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The relative resistance of a filament vs. a conductor

  1. Jan 21, 2015 #1
    1. I'm a middle school science teacher with no science background, and I'm trying to avoid teaching any misconceptions on this topic! However, I don't know that I grasp what's going on here myself. I'm trying to explain rationale for why, on a circuit diagram, the symbol for a resistor may sometimes also indicate a load.

    My question is: would a light bulb filament have a lower resistance than any conductors in a circuit? Or would it be greater?

    Here's what I think I know so far:

    -Materials with lower resistance give off more light because electrons move through these materials more freely and the kinetic energy of their motion is transformed into thermal energy.

    -Conductors have low resistance than other materials because they allow electrons to move through them freely.

    3. At first I assumed that tungsten would have a higher resistance--this seemed like the only reason why it would be classified as a "resistor" on a diagram. But if this was true, wouldn't the copper wire or other conductors in the circuit be glowing even more brightly than the tungsten? Logic would seem to suggest, then, that tungsten would have a much lower resistance than the conductors. But then why is it classified as a resistor?

    If I've gotten anything wrong, please disabuse me. Any clarity would be much appreciated.
     
    Last edited by a moderator: Jan 21, 2015
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  3. Jan 21, 2015 #2

    Quantum Defect

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    You can do some interesting experiments with a flashlight bulb, some batteries and a multimeter.

    (1) Measure the resistance of the flashlight bulb by itself with the multimeter.

    (2) Make a circuit with a fresh AA battery and the light bulb.
    (a) measure the current flowing through the circuit -- using the ammeter function of the multimeter.
    (b) calculate the resistance of the light bulb (V = i * R) Use the voltage of the battery.

    (3) Make a new circuit with two AA batteries, etc...

    You'll see that as the filament in the light bulb glows brighter, the resistance goes up. This is normal behavior. Electrical resistance goes up with increasing Temperature.

    The power lost in a resistive element is P = i^2 * R

    Filaments in light bulbs are very thin. Thin wires have higher resistance than thicker wires. The resistances of any conductors in a typical circuit are going to be very, very small. Heating elements, light bulbs, etc. will have higher resistances.
     
  4. Jan 21, 2015 #3

    berkeman

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    Welcome to the PF.

    The filament in an electric light bulb is just a conductor with some special properties that let it carry enough current to glow brightly and not burn up quickly. See the "Electric Light" section of the wikipedia article on Edison, for example: http://en.wikipedia.org/wiki/Thomas_Edison

    You can make a light bulb with copper or other conductors as the filaments, but they just burn out more quickly. :-)
     
  5. Jan 21, 2015 #4

    rude man

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    No.
    "Glow" is basically proportional to the temperature of the filament. This in turn is mainly a function of the power dissipated in the filament. The power dissipated in a filament is V2/R where R is the resistance of the filament and V the applied voltage.
    If copper, which has a greater conductivity than tungsten, were used as a filament in a light bulb, the filament would glow very brigthtly indeed - for a fraction of a second - then open up so no further current is conducted. The tungsten filament has a higher resistance such that, when excited by a 115V source such as obtains in a U.S. or Canadian home, will conduct enough current to glow, but not enough to self-destruct.

    A resistor resists current in the sense that current I = applied voltage V divided by resistance R. So for a given applied voltage, the greater the resistance, the smaller the current, the smaller the power dissipated, and the less glow obtains.
     
  6. Jan 21, 2015 #5

    Why doesn't copper self destruct, then, when used in other parts of the circuit?

    Thanks for your help!
     
  7. Jan 21, 2015 #6

    Quantum Defect

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    If you put enough current through it, you can make copper self destruct in the same way. It is scary.
     
  8. Jan 21, 2015 #7

    rude man

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    Because the voltage across the copper wiring is very small. Take a 100W light bulb, for example. 99% of the 115V will be across the light bulb, and only 1% might be across the copper wiring.Reason: the wire R is much less than the R of the bulb. So the power dissipation of the bulb will be 100 x 0.99 = 99W but the wiring will only see 1% of 100W = 1W which is not nearly enough power to melt the copper wire. Still, household wiring is fairly thick so even several 1000W loads will not produce enough wire heat to severany wire.
     
  9. Jan 22, 2015 #8
    So what I'm understanding is a filament needs much lower resistance than the wires. Why, then, is a filament considered a "resistor" when its added to the circuit. It seems like wires should technically be resistors as well--even more so than loads.
     
  10. Jan 22, 2015 #9

    rude man

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    I'm afraid you misunderstood and misread. I stated that filament resistance is, and must be, much greater than wire resistance. To wit:
    Reason: the wire R is much less than the R of the bulb.
     
  11. Jan 22, 2015 #10

    haruspex

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    Think about is what is the same for the filament and the wires in the circuit: it's the current, not the voltage. You cannot easily compare the power dissipation in them using P=V2/R because they are being subjected to different voltages. So instead use P=I2R, where I is the same for both because they are in series. Now you can see that the higher resistor dissipates the greater power.
     
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