I'm doing a low voltage project, and not use to the amperage etc. the smaller wires. When calculating the amperage, I'm assuming the math technique is the same. I have 1 alarm that is 15 watts. the voltage is 12 volts. Therefore 15 watts divided by 12 volts gives me 1.25 amps. This is the right formula no matter what the voltage etc. is, isn't it? I need a little assurance here:; Thanks ..

Ohms law applies at low voltage.

That's not Ohm's Law, however...

Ohm's law applies, but when working with LV, wire and connection resistance comes into play. I assume your device consumes 15 watts over a range of voltage say from 11 to 14 volts applied at the device.

Just as an FYI: if your project is security alarm or fire alarm there are other requirements for the load calc.

Also be a bit careful about the length & size of the wires. #18/20/22 alarm wiring is often very long runs and can add a lot more resistance than the 14/12/10 we normally use.

A certian length of say 14 gauge wire will have the same voltage drop at the same amps no matter what the supply voltage is. However, that voltage drop will be a bigger percentage of the supply voltage when the supply voltage is lower. Thus, what is an acceptable drop at 120V would be bad when the supply is 12V. 12 volt light bulbs will look dimmer than 120V bulbs seeing the same voltage drop. This is a big reason why POCOs use high voltages for long distance transmission.

That's not Ohm's Law, however...

Isn't it derived from ohms law though??. Seem to remember someone proving that one time or another?

Ohm's Law is E = IR.

Watt's Law is P = IV.

My "Ugly's book" says that the three basic Ohm's law formulas are: I=E/R, R=E/I, and

E= IxR

I = Amperes

E = Volts

R = Ohms

P = Watts

Maybe Ugly's wrong

Thanks anyway for the input... Steve

George Simon Ohm.(March 16,1789-July 6,1854)

E=IR, I=E/R, and R=E/I are all Ohm's Law, just stated in different ways.

Watt's law is W=IE(orV) or W=I^2V (that's "I" squared)

Ohm's Law is E = IR.

Watt's Law is P = IV.

This is correct as V is the same as E. It is your (EMF) ElectroMotive Force or Volts. So yes that is correct.

Sorry about going multilingual there.

Indeed, E == V. (Please hold the remarks about being a cunning linguist.)

By reading some of the responses I thought I would chime in with a little more.

In school they have shown us 12 different formulas for finding whatever it is you want to find. There are several variations and now they are gradually going into inductance and capacitance. Doesn't really apply here though.

Anyway the basic that everyone has listed is:

E = I x R

I = E / R

R = E / I

then we have:

P = I x E

I = P / E

E = P / I

from these two formulas above they have shown us how to break these down to get 6 more formulas.

E² = R x P

R = E² / P

P = E² / R

and then we have:

P = I² x R

I² = P / R

R = P / I²

They all work if you give them a shot. I know it has helped is school by being able to go through the "backdoor" on some problems. Enjoy.

* edited because I can't spell this early in the morning*

[This message has been edited by GA76Apprentice (edited 12-12-2004).]

By the way,

There is no such thing as the common thinking "Watts Law".

As any free thinking person will tell you, James Watt invented, among other things, the Steam Engine, the term "Watt" and it's multipliers, was applied after his death.

It was actually Edison that came up with the Power Calculation that we call Watt's Law.