A certain resistor has a resistance of 1.484 omega at 20 degrees celsius and a resistance of 1.512 omega at 34.0 degrees celsius. What is its temperature coefficient of resistivity?
ok..I used a temperature equation but I can't get the right answer. so maybe I am doing something wrong
I will be happy to check your work.
ok I used 1.512 = 1.484[1+ alpha (34-20)] and the answer in the back of my book isn't the answer I am getting
That is the right equation.
Then....
alpha= (1.512-1.484)/(14*1.484)
check my thinking.
ok yea! I got it now..I just made a stupid mistake with my algebra! Thank you!
Can I ask what's the answer? Thankyou :)
To calculate the temperature coefficient of resistivity, we use the formula:
α = ΔR / (R₀ * ΔT)
Where:
- α is the temperature coefficient of resistivity.
- ΔR is the change in resistance (R₂ - R₁).
- R₀ is the initial resistance.
- ΔT is the change in temperature (T₂ - T₁).
In this case, we need to find the temperature coefficient of resistivity using the given resistance values at two different temperatures (20°C and 34.0°C).
ΔR = 1.512 Ω - 1.484 Ω = 0.028 Ω (change in resistance)
R₀ = 1.484 Ω (initial resistance)
ΔT = 34.0°C - 20°C = 14.0°C (change in temperature)
Now, let's plug these values into the formula to calculate the temperature coefficient of resistivity:
α = 0.028 Ω / (1.484 Ω * 14.0°C)
Simplifying further:
α = 0.028 Ω / 20.776 Ω°C
Finally:
α ≈ 0.00135 Ω/°C
Therefore, the temperature coefficient of resistivity for the given resistor is approximately 0.00135 Ω/°C.