Direct Current Circuits

In the above 3 pictures Professor Mason showed up what would happen when a switch was changed in different electrical setups. In the top setup the switch was initially off and Professor Mason turned it on. We thought that the two bulbs would remain unchanged and that the center one would turn off since there was 0 potential energy in the center of the setup. In the bottom setup we see that both bulbs stayed the same when the switch was turned on since the same amount of current will reach both bulbs. 

In the top two pictures we first set up batteries and bulbs to show which setups would produce the brightest bulbs and which setups would produce the dimmest bulbs. We see those results on the top left table. The symbols on the left were the options we had when placing the dial on the multimeter, we used Direct Current. We also found out that brightness depends on current, voltage, and power. The battery, wire, and resistor setups were used to determine the answers in the picture below.

We used the 2 battery setups in the 2 prior pictures to determine these answers. For the first setup we placed the resistors in series and found out that the voltage would split but the current would remain the same. In the second setup we placed the resistors parallel to each other and found the opposite to be true, the voltage would remain the same and the current would split.  

In the picture above we were able to calculate resistance based on the colors on a resistor and also with the use of the multimeter. We found that the color system was within it's margin of error.

In the two pictures above we calculated total resistance over parallel and series. For series we simply add the resistances. In a parallel setup though we must go with 1/R for each, add them together, and take the inverse of that to find the total resistance.

This picture shows us the loop rule, the current rule, and how to find the total by summing it and equating it to zero.

Electric Potential

He we showed multiple ways to derive electric potential and an observation location, point P. First we do it when P is at a distance X from the center of the ring, then we move P to a distance of X from the top of the ring. We knew that V = kq/r so with this we found dV = kdq/r, and in this situation r is equal to sqr(x^2 + a^2). For the second point we found that cos(theta) was equal to x/r and with this we determined that V = cos(theta)kq/x.

The above picture shows how we calculated the electric potential yet another way, with the equation shown on the left, integral from a to x of E(electric field) dot change in position (ds). E is only from the x component in this so we take Ex to equal kq/r * cos(theta) (r is equal to sqr(x^2+a^2) and we use cos(theta) since we only care about the x value in this). We see yet again that the potential is equal to kq/sqr(x^2+a^2).

 

We worked out what the net electric potential was from a rod to an observation point which was 15 cm above it.

In this excel file we first found the Vnet when there were 20 segments on a ring given the x and a values (x is the distance from the ring and a is the radius of the ring). R, the radius, is the square root of the sum of x squared and a squared. dQ was just the given q divided by 20. To find dV we multiplied the k and dq values and divided by R, then we multiplied by 20 to find Vnet of the ring. The second part of the excel file shows the net Q value and dV values from an observation point that was 15 cm away from a rod. The Y value is the same in each case but the X values varied.

For this lab we used a multimeter to find the difference in potential from a negative charge (ground) to another spot on the black conductive paper. We used a voltage regulator to set the potential of the positive and negative pins at 15 Volts. Harrison placed the black pin on the ground thumbtack and used the red pin on various points on the paper and Mario read the measurements while I recorded. We repeated this process 9 times to the left of the ground pin and 6 times to the right of the ground pin. In the lab book we calculated Work with the equation W = Vq. 

Python Activity: Determine the Electric Potential of the 3 charge arrangement you did in class.

Python Challenges:

Challenge 1: Electric Potential on a ring around an arrangement of point charges.

Original arrangement of charges completed in class.

Picture of the loop used to create the circle and the labels for the points on the ring.

Electric Potential on a ring that surrounds the arrangement of charges.

How this was achieved: I created a while loop that would first create an orange sphere as a visual representation of the location in which we were measuring the electric potential, similar to how we did it in class. Then I repeated the steps we did in class of find the potential of each charge, then adding them together.  The loop would run based on a theta variable, which was the angle starting from 0 going to 2*pi. The radius of the circle used was 5 units. I was able to set each location sphere in rectangular form : (R*cosTHETA, R*sinTHETA, 0). The increment of pi/18 was used to create a solid circle. The only mistake I made was I forgot that this was based on floating point, and I could not use the syntax: while theta != 2pi. To fix this I just changed it to: while theta < 2pi.

Challenge 2: Electric Potential Between 2 uniformly charged Rods and 3 Point charges at 4 observation locations.

For this challenge I decided to create each rod inside of a loop. Each point in each rod had a charge of 1e-9 for the left and -1e-9 for the right. I had to calculate the potential for each observation point and add it to a summation variable for each spot that I would print out at the end of the loop. I decremented the loop by 0.2 to ensure I had plenty of spots to emulate a rod.

