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The Hot Resistance Of the filament

  1. Apr 24, 2006 #1
    Hey so im stuck on a 2part question, it goes as follows:

    A tungstan filament in a light bulb is rated 75w for a 230v supply. The resistance of the filament at 20*C (*C means degrees celcius) is 68 ohm.
    Assume a constant temperature coefficient of 0.005.

    Calculate:
    A) The hot resistance of the filament

    Ok so I have no idea what temperature to use for "hot" so I used 100*C (I mean thats hot!:tongue2: )

    Then used the equation R1=R0 (1+0.005x100) - Where
    R1= Resistance at 100*C
    R0= Resistance at 20*C
    0.005= Temperature Coefficient
    100= 100*C "Hot" temperature.

    If you work this out it gives you a resistance of 102ohm @ 100*C.
    I think that bit is right but is the 100*C right?


    I have a second part to this so I'll put it in a different colour to help....

    B) The operating temperature of the filament?

    Well for this one I was going to use the formula (T1= R1/R0-1 All Divided By 0.005)

    But turns out that gives you an operating temp of 100*C .... yea im lost and have googled looking for operating temperature formulas...Help!



    Thanks People!!
     
  2. jcsd
  3. Apr 24, 2006 #2

    Kurdt

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    You have two equations with two unknowns. Rather than try to solve one at a time try solving them both at the same time.
     
  4. Apr 24, 2006 #3
    Ive got lost again - But hey I got a new resistance this time - The "Hot temperature resistance" maybe?

    I used the R1= 68(1+0.005x20) = 74.8

    But I still cant figure out how to find the operating temperature with that last part solved.

    A formula for this would help LOTS!
    ....Hints are just plain annoying at this time of night (NZ 12:55am)
    And yes I have left this assignment to the last minute...

    Thanks!
     
  5. Apr 24, 2006 #4

    Kurdt

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    Fair enough I can say nothing about whether the formulas are right because its so long since I've done anything with electrical circuits but assuming they are correct the mathematical way of solving them is by simultaneous equations. Try solving this:

    R1-T1 = 1/k, where k=0.005 alongside this,

    T1/R1 = 1/(k*R0-k)

    Unfortunately i can only give hints as part of the rules of this forum, and with good reason because if you don't get there by yourself (albeit with a little help) then you won't have learned anything.
     
  6. Apr 24, 2006 #5
    The lamp is rated 75 watts at 230 volts. Do you know how to calculate the resistance of the lamp with these values (i.e., the hot resistance of the filament) ?
     
    Last edited: Apr 24, 2006
  7. Apr 24, 2006 #6

    Hootenanny

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    From what I can remember the resistance of a conductor is given as;

    [tex]R = R_{0}(1 + \alpha \cdot \Delta T)[/tex]

    Where [itex]\alpha = k[/itex] in kurdt's example.

    ~H
     
  8. Apr 24, 2006 #7

    Kurdt

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    If that is the case Hoot, I hope the second equation involves a delta T too!
     
  9. Apr 24, 2006 #8

    Hootenanny

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    Perhaps hammie's suggestion of using;

    [tex]P = I^2 R[/tex]

    to calculate the running resistance would be a more logical step, in light of the lack of a delta T in the second equality.

    ~H
     
  10. Apr 24, 2006 #9
    I was thinking of

    [tex]P = V^2 /R[/tex]
     
  11. Apr 24, 2006 #10

    Hootenanny

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    I apologise, I thought the current was given :blushing: . If we apply ohm's law to my original equation to eliminate I, we can obtain the above equation.

    ~H
     
  12. Apr 24, 2006 #11

    Gokul43201

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    That's a thing you can NOT do! You are not given the temperature hot, and just choosing a random value is not the way to solve the problem. You need to look carefully at ALL the data provided and see how you can arrive at the required quantity from known things.

    And just to let you know how wrong your assumed value (100 deg C) is, let me tell you that the typical filament temperature is about 2500 deg C. That's an error of about 3000% in the temperature change !

    That is an incorrect equation. The correct equation is given in post #6. The temperature in the equation is not the "hot" temperature, but the temperature change. In any case, you do not use this equation in part A.

    No, you should work this out using the method described by hammie.

    Well, you assumed an operating temperature of 100 deg C to calculate the first part. Naturally, if you use the same equation as you used there, you will merely extract that same number.

    If anything, this second part should have told you that you do not want to assume a value of the temperature in part A. Nevertheless, if you correctly solve part A as suggested you can use the equation in post #6 to find the temperature change.
     
    Last edited: Apr 24, 2006
  13. Apr 24, 2006 #12

    Kurdt

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    Glad to see I was of zero assistance there :blushing:. I'll leave my patheic knowledge of electrical circuits on the sidelines in future :biggrin:.
     
  14. Apr 24, 2006 #13

    Hootenanny

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    I may come with you. Nevermind, we can eat all the half time oranges!:devil:

    ~H
     
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