Day 12 - electric potential

We were required to make the light bulbs as dim as possible with the given equipment. Ours was the dimmest out of the entire class such that it could almost not be seen. I was not sure if this set up would work initially, but jeff had the same idea as I did and we went through with it. It should be noted that Bianca did not come up with this idea.







Change in temperature by electric heating element.

For this experiment we used a coil heater with a length of 42 centimeters. We attached it to a battery of 4.5 volts and let it run for 10 minutes. We needed to find the change in temperature based on the length of the heater, as well as the time and voltage, which is calculated below:

We found the resistance of the heater by multiplying the power by the length, and then dividing by the area. We then used the resistance to find the current based on Ohm's law. The relative uncertainty was found by seeing the deviation in the resulting power by plugging in two different values. The power was easy to solve because we knew both the voltage and current. 

We then proceeded to perform the same experiment with twice the voltage, and the graph is shown below. The red line is 4.5 V and the blue line is 9V. For the 4.5V test the temperature changed from 23 to 25 celsius. For the 9V test it changed from 24.6 to 32.1, which was much more significant. The rate at which the heat increased was more than double, even though the value for the voltage was only doubled.

It can be proven theoretically why the temperature increased nearly exponentially This is because:

delta T = Pt/mc = (V^2*t)/(RMC)

We substituted the definition of power being equal to V^2 divided by R, and then when you plug in a value of V that is double than a previous initial value, it is increased more than two-fold because its value is squared after it has been doubled in respect to the temperature change equation.