BimmerJeeper

Watts is the amount of power the component draws.
Like 60W bulbs.

Voltage is the "pressure" that is pushing the electricity through the wires.
On cars, you have a 12V system.
In homes, you have 120V.
So, a home system can power a much higher wattage device.

Amps is the current or flow rate.
In cars, this relates to the size/gauge of the wires.
4 AWG battery cables are thicker than OEM.
This means these cables have a higher capacity.
1) Do you say these are "4 amp" since it's 4 AWG? Or do you just say they are higher amp cables.
2) They require LESS voltage since there is less resistance?

94XjSport94

Eh those definitions are right more or less.

No, it's 4 AWG wire. Since 10 gauge MTW wire can handle 30 amps it's not called 30 amp wire.

It doesn't require less voltage, it's supposed to handle more current and can handle bigger voltage. Plus there is more wire meaning more resistance on longer runs of wire.

You're not going to be pushing more than 12 volts to the starter or main power distribution block.

BimmerJeeper

Ok, so a certain gauge of wire is not called by its amperage rating, but the amp rating is just a specification value. Like 10 MTW will be labelled as "30 amps" ?

94XjSport94

I'll read the spool of wire at work and type what it says tomorrow for you.

We have a chart at work somewhere that gives the maximum values about wire sizes and other random info.

10 gauge might not be listed at 30 amps since that would be considered its max, i'll check on it though.

5-90

True - voltage, current, power, and resistance are all mathematically related in a closed electrical system. cf: "Ohm's Law," "Watt's Law."

Essentially true.

Less true. Most home service is rated for something like 100A delivery. For the whole house. Typical loads are 30A or less, measured at 110VAC (home mains are 220VAC, with two "hot" legs, split two to 110VAC buses for most circuits. 220VAC circuits - electric dryer, electric stove/oven - use both "hot" legs.)

Starter motor current in our 6-242-powered Jeeps, for instance, runs up around 160-180A without starter motor load (called "no-load current.") Actual loaded starting current is somewhat higher, coupled with engine management loading. Actual starting current for the 5-20 seconds you're cranking an engine to start it can be up around 300A (figure 160-320A for typical gasoline engines, up around 400A or so for most common Diesels.)

Winches can go even higher - 350A would be a low draw for a winch!

(And, know that "house current" and car power are different animals - your car runs on 12VDC - direct current, while your house uses 110/220VAC - alternating current. Direct current runs at a constant level with respect to the ground ("zero voltage,") while alternating current runs on a regular cycle above and below the zero voltage reference (110VAC will alternate from +110V to -110V, with respect to ground, alternative 60 times per second, or 60Hz. AC is more efficient for long-range electric transfer than DC. AC can also be stepped up and stepped down using a transformer, and higher voltages are more efficient for long-range transmission as well. DC can't be run through a transformer on its own. Yes, you can find AC in your car - the 'alternator' gets its name because it generates alternating current, which is rectified to DC by the diode packs in the back of the case before it gets distributed to the rest of the vehicle.)

1) "AWG" stands for "American Wire Gage" - #4 wire would be "4AWG" or "4 gage." Wire isn't called out by ampacity. Wire gages can run from up around 0000AWG ("4/0 AWG") for very large wiring, down to about a 32AWG (most common small wire gage - but I'm sure they go smaller) for magnet wire. Smaller numbers mean larger wire.
2) With a larger conductor cross-section, you have less conductor loss. The voltage requirement at the consumer (lamp, electric motor, whatever) remain the same, but you can run a source voltage closer to the actual consumer requirement because you lose less voltage in between.

Let's expand your water analogy:
- Voltage = Pressure. This is the potential for current flow.
- Current (Amperes) = Flow rate. This is the actual flow of electricity, essentially the number of electrons passing a given point in a given time
- Power (Watts) = Work done. This is the lifting, rinsing, flushing, whatever done as a result of the pressure and flow rate (Volts and Amperes.)
- Resistance = resistance. Either way, it's a reduction in flow rate. For electricity, that can be a special wire alloy or a carbon track. For water, that's a reduced orifice in the flow path (say, a 1/8" orifice in a 3/4" pipe, which poses a significant resistance to flow.)

MATHS:
Ohm's Law - E=IR
Watt's Law - P=IE

E = Voltage (EMF, or Electromotive force)
I = Current (Amperes. I'm unsure of what the "I" stands for on its own.)
R = Resistance (in ohms.)
P = Power (Watts.)

Both of those formulae may be rearranged or combined using regular rules of algebra. Examination of these will show the relationship between voltage, current, power, and resistance (all four quantities are present in any electrical circuit.)

BimmerJeeper

Question about Watt's Law - P=IE

3 ways to look at it:
P = I x E
Power = Current x Voltage
(watts) = (amps) x (volts)

Watts is normally fixed. (60W bulb)
Amps can vary based on wire gauge.
Volts is usually fixed by the power supply you are using (12V car or 110V home)

So, if a 60W bulb is attached to the same voltage supply, but switched to a lower current wire (lower amp wire) then the bulb will get choked and draw less than 60W? OR does it always draw 60W no matter what? And the lower amp wire will not require more voltage to push the needed electricity through? Is there usually excess voltage on tap to handle situations where the wire is rated too low amps?

