 GMT, (Doug Miller)
wrote:

In article , Metspitzer wrote:
On Sun, 30 Aug 2009 20:58:38 GMT, (Doug Miller)
wrote:

In article
, JIMMIE
wrote:
It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total circuit.
Adding resistance to a circuit changes voltage, not current.
I can't believe what I am hearing.
Oh? What's the problem?

If you add resistance in series to the destination load (which is what
higher wire resistance does), the *total* R seen by the source is
higher. Assuming a constant V at the source, this means I (current)
through the wire will be lower.

At the *load*, V will be lower also (often called the "IR drop" of the
wire) -- the load is now part of a "voltage divider".

So yes, adding resistance changes both voltage (away from the source)
and current.

Another way to think of it -- if the V across the final load (constant
R) is lower, the current (I) must also be lower. Circuit analysis can
be kind of fun -- you can often approach it from multiple perspectives
and get the same answer.

Hi,
R is lower and I is lower?
Where does it come from?


This assumes a constant resistance load, which a light bulb isn't
completely, but is sufficient for this purpose.

Josh

In article ,
(Doug Miller) wrote:

If you want to limit the current passing through an LED, you put a resistor
in
*parallel* with it, not series.

Surely you jest. The resistor absolutely goes in series.

I've read so many erroneous claims in this thread I hardly know where to
start setting the thing straight. In a purely resistive circuit, whether
it's AC or DC, ohm's law rules. You cannot possibly change one without
affecting another.

For example, a long length of wire supplying a light bulb adds
resistance to the (series) circuit. The first thing this does is reduce
the circuit current, by increasing the total circuit resistance while
the supply voltage remains constant.

The second thing it does (simultaneously of course) is *split* the
supply voltage between the wire and the light bulb. So the intended load
(light bulb) gets less voltage. Some of the voltage is dropped (lost)
across the resistance of the wire.

Like Arnold, "I'll be back," but I've had trouble posting these last few
days due to ISP issues, so am limiting my postings severely in order to
stay connected at all.

"Smitty Two" wrote in message
news
In article ,
(Doug Miller) wrote:

If you want to limit the current passing through an LED, you put a
resistor
in
*parallel* with it, not series.

Surely you jest. The resistor absolutely goes in series.

I've read so many erroneous claims in this thread I hardly know where to
start setting the thing straight. In a purely resistive circuit, whether
it's AC or DC, ohm's law rules. You cannot possibly change one without
affecting another.

It is very funny to read many of the past posts. Unless it is something
very unusual, the resistor does go in series with a led , not in parallel.
When using a long extension cord of small wire to power a device, you drop
the voltage at the far end. This usually causes the device to use less
current. There are special cases such as motors where there is not enough
power at the motor so it can run at its rated speed and will burn up.

David Nebenzahl wrote:
On 8/29/2009 2:49 PM EXT spake thus:

Also, just as a long wire needs to be a thicker gauge, compared to a
normal length of wire, to carry a fixed amount of amperages,

Saying "a fixed amount of *amps*" would do.

a very short length of wire can be rated to carry a larger amperage
at a smaller gauge than normally used.

That's not true. Conductors are rated at a certain current regardless of
their length.

Agreed. That was probably the nuttiest comment in this thread.

This is the rational used on appliance cords, and the internal wiring
in appliances. I have seen a formula somewhere that will determine
the exact gauge needed for a given length at a specific amperage.

As others have said, length is a consideration only to limit voltage
drop. Length has no relation to the current carrying capacity of the wire.

I was going to bring up the aspect of cords too, as our 20-amp circuits
have cords plugged into them that are rated at far less than that,
creating another potential source of fire.

The code is rather pragmatic. #18 fixture wires on a 20A circuit are
not a problem because the lamp socket can not have a bulb over the wire
rating. (Someone could screw in an adapter-to-plug and run a space
heater - the rules of natural selection would then apply.)

Apparently there are not major problems with overloading #18 extension
cords on 20A circuits - dead bodies are very effective at promoting code
changes, probably also effective at UL. If I remember right, a
consideration was that the available fault (short circuit) current at
the end of the #18 cord is high enough so a short will give
"instantaneous" operation of the circuit breaker/fuse (not time delay).
(That is not likely true for Christmas tree lights.) AFCI breakers may
also help (but the selling point was cords that had been abused, like
walking on them).

