# Wire Gauge ratings

Discussion in 'Electronic Basics' started by Jamie, Mar 7, 2006.

1. ### JamieGuest

I have a question about wire gauge ratings. Why do wire gauge charts
show how many amps that the wire is able to carry, and not the wattage
that the wire can dissipate safely? Resistors are rated in watts, not
amps.

I would think that if I had a length of wire that was just barely
capable of conducting a charge at the rate of 10 amps at 12 volts, that if
I were to hook up 120 volts to the same piece of wire and attempted to run
THAT charge at the rate of 10 amps, that, since P=I*E, the wire would be
dissipating heat much more rapidly and could overheat.

Why isn't voltage taken into account in determining the right diameter
for a conductor in a circuit?

- Jamie

The Moon is Waxing Gibbous (55% of Full)

2. ### Tim WilliamsGuest

You certainly can, although it varies by insulation (plastic insulates
thermally as well as electrically!).
When you have several variables appearing in a problem, you can't just take
two and multiply or divide; you have to pick and choose. You picked the
wrong ones -- and that's why it doesn't work.

You may ask that -- well, there are only two numbers in the question!
True enough, but the others are hiding in the problem. When current flows
through a wire, its resistance develops a voltage, and *this* voltage
multiplied by the current equals the wattage.

The line voltage does not appear across the wire itself. It appears outside
of it, where there are insulators, so we could care less.

-- Wires do, in fact, have a voltage rating, but that relates to the voltage
the *insulation* is rated to handle, under a variety of conditions
(temperature, solvents, crush, etc.), as rated by the manufacturer.
Obviously, bare conductor cannot have a voltage rating; it would be
completely arbitrary.

Tim

Good Lord

4. ### John FieldsGuest

---
OK, lets say that you've got a 100 foot extension cord and it's
rated to safely dissipate 10 watts. What are you going to do with
that information? More importantly, how do you know what you can
plug into it safely?

On the other hand, if it's rated to safely carry 10 amps all you
have to do is look at the nameplate of what you're going to plug
into it to know how much current the extension cord will have to
carry.
---

---
You forget that the wire isn't supposed to be a significant portion
of the load. Using our 100 foot extension cord as an example, if it
can safely dissipate 10 watts with 10 amps going through it, then
the voltage dropped across it will be:

P 10W
E = --- = ----- = 1V,
I 10A

with the balance being dropped across the load, and its
(the extension cord's) resistance will be:

E 1V
R = --- = ---- = 1 ohm,
I 1A

and the circuit will look like this:

E1
|
[R1]
|
+----E2
|
[R2]
|
GND

Where E1 is the supply voltage, E2 is the voltage appearing across
the load, R1 is the resistance of the extension cord, and R2 is the

So, let's take a look at what actually happens with a load hooked up
to our extension cord.

First, if we have a 120V supply feeding a load which draws 10 amps,

E 120V
R --- = ----- = 12 ohms
I 10A

and our circuit will look like:

120V
|
[R1] 1R
|
+----E2
|
[R2] 12R
|
GND

Now, since the extension cord and the load are in series, that's a
total of 13 ohms, so the current the extension cord will have to
carry will be:

E 120V
I = --- = ------ ~ 9.2 amperes
R 13R

there because of the ~ 1V drop in the cord.

On the other hand, if you wanted to treat the extension cord like a
resistor, all you'd have to do would be to short out the socket end
of it and plug it in.

Since the cord looks about like an ohm, when you first plugged it in
you'd get:

E 120V
I = --- = ------ = 120 amperes
R 1R

through it, and the power it would be dissipating would be:

P = IE = 120A * 120V = 14,400 watts!

---
Because it's largely irrelevant since the bulk of the voltage is
supposed to appear across the load. The diameter is important
because that, in conjuction with the length of the wire will
determine its resistance and that, in turn, will determine how hot
the wire gets with a specified current running through it.

5. ### ChrisGuest

Good morning, Mr. Fields. If I might ask the master if he'd like a
second cup of coffee?

E 1V
R = --- = ---- = .1 ohm,
I 10A

and then through to:

E 120V
I = --- = ------ = 1200 amperes
R .1R

and

P = IE = 1200A * 120V = 144,000 watts!!!

(Extra exclamation points added -- it only makes your point to the OP
more relevant.)

I never post before the second cup of coffee anymore. I used to do
this almost every morning.

Cream or sugar, sir?

Cheers
Chris

7. ### ehsjrGuest

P = I*E which is also equal to I^2*R

In the math, voltage is automatically taken into account:
E = I*R. Substitute I*R for E in your power equation and you get
P = I*(I*R), which is more commonly written P=I^2*R. That means
power is equal to current squared times resistance.

know the source voltage when making a wire table.
They know the resistance of the wire and can determine
how many watts will be produced in it for a given
current through it. They know how many watts the wire
can safely dissipate, so it is simple to rate the
wire in amps.

Ed  