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Calculating amperage and wattage?

So I'm doing an energy inventory on our house to see how much electricity we use, and maybe see if we could cut down on some costs. 
Being Christmas right now, we have a long multi-string chain of lights going across our living room. I'd say it's about 50 feet long. They are just regular incandescent lights. typical normal size, no LEDs or flashers or anything. When you plug them in, they come on, and that's all they do. I wanted to figure out their wattage, so I started with finding the amperage. My multimeter isn't very good when it comes to measuring amps, so I decided to use a roundabout method by finding the resistance of the circuit. The resistance came out to be 13.3 ohms. Plug that into ohm's law and you get around 9 amps of current and therefor, about 1080 watts of electricity. That sounded right to me...after all, it is a long string of lights, and even though they don't individually suck up a lot, I figured it adds up. 
I tried this same method with a single bulb that I knew to be around 300 watts. The resistance came out to be about 4.5 ohms giving a whopping 26.7 amps with a wattage of around 3200 watts! That can't be right. I want to know what I'm doing wrong. I'm pretty sure it can't be my multimeter because I just bought it brand new about a month ago. What am I doing wrong?

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lemonie5 years ago
Does it still work if you fit a 1A fuse to the plug?

L
Light bulb filaments are more resistive when hot. So measuring the cold resistance is going to give you a P=V2/R that's a lot bigger than it really is.

Have you seen that Kill-A-Watt(r) power measuring gizmo. The price is typically 20 to 30 USD, at the time of this writing.  Example:
http://www.homedepot.com/buy/electrical/electrical-tools-accessories/kill-a-watt-83064.html

I've got one, and that little tool is totally worth it.  That tool, or one like it, is  kind of what you want for your power audit.  It tells you the real, actual amount of power flowing through a cord.  It even works on loads with a large reactive component; i.e it figures out the power factor for you, and includes that in its calculations.  Basically it' will just tell you what you want to know, without you having to do any math.  I mean  besides adding up all the KWs, or KW*hr/day for all your loads taken together.
iceng5 years ago
Easy enough go out and time your watt-meter wheel (itsi black line) for a 100 turns before and then repeat again after.  The difference can be worked back to KW and Amps.

A
When the filament is cold, its resistance is far lower than when it is at operating temperature and emitting light, so the initial current draw and therefore power will be higher than when running.

Being indoor lights, the power of each individual bulb is unlikely to be above 10W, as 100W will fully light a small room.

Have you tried measuring the current through a single bulb in the string?
tylervitale (author)  The Skinnerz5 years ago
It isn't really a small room, it's the living room and the kitchen. But it doesn't fully light either of them. You have good point. The resistance would be much higher during operation.
I haven't tried measuring a single bulb in the string, but given what you just said, I don't see why it should make a difference...
or quickly measuring the resistance after the hot bulb is removed...Still not white hot, but a lot closer than a cold bulb.

frollard5 years ago
The wattage of each bulb should be written somewhere on the base of the bulb.