You might be upset with motor circuits, that can have a circuit breaker
significantly larger than the wire ampacity. Welder circuits even bigger
difference.

--
bud--

On 8/31/2009 7:32 AM Ralph Mowery spake thus:

"Smitty Two" wrote in message
news
In article ,
(Doug Miller) wrote:

If you want to limit the current passing through an LED, you put
a resistor in *parallel* with it, not series.

Surely you jest. The resistor absolutely goes in series.

I've read so many erroneous claims in this thread I hardly know where to
start setting the thing straight. In a purely resistive circuit, whether
it's AC or DC, ohm's law rules. You cannot possibly change one without
affecting another.

It is very funny to read many of the past posts. Unless it is something
very unusual, the resistor does go in series with a led , not in parallel.

Thank you. Just to show those doubters that I'm not a complete idiot,
yes, I know that the current is the same at all places in a series
circuit. Inserting a resistor doesn't change this: whatever current
flows through the resistor also flows through everything else. But a
resistor in series with a LED does indeed *change the total current* and
reduce the current through the device, which is why they're used: so the
LED doesn't burn up. That's because the *total resistance* of the
circuit has been changed. A simple calculation using Ohm's Law will show
this.


--
Found--the gene that causes belief in genetic determinism

On Mon, 31 Aug 2009 02:47:34 GMT, (Doug Miller)
wrote:

In article m, David Nebenzahl wrote:
On 8/30/2009 1:58 PM Doug Miller spake thus:

In article
,
JIMMIE wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total
circuit.

Adding resistance to a circuit changes voltage, not current.

Changes current, too. E.g., a current-limiting resistor in series with a
LED, sized so that the LED won't draw excess current and burn itself out.

Wrong.

If you want to limit the current passing through an LED, you put a resistor in
*parallel* with it, not series.


I guess you didn't try that.

The resistor in parallel wastes power, and will have no effect on
current through the LED (other than if it creates excessive voltage
drop, a really inefficient way to dim the LED).

With no resistor in series, current (from a constant voltage source
greater than about 2V) will approach infinity, until the LED is
destroyed. I've seen that happen.

Total current in the circuit remains the same.

Not when you add series resistance.
LEDs regulate the voltage across them, unless the current is
excessive. Since that voltage is around 2V, a 120V source would
require a really big series resistor. To avoid that, 120V would
usually be applied to a series string of multiple LEDs (in series with
one smaller resistor).

A string of 50 LEDs (common in holiday lights) using 20mA from a 120V
source would require a resistor of 1000 ohms (that's assuming 2V
voltage drop on a LED). There would be 2V across each LED and 20V
across the resistor. Use multiple strings in parallel for more light.
Current is 20mA and power dissipated by the resistor is .4W.

Changing that resistor to 2000 ohms will change the current to 10mA.

Changing it to 500 ohms will change the current to 40mA. The resistor
will need to handle .8W. You may need a bigger resistor.

--
Mark Lloyd
http://notstupid.us

"How could you ask me to believe in God when there's
absolutely no evidence that I can see?" -- Jodie Foster

On Sun, 30 Aug 2009 18:25:36 -0700, David Nebenzahl
wrote:

On 8/30/2009 1:58 PM Doug Miller spake thus:

In article
,
JIMMIE wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total
circuit.

Adding resistance to a circuit changes voltage, not current.

Changes current, too. E.g., a current-limiting resistor in series with a
LED, sized so that the LED won't draw excess current and burn itself out.

I had one do that once. It was a surprisingly loud noise for something
that little. Half the plastic case disappeared.

BTW, one time I saw a working LED with no apparent series resistor. It
was in a little flashlight powered by 2 button cells (CR2032 IIRC).
This seeming impossibility seemed to make use of the series resistance
of the battery, which couldn't supply too much current to this LED.
--
Mark Lloyd
http://notstupid.us

"How could you ask me to believe in God when there's
absolutely no evidence that I can see?" -- Jodie Foster

On Sun, 30 Aug 2009 20:58:38 GMT, (Doug Miller)
wrote:

In article , JIMMIE wrote:

It does not affect the current that the wire COULD carry or the
AMPacity. A long cable, compared to a short cable, will the current
in the total circuit because it adds resistance to the total circuit.

