# Temperature and Resistance

1. Mar 7, 2009

### roxychc43

1. The problem statement, all variables and given/known data
We did a project in our physics class that involved a power source, a light bulb, connecting wires, a voltmeter, and an ammeter. All items were connected in a series. We were asked to raise the voltage on the power supply and measure the current across the light bulb after increasing the voltage. We were then asked to determine the mathematical expression that describes the change in resistance as a function of temperature (but we didn't measure temperature).

What is the mathematical expression?
What is the Theoretical expression the describes that relationship?
Determine the experimental resistivity of the bulb filament and compare the theoretical resistivity to your measured value.

I don't understand how we are supposed to get that "measured value" and then compare it to the theoretical resistivity... Can you explain how to get all of these calculations?
2. Relevant equations

3. The attempt at a solution

2. Mar 7, 2009

### Nyme

First: be careful on whether writing "resistance" or "resistivity", since these two are different quantities. Resistance is measured in ohms, and concerns a whole electrical component. Resistivity is the resistance of a material. In the following equations you can replace $$\rho$$ (resistivity) with $$R$$ (resistance) if needed.

Temperature dependence

The change of resistivity depends on temperature change, original resistivity and a temperature coefficient. The formula is very similar to the thermal expansion formula.

$$\rho_{new} = \rho_{original} (1 + \alpha \Delta T )$$

where
$$\rho_{original}$$ is the resistivity at 20 celsius
$$\rho_{new}$$ is the resistivity after temperature change
$$\alpha$$ is the temperature coefficient (find a table in http://en.wikipedia.org/wiki/Electrical_resistivity#Table_of_resistivities" or another source)
$$\Delta T$$ is the change in temperature

You can obtain the change in resistivity by subtracting the $$\rho_{original}$$ from that, yielding

$$\Delta \rho =\alpha \rho_{original} \Delta T$$

Determining experimental resistance/resistivity

Since you had a voltmeter and an ammeter, you know the voltage and the current of the system. You can calculate the resistance of the system simply.

$$R = V / I$$

To calculate the resistivity of the filament, you need to know its length and cross-sectional area.

$$\rho = R A / l$$

where
$$\rho$$ is resistivity in ohm meters
$$A$$ is the cross-sectional area (i.e. the "thickness" of the filament) in square meters
$$l$$ is the length of the filament in meters

If the filament is made of tungsten, the theoretical resistivity at 20 celsius is 5.60 x 10-8 ohm meters, which you can obtain from a table (like the Wikipedia one I mentioned before).

If it was totally resistance you were talking about, I would assume by theoretical resistance they mean the one you would get by combining the power = voltage x current and resistance = voltage / current equations, yielding

$$R = V^{2} / P$$ (try it yourself)

You know the voltage for sure (by controlling through power supply) and the power (i.e. you should know if it is 60 W lightbulb or something else). As such, you get a theoretical value. But because the high temperature of the filament actually increases the resistivity and thus resistance, you will notice that the experimental resistance is higher than the theoretical one. You find the experimental value by using resistance = voltage / current with the values you got from the voltmeter and the ammeter. You can see more consequences of the increase in resistance by looking at the equations. Just remember that your voltage is constant because you select that from the power supply.

I hope I made some sense.

Last edited by a moderator: Apr 24, 2017
3. Mar 7, 2009

### wywong

The relationship between temperature and radiation power is given by the Stefan–Boltzmann law: http://en.wikipedia.org/wiki/Stefan-Boltzmann_law. Since the filament temperature is many times the room temperature, we can ignore the room temperature and assume the following equation:

Power = $$VI \propto T^4$$

So in theory the quantity $$(VI)^\frac{1}{4}$$ should have a linear relationship with $$T$$ which in turn has a linear relationship with $$R=\frac{V}{I}$$.