Temperature coefficient of resistance

In summary, the conversation discusses a homework problem involving calculating the coefficient of resistance (a) using the known variables of resistance (R), temperature (T), and the intercept (c). The attempt at a solution involves finding the gradient and intercept, and then using them to solve for a. However, a small mistake is made and the conversation ends with the realization that the temperature must be converted to kelvin before calculating the coefficient.
  • #1
Philip Wong
95
0

Homework Statement


ok we did a temperature coefficient of resistance at class, we need to calculate the coefficient of resistance (a) ourselves.
known variable:
R = resistance
ln R = log of resistance
T = tempreature
a = coefficient of resistance
c = the intercept


Homework Equations


ln R = aT+c
i.e. R= k exp(aT), where k=exp(c) is a constant.


The Attempt at a Solution


at first I thought I could calculate a as if it is gradient, and therefore worked out the intercept (c). But since a is the coefficient of resistance it didn't make sense to me to do it this way. So I'm lost
Anyways this is my working:
gradient = delta ln R/ delta T
= (lnR_1-lnR_2/T_1-T_2) = -.912185/44
= -0.02682897
therefore intercept (c) = -.912185 - (-.02682897 * 44)
= 10.8925618

put it back to the equation
lnR = a*T+10.893
given T = 64, its correspond ln R = 4.957 (we measured these ourselves, assume correct)

so:
4.957 = a*64+10.893
-5.936=a*64
a=-5.936/64
a=-0.0927

put everything back to original equation:
ln R = -0.0927 T + 10.893

is this even correct??
thanks
 
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  • #2
if my above working was correct, here is the second part of my question.
during the experiment we only measure the resistance every 2C drop (from 100C to 30C). I've plot a ln R vs T graph, how can a determine the a at temperature of 65C?
 
  • #3
arha! I was right in general, but had made a small mistake! I forgot to convert temperature into kelvin before I calculated the coefficient. when I convert it, everything looks more reasonable and correct!
 

1. What is the temperature coefficient of resistance?

The temperature coefficient of resistance is a measure of how much a material's resistance changes with a change in temperature. It is represented by the symbol α and is typically given in units of ohms per degree Celsius (Ω/°C) or ohms per degree Kelvin (Ω/K).

2. How is the temperature coefficient of resistance calculated?

The temperature coefficient of resistance is calculated by taking the ratio of the change in resistance (ΔR) to the change in temperature (ΔT) and multiplying it by the initial resistance (R0) and the reference temperature (T0). This can be expressed as α = (ΔR/R0)/(ΔT/T0).

3. What is the significance of the temperature coefficient of resistance?

The temperature coefficient of resistance is an important characteristic of a material and can affect its performance in electronic circuits. It can help predict how a material's resistance will change with temperature, allowing for proper selection and design of components that will operate reliably under different temperature conditions.

4. How does temperature coefficient of resistance vary among different materials?

The temperature coefficient of resistance varies among different materials and can be positive, negative, or close to zero. For example, metals typically have a positive temperature coefficient, meaning their resistance increases with temperature, while semiconductors can have a negative temperature coefficient, causing their resistance to decrease with temperature.

5. Can the temperature coefficient of resistance be constant for a material?

No, the temperature coefficient of resistance is not constant for a material and can vary depending on the temperature range. It can also be affected by impurities, crystal structure, and other factors. Therefore, it is important to consider the temperature coefficient of resistance over the operating range of a material to accurately predict its behavior in different temperature conditions.

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