Parsnip

well look at what the math says
60w=A*90V
this will give you 2/3A, or .66 repeating amps
you will get more resistance over a shorter wire, because it cannot handle the current your trying to push through it. thats why if you wired up a sub/amp with 16ga, the wire would get excessively hot and burn, because energy, like matter, cannot be created nor destroyed, it can change form. in this case, the excess energy is released as heat from the resistance, untill the wire burns.

5-90

The bulb will always try to draw 60W, because that's what it's rated for. If the wire can't handle it (too small in cross-section/too long for cross-section,) then you have problems.

Note that both length and cross-section are important values. One need merely to look at a chart of ampacity v. wire gage v. length to see that this is so. A longer wire will need to be larger in cross-section for a given current-carrying capacity (ampacity,) simply because heat will be greater otherwise. This is due to the increased internal resistance of the wire (due to the extra length.)

Paradoxically, a larger wire gage reduces internal resistance, because there is more conductor area available to transfer electrons (electricity,) and increased surface area to safely dissipate heat (there is always heat resulting from current flow. Can't get away from it. However, you can work with it, and dissipate heat using enough surface area that you don't have insulation failure - and now much more heat than you originally intended (read: electrical fire).)

The resistance of the bulb does not change. The voltage supplied to the bulb changes, which means that required current changes - refer to Watt's Law (which you posted) and Ohm's Law (q.v.) for information on that. The automotive electrical system may be considered "closed" for purposes of calculation, which fixes the relationship between quantities expressed in a circuit (voltage, current, power, and resistance.)

BimmerJeeper

Wait, something doesn't make sense. If the forumula is true, then (watts) = (amps) x (volts) But in the case of a 60W bulb in a 110V home, if watts and voltage are constant, then amps has to be fixed also. So, what happens when you switch the wire to a lower amp wire? One of the other values much change for the equation to still work.

60w = (amps) x 90V.
So, amps = 2/3

60w = 2/3 amps x 90V
So, in this case, you have to have a 2/3 amp wire.
But, what if you change the wire to 1/2 amp?
Now, what?

60w does not equal 1/2 amps x 90V
So, which one will change?

Parsnip

They won't change, ohms law is still in effect, the outcome is that the wire will dissapate the remainder of amperes that it can't carry as heat, untill the wire burns and the circuit no longer is complete. The light for example will still try to draw 60 watts untill the circuit is broken.

5-90

No.

P=IxE. Power (Watts) equals current (Amperes) x Volts.

We fix the left side of the equation at 60W, then plug in the two different values for voltage.

60W = Ix12V
60W/12V = I
5A = I
60W = 5A @ 12V (AC or DC are irrelevant here.)

60W = Ix110V
60W/110V = I
0.(54)...A = I (5/9 Ampere.)

Therefore, at the higher voltage, less current is needed to give the same power - 60W.

BimmerJeeper

So, in the example using
60w = 2/3 amps x 90V

If you change to 1/2 amp wire,
either you have to reduce to 45W
45w = 1/2 amps x 90V

or you increase to 120V
60w = 1/2 amps x 120V

But, you're saying neither happens...
and that it's still acting as a 2/3 amps current,
b/c that is what the 60w and 90V demands.

But, since the wire can only handle 1/2 amps current,
how can it still act as a 2/3 amp current?
You're saying the rest of the amps current is thrown off as heat.
Effectively, the wire is still functioning as 2/3 amps
(with 1/2 amp going thru the wire, and 1/6amps being given off as heat current.

Is that right?

I was referring to the situation where voltage and power are fixed.

Parsnip

It will function that way for a short time untill the wire burns because it's not rated for the amperage. It's flowing the 2/3 amps, but the wire will heat up from the resistance because the cross section is too small for the amount of current.

Take for example the water analogy. If you were trying to flow too much water through a pipe that could not handle the pressure the pipe would hold only for short time before bursting.

salad

When trying to jam more current through a conductor than it can handle you have a couple problems. The first, as mentioned, is that a conductor rate for 5 amps simply cannot carry 10 amps while maintaining a low resistance. As there are fewer opportunities for power to move along the wire (due to limited material) this extra power is discarded as heat. This works on the same sort of idea as mechanical friction, light bulbs and electric heaters rely on this principle. The secondary problem is that as a result of that heat, the conductor's resistance increases further and it cannot deliver full voltage any more. This is called voltage sag and can be observed many places in Jeeps and in home during brown-outs.

The exact math is beyond me (http://en.m.wikipedia.org/wiki/Elect...d_conductivity) but essentially your 60W bulb trying to draw extra power will in the end be receiving less power overall.

ctjeep

Wow - a typical internet pile of truths, 1/2 truths and just bs. . .

Let's take a simple one.

For simple math let's take a resistive device rated at 60 watts at 120 volts. The current drawn by the device = volts/watts=1/2

The resistance of that device
=voltage/current (ohm's law) = 240 ohms.

The resistance of the device doesn't change.
So if we reduce the voltage from 120 to 90 the current will be 90/240=.375 amps

The wattage that device now consumes
= volts*amps=90*.375=33.75 watts

So by cutting the voltage delivered to the device by 25% (120 to 90 volts) we reduced the power by about 1/2.

Note: this example is for resistive devices, not motors.

btw DC current is current always flowing in the same direction. AC current reverses direction by its frequency, now measured in Hertz (cycles per second)