Adding resistance to a circuit changes voltage, not current.

This is true if the circuit is powered by a constant-current source,
something unusual.

There may be some confusion here between actual current and current
CAPACITY (of a conductor).

On Mon, 31 Aug 2009 03:47:35 GMT, (Doug Miller)
wrote:

In article , Josh wrote:


If you add resistance in series to the destination load (which is what
higher wire resistance does), the *total* R seen by the source is
higher. Assuming a constant V at the source, this means I (current)
through the wire will be lower.

No, it doesn't. It means *voltage* on the other side of the resistance will be
lower. Current is the same at all points in a series circuit.

Yes, I agree. I could have worded that better -- I meant "lower than
it would have been without the extra resistance in series", not "lower
at this point in the circuit than at some other point".

The earlier post I was responding to claimed that the current would
not change if the resistance of the wire changed. That's what I was
trying to refute. You and I both agree that the current at any point
in one circuit will be the same as any other point in that (series)
circuit.

I suspect a lot of the next comments stem from that unclear wording;
here goes...


At the *load*, V will be lower also
^^^^
You misspelled "only".

Yes -- again, I meant "The V across the load will be lower than it
would have been without the extra series resistance, just as the I is
lower than it would have been without the extra series resistance"


(often called the "IR drop" of the
wire) -- the load is now part of a "voltage divider".

So yes, adding resistance changes both voltage (away from the source)
and current.

No, it doesn't.

Adding resistance to the wire changes the voltage across the load, and
thus the current through the load (and the rest of the circuit). It
doesn't change the voltage across the source (until you consider
battery internal resistance, regulator quality, and other factors).



Another way to think of it -- if the V across the final load (constant
R) is lower, the current (I) must also be lower.

Wrong again. Current is the same at all points in a series circuit.

I agree. But it's not the same as it would be in a *different*
circuit, one with more or less resistance in series.


Circuit analysis can
be kind of fun -- you can often approach it from multiple perspectives
and get the same answer.

Evidently you've found multiple ways to get the same wrong answer.

I've never been accused of getting things right the first time :-)


Back to Circuit Analysis 101 for you, and this time pay attention when the
instructor discusses Kirchoff's Current Law.

This assumes a constant resistance load, which a light bulb isn't
completely, but is sufficient for this purpose.

You're assuming a *lot* of things; unfortunately, almost none of them are
correct.

Hopefully I've reworded my statements in a way that Kirchoff would
agree with -- it's been 18 years since my Circuit Analysis classes; I
have an ECE degree but do logic design/architecture for a living, so
rarely get to break out the Ohms law these days; the Fourier
transforms scared me away from analog design for good :-)

Josh

On Sun, 30 Aug 2009 11:50:05 GMT, (Doug Miller)
wrote:

In article , dpb wrote:
David Nebenzahl wrote:
....
That's not true. Conductors are rated at a certain current regardless of
their length.
....
Not exactly so--look at the voltage drop tables; at a given voltage the
drop becomes excessive at a minimum conductor size and a larger
conductor is required.

Yes, exactly so. Voltage and current are not the same. A long conductor *does*
cause voltage to drop, but it does *not* affect current.

It may not affect current, but it sure as heck affects power. The
current stays the same, being a series circuit, but the voltage is
shared between the wire and the device. - hense power to the device is
reduced.

On Sun, 30 Aug 2009 11:50:05 GMT, (Doug Miller)
wrote:

In article , dpb wrote:
David Nebenzahl wrote:
....
That's not true. Conductors are rated at a certain current regardless of
their length.
....
Not exactly so--look at the voltage drop tables; at a given voltage the
drop becomes excessive at a minimum conductor size and a larger
conductor is required.

Yes, exactly so. Voltage and current are not the same. A long conductor *does*
cause voltage to drop, but it does *not* affect current.
I just rethought my answer and I was wrong. It not only drops the
voltage but it DOES drop the current, because the total resistance of
the circuit is higher. With a fixed supply voltage the current WILL
drop. Combined with the voltage drop it is a double whammy on the
power.