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Yup Posted by Ari [Email] (#2847) [Profile/Gallery] (more from Ari) on Wed, 13 Nov 2002 09:04:12 In Reply to: So...just to be sure on this., Jeff Cunningham, Tue, 12 Nov 2002 13:22:25 Members do not see ads below this line. - Help Keep This Site Online - Signup |
Yes, Watts = volts x amps (so watts/volts = amps)
But to add- you said you'd measure the voltage at the circuit ground, compared to the voltage at the battery, then wrote (13.5-12).
There are two 'wires' connecting the load (e.g. bulb) to the battery - the power wire and the ground through the chassis. If you measure the voltage across the bulb, say you get 12 volts. From your calculations, yes there is 0.3 ohms of resistance. But not all of it is in the chassis. Some (probably most) is in the wire from the positive side of the battery to the light. The chassis tends to be a pretty good conductor. The line from the battery to the light isn't as good, especially since that wire isn't a straight piece of wire - it's a few pieces of wire, a relay, and a bunch of connectors. Each has different resistances, and they all add up.
I will use a good voltmeter to read from the battery negative post to the chassis of the car. This should be well under 100 millivolts, with the car running and all the loads you're interested in ON - lights, fans, etc. I would then measure from the chassis to the ground line of the load. I would expect this to be less than a few tens of millivolts for a small (5 amp or less) load. Any more than that, and the ground connection bears cleaning. It's reasonable to see up to a volt for a voltage drop on the positive side, especially when there may be switches and relays in the path.
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