Multimeter Readout

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March 16th, 2009 admin Leave a comment Go to comments

Multimeter question?

I have a 60W bulb that says 130V on it. I’m assuming that means that when 130V is applied to it, the bulb consumes 60W of power (shouldn’t it be 120V, though?). Rearranging the equations V = IR and P = IV to get the resistance in terms of voltage and power you get R = V²/P. 130V²/60W ≈ 280Ω.

So here’s the problem. When I turn my multimeter to the 2000Ω setting and touch the two wires to the light bulb, one to the bottom and one to the side, I get a readout of 22.

Your calculations are correct; the resistance is about 280 ohms when the filament is hot. Normally the resistance is much lower when it is cold, which means you get a higher current when you first turn on the light. That’s why bulbs usually burn out when they are turned on, instead of after they have been on for a while. I think that the 130 V is just a maximum rating, and 60 W is the actual power you get at a standard voltage of 115 to 120 volts. The ohmmeter reading sounds about right; the resistance of lamp filaments is very nonlinear. By the way, the fact that the bulb is normally used on AC doesn’t affect things very much. On the other hand, if you check the resistance of a transformer with the meter, you will also get a much lower reading than you expect, because the inductance of the transformer creates a lot more impedance for AC than the DC resistance measured by the meter.