Resistivity Coursework problem

In summary, the conversation is about a student asking for help with their physics coursework on measuring the resistivity of Constantan. They provide their readings of Current = 0.33A and Voltage = 2.13V and ask for assistance in finding any errors. They also mention using multimeters with specific settings and questioning if this will affect the readings. Another person suggests verifying the current and voltage data to find any discrepancies.
  • #1
emikkime
3
0
OK doing some physics coursework at the moment where I have to measure the resistivity of Constantan. Anyway, for those who don't know, the resistivity of constantan is 49x10^-8.

Anyway the formula for those who don't know for resistivity is...

RA/l

So anyway at length of 100cm, with a diameter of 0.464mm, I got these readings in a circuit. Current = 0.33, Voltage = 2.13, which gives the resistance as 6.45. (I did repeat readings to get an average but this is just one reading as an example.)

Anyway after finding the Cross-sectional area of the constantan by doing A=pi*r^2. Now to do this, I halve the diameter to get 0.232mm. This is the same as 0.000232m (metres). So after squaring this I get, 5.3824^-8. After multiplying this by pi, I get the cross-sectional area to be 1.69...^-7. Multiply the cross-sectional area by the resistance (6.45) which gives 1.09...^-6. Finally dividing by the length (1metre), which leaves the same answer.

I know this is a lot to read, and I'm asking a lot for help, but please, can you find what I've done wrong, or anything?

I really appreciate all help :smile:
 
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  • #2
Current = 0.33, Voltage = 2.13,

What are the units? check the units
 
  • #3
Not quite sure now. Using multimeters at college, I think the ampmeter was set to "10A" and the voltmeter on "20V"

How would this affect the readings?

Would this mean that I'd have to halve my readings of p.d. before putting them into the formula R=V/I to work out Resistance. If yes, then that would make sense, as I did a quick test and using a halved p.d. gives a final value of resistivity very close to the real value.

Thanks in advance.
 
Last edited:
  • #4
If you use the given value of [itex] \rho [/itex] and the length and area of your wire to compute an expected resistance you will find a signifiant difference. You need to verify your current and voltage data.
 
  • #5
So you're guessing that it's my current/voltage readings which are incorrect somehow?
 

Related to Resistivity Coursework problem

1. What is resistivity?

Resistivity is a measure of a material's tendency to resist the flow of electric current. It is represented by the Greek letter rho (ρ) and is measured in ohm-meters (Ω⋅m).

2. What is the formula for resistivity?

The formula for resistivity is ρ = R x A / L, where ρ is resistivity, R is resistance, A is cross-sectional area, and L is length.

3. How is resistivity related to conductivity?

Resistivity and conductivity are inversely proportional to each other. This means that materials with high resistivity have low conductivity, and vice versa.

4. What factors affect resistivity?

The resistivity of a material is affected by factors such as temperature, impurities, and cross-sectional area. Higher temperatures and impurities increase resistivity, while a larger cross-sectional area decreases resistivity.

5. How can resistivity be measured in a laboratory setting?

Resistivity can be measured in a laboratory using a device called a multimeter. The multimeter measures the resistance of a material and with the given dimensions, the resistivity can be calculated using the formula ρ = R x A / L.

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