Electric Potential Energy and Work

The two pictures above show the arrangement needed to ensure the brightest light from the bulbs. The bulbs each are connected to one end of the battery arrangement, and the other end of the respective bulb holders are touching with 1 cable running to the other end of the battery. In this arrangement the bulbs are getting the required energy at the same time rather than one getting more than another, which would happen in a linear arrangement where one bulb gets most of the energy first.

This arrangement shows the least brightest light. The battery are placed parallel to each other and one bulb receives the little bit of energy before the other bulb, resulting in one bulb being barely lit and the other not being lit at all.

The picture above is our illustration of the bulb and battery arrangements shown above. The top pictures are Mario's artistic representations while the bottom two pictures are more scientific representations of the bulbs and batteries. The batteries are represented by 2 parallel lines, the shorter being negative and the larger being positive and the bulbs are represented by swirls.

This is an experiment Professor Mason did in class in which he heated up water with a cup of noodles heater (water heater) that was hooked up to a voltage regulator.

This is the graph of temperature vs. time of the experiment mentioned above.

He doubled the voltage and repeated the experiment. As seen doubling the voltage more than doubled the temperature meaning that there are some other factors involved which accounted for the sharp rise in slope.

The picture above shows our attempt at figuring out why the slow in the above experiment jumped 7 times higher. We determined that the change in resistance was so negligible that we could consider it constant. So because of that we could determine that doubling the voltage would double the current, meaning that the power supplied would quadruple. In class we showed a near 7 time jump in slope, so there was probably a mistake in the experiment since the change should've been closer to 4 times larger.

In the above two pictures we calculated the amount of work needed to get the cart from point to point C. We determined that any path taken would result in the same amount of work.

We also symbolically showed the work needed to get from one point to another in an electric field. With distance D, A had the least amount of work since it was parallel to the field, B had the most since it was perpendicular to the field and C was in between since it had an angle less than 90 degrees to the field.

The picture on the right shows that a circle would have equal potential lines when a is going out at all possible angles from a central location. On the left we derive the formula for a change in voltage from an infinitely large distance (which would end up being zero) to a set distance r.

The picture above illustrates what we though the code provided by Professor Mason would produce.

The picture above shows my arrangement of charges (blue, red, yellow) and 3 locations (green spheres) where we observed what the electric potential would be. This was achieved by multiplying each charge by 1/4*pi*epsilon not and dividing that by the distance between the observation location and the actual charge. The three were then added together to find the total electric potential at each location.

This work shows that the net electric potential was zero for 2 equal but opposite charges.

The picture above shows what we though the difference between plagiarism and collaboration was. Professor Mason brought this up while talking about labs to ensure we knew what not to do when creating our blogs.

Current and Resistance

In the picture above we light the bulb by ensuring that the cable is touching one end of the battery and the side of the bulb and that the bulb's bottom is touching the other end of the battery. We could have also lit the bulb by touching the cable to the bottom of the bulb and placing the side of the bulb on the battery. This works because the charge goes from the battery into the bulb and then flows out back into the battery. This keeps the flow of energy going.

We show the same thing we did above but with 2 batteries.

This is an electroscope. You touch the top bulb with an object and the two plates inside the box will move if the object has a charge.

In the video above Professor Mason shows us how the electroscope works. He introduces a charge to the rod by rubbing it with animal fur, then touches the electroscope and we see the plates move, confirming that the rod has a charge.

The picture above illustrated 2 methods where the bulb could be lit by a battery and 2 methods where the bulb would not be lit. The right most picture was something we accidentally did at first, without the fire of course. The bottom picture shows how to make the bulb light with 2 batteries.

We assumed that nothing would happen if we connected the positive end of the battery to the electroscope and we were correct. We also explain why a close circuit connection is required to light the bulb. Electrons flow into the bulb but also flow out, ensuring that it does not get full.

In the above picture we use a 4 amp bulb and battery with an old style multimeter and we illustrate the same thing we did above by creating a close circuit connection in which the bulb could light. We wanted to determine if flow of energy outside of the bulb would match flow of energy to the bulb, so we tested it at both ends, and in both situations the meter read 4 amps, meaning that they were equal, there was no energy lost.

We assumed that the charge of an electron, cross sectional area, drift velocity, and number density were required to find the charge. In the picture above we were able to find drift velocity with some givens.

Professor Mason adds a current to a wire typically found in a toaster. We wanted to see if there was any relationship between current(Amps) and potential energy (volts).

The picture above is the graph of the experiment Professor Mason performed in class. There is a linear relationship between current and potential energy.

Professor Mason also tried the same thing with the light bulb, but unfortunately the bulb kept blowing.

This is what we assumed the graph would look like and we were correct. We also assumed that the opposite relationship would be linear, which was also correct.

We determined that if the diameter of a cable goes up the resistance goes down. Resistance also goes up with length. We also determined that material used also factors into how much resistance there is.