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Calculating the heat in the circuit when adding resistances

  1. Jan 30, 2013 #1
    Hi everyone! I've got to problems I need your help with, I would really appreciate it if you could help me.

    1. The problem statement, all variables and given/known data

    A conductor made out of copper (Cu) has a resistance of 4Ω in 20° Celcius. When current flows in the conductor, it's temperature rises to 220°C. In parallel with the resistance, we connect a Rx resistance. How should the value of the Rx resistance be, compared to the original resistance, so that a greater heat can be released in the Rx? For Copper: α=0.0039, ρo=1.7x10-8

    2. Relevant equations
    Q=I2Rt; Q=(U2/R)t




    3. The attempt at a solution

    So since the resistances are connected in parallel, I tried using the first one but I don't have the time, do I have to build the solution in form of a ratio?? And also, why do we need the ρo of Copper if the R=Ro(1+αΔt) formula is used, in my opinion? Can anyone please help me with this?

    Thank you :)!
     
  2. jcsd
  3. Jan 30, 2013 #2

    haruspex

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    You don't need the time - just treat it as a rate (Q/t). Quite possibly you don't need ρo, but if so it will just cancel later. Please post your attempt.
     
  4. Jan 30, 2013 #3
    Ughh, thanks for your reply, although I think I'm stuck again...

    I separated R from the Q=I2Rt, which gives R=Q/(I2xt).
    I treat the two resistances as a rate (Rx/R), which is Rx=Q1/(I2xt) and R=Q/(I2xt) but here's where I'm stuck, the problem doesn't ask for a specific heat, it just says: "So that the heat that would be released in Rx can be greater"... What do I do here? Do I just continue solving the rate, going to another rate of R=Ro(1+αΔt)? Wouldn't that give the same in both parts of the refraction??
     
  5. Jan 30, 2013 #4

    haruspex

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    The answer will be in the form of an inequality. Rather than using <, >, and running the risk of getting them reversed somewhere, probably simplest to calculate the resistance that would make them equal, then figure out how to write the inequality.
     
  6. Jan 30, 2013 #5
    Should I calculate the resistance from R=Ro(1+αΔt) or R=Q/(I2xt)?

    Either way the symbols just cancel themselves and I am stuck with unknown ones again...
    Q/(I2xt)=Q'/(I2xt). The current and time are constant, right? So they cancel themselves and I end up with Q=Q'... Same thing goes for the other formula...

    So if I do it in form of an equation from the formula of heat: Q=I2xRxt, it will be:
    Q'>Q -> I2xR'xt>I2xRxt, the I and t cancel themselves, so R'>R, I'm left with that, is that really it, to get a higher heat I need a higher resistance, so Rx has to be higher? It just sounds too easy :S..
     
  7. Jan 30, 2013 #6

    haruspex

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    You don't know the current or the power, so R=Ro(1+αΔt) is the only option (for the hot copper resistor).
     
  8. Jan 30, 2013 #7
    Okay, I understand. Is this okay?

    Ro'(1+αΔt)>Ro(1+αΔt)
    Ro'x1,78>4x1,78
    Ro'>4

    I don't know how accurate I am, but the problem is asking me for the Rx resistance, which I haven't found, right?...
     
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