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Joined: Oct 2002
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S
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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 ..

Joined: Jul 2002
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W
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Ohms law applies at low voltage.

Joined: Oct 2002
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Thought so; Thanks ....

Joined: Jul 2004
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That's not Ohm's Law, however...

Joined: Oct 2003
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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.

Joined: Aug 2001
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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.

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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.

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Quote
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?

Joined: Jul 2004
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S
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Ohm's Law is E = IR.

Watt's Law is P = IV.

Joined: Oct 2002
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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 [Linked Image] Thanks anyway for the input... Steve

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