Calculating the heat in the circuit when adding resistances

  • Thread starter Thread starter satycorn
  • Start date Start date
  • Tags Tags
    Circuit Heat
AI Thread Summary
The discussion revolves around calculating the resistance value of Rx needed to ensure it releases more heat than a copper conductor with a resistance of 4Ω at 220°C. Participants clarify that the heat equation can be simplified by treating the resistances as rates, leading to the conclusion that for Rx to release greater heat, its resistance must be higher than that of the original conductor. The formula R=Ro(1+αΔt) is emphasized as the appropriate method to determine the new resistance due to temperature changes. The conversation highlights the importance of understanding the relationship between resistance, heat, and temperature in parallel circuits. Ultimately, the conclusion is that a higher resistance for Rx is necessary to achieve the desired heat output.
satycorn
Messages
6
Reaction score
0
Hi everyone! I've got to problems I need your help with, I would really appreciate it if you could help me.

1. Homework Statement

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

Homework Equations


Q=I2Rt; Q=(U2/R)t




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 :)!
 
Physics news on Phys.org
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.
 
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??
 
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.
 
haruspex said:
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.

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..
 
satycorn said:
Should I calculate the resistance from R=Ro(1+αΔt) or R=Q/(I2xt)?
You don't know the current or the power, so R=Ro(1+αΔt) is the only option (for the hot copper resistor).
 
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?...
 
Back
Top