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Not Understanding this AC Circuit!!

Discussion in 'General Electronics Discussion' started by Gillesfizzog, Mar 24, 2014.

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  1. Gillesfizzog


    Jan 14, 2014
    I have problem im trying to understand and my teacher here at my college is puzzled aswell, we must be missing something. Here it is.

    My goal was to calculate the power factor of a 60 watt light bulb, i think the bulb is an incandescent but the reading im getting with my multimeter makes me think thats there is a coil in it making it an RL circuit. Heres why?

    First I tested the voltage at my outlet it was 122 volts.
    Then I tested the current that the bulb is drawing and it measured 0.45 amps

    Now correct me if im wrong, but from what i learned, in a purely resistive circuit, the bulb should have a resistance of 271 ohms. (122v devided by 0.45 amps)

    However, when I test the ohms of the light bulb, it measured 20.5 ohms. For this reason, my teacher thinks its an inductive load.

    If it is an inductive load, that means that the calculated power factor would be 7.5% This seems wrong because its such a low efficientcy??? and combine that with the fact that most of the energy used is converted to heat!!! Is this bulb really that inefficient? :eek:
  2. hexreader


    Apr 21, 2011
    The resistance of a light bulb is very different when cold, compared to when hot (assuming old-fashioned coiled-coil light bulb).

    Your calculation of resistance-when-hot = 271 Ohms is probably correct.

    As far as I know, inductance is negligible, and at 60 Hz would have little measurable effect.

    Tell your teacher from me - he/she is wrong :)

    ...and yes, incandescent light bulbs are inefficient, which is why they are going out of use..

    EDIT: maybe post a picture of the light bulb, so we know what sort we are talking about - just in case I owe your teacher an apology.
    Last edited: Mar 24, 2014
  3. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    Hexreader is right.

    You'd need a 2-channel oscilloscope (with isolated power, e.g. battery powered) to measure voltage and current (using a small shunt resistor) simultaneously. You can tell whether a load is resistive, capacitive or inductive from the phase shift between current and voltage.
  4. Gillesfizzog


    Jan 14, 2014
    Haha thanks for the replies i think thats it. it looks like a regular incandescent bulb. My teacher mentioned that it could be temperature aswell but then when i left the light on for a while to get hot and took it off as fast i could to get the resistance it was only 22.5 ohms, but even from that short amount of time it had went from really hot to a little hot.
    Last edited: Mar 24, 2014
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    If you had access to a high voltage DC power supply you could confirm that the lamp draws the equivalent current from 122V DC as it does from 122V AC.

    That would be a better (although probably not safer) way to calculate the resistance to DC vs the reactance to AC.

    I think you would find that any difference would be well within the tolerances for your experiment.

    If you have access to a variac, you can measure the current as you increase the voltage and graph one against the other. This will allow you to graph the characteristic curve for a filament lamp (which you might preview here)

    I have to ask though, why is your teacher letting you play around with mains voltages. You can see exactly the same thing with a low voltage (12V, say) lamp and it's well within the voltages you can get from regular power supplies. (And they are less likely to kill you)
  6. Gillesfizzog


    Jan 14, 2014
    I was working on it at home mostly , but it wasnt his idea, i was just testing something in the shop as safe as i could while we were supposed to be working on other non live projects. i guess i should probably do it at home.
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