Day 16: RC Circuits

Quantitative Measurements on an RC System

For the following experiment we were asked to create an RC circuit with a resistor, power supply, and capacitor in series (this had to be done otherwise the charge time was too rapid to see). We hooked up the capacitor to logger pro using a voltage measuring lead. However, before this we made a few predictions. Primarily that the relationship between Potential (Voltage) and time would be inversely related as seen below. After logging all of our data we had the following graph. 

By looking solely at the Discharge portion (pink line) we came up with a few conclusions. The first conclusion we came up with was that B in the natural exponential equation should be fairly close to 0. We also concluded that the A value should be our close to our original 4.5 Voltage (from the power supply). Our final conclusion was that C was our charge time which should have been the same for both Discharge and Charge. Our ideas are summarized below.

After finding our experimental value for the charge time, C=0.003620 sec., we theoretically found what the charge time of circuit should have been. In order to do this we first used a multimeter to determine the resistance of our resistor, R=2.15 kΩ. We next used the relationship we discovered: t=1/RC where R=Resistance and C=Capacitance to solve for our theoretical value of charge time. What we found is that theoretically the charge time should have been .00465 sec. That means there was a 22% error. This large error could be due to the fact that in our theoretical value we took the capacitors value straight off the labeling and it may have in fact been lower or higher. 

Day 15: Capacitors

Homemade Capacitors

For this experiment we used two pieces of aluminum foil and an old physics book to create a capacitor. The pages of the book acted as the separation distance between the "two plates" (i.e. sheets of aluminum foil). By connecting the two sheets to a multimeter we were able to measure the capacitance. As area remained the same the only changing factors were d (distance) and C (capacitance). We took three different measurements for 1, 10, and 20 pages. We graphed our results below and found there to be an inverse relationship between distance and capacitance. This means that as the distance between the plates increases the capacitance decreases. We also solved for Kappa for the first trial and got a value of 0.367, this is off by a factor of 10 from the accepted 3.5 kappa value for paper. 

Capacitors in Series and Parallel

For this experiment we were given two capacitors. We first measured the capacitance of each capacitor and noted it below. We next set up the capacitors in series and in parallel as seen above. After measuring the capacitance we came up with two conclusions. The first conclusion was that in parallel the capacitance of the two capacitors was simply the sum of their respective capacitance. This means that in parallel Ctot=C1+C2. The second conclusion we came up with is that in series capacitors add up inversely: 1/Ctot=1/C1+1/C2. This can be seen below.

Day 14: Series Circuits and Resistors

Voltage

In the following experiment we set-up the required series and found in both cases that the voltage across the resistors (batteries) in series added up to the voltage of the source.  As for the current, it remained constant in each area tested.

Parallel Circuits

When we set up the bulbs in parallel currents 2 and 3 added up to current 1. The voltage in this case remained constant along each resistor (light bulb). 

Decoding and Measuring Resistors

Given the above pictured resistors we determined the coded resistance using the provided key. We then measured the true resistance and placed our results in the data table above. Our results concluded that the carbon resistor had the best result. It's discrepancy between the coded and measured resistance was less than 1%. The blue resistor that was supposed to have a much better tolerance rating in fact proved to perform much worse with a 4.76% discrepancy. Overall, the carbon resistor is a much better purchase than the blue resistor. 

The Equivalent Resistance for a Network

Given the above circuit we were asked to determine the total resistance of the system. We first calculated each section and estimated that our theoretical total resistance should be 52.17 Ω. We next set up the desired circuit as follows:

After measuring the resistance we got an experimental value of 51.9 Ω.

Applying the Loop Rule Several Times

Given the following circuit we developed 3 equations to help solve for the current across the noted resistors. Using these equations we were able to use substitution and elimination to solve for each current. 

We first substituted equation number 3 (current eq) into equation number 1 (top eq with ε1). This resulted in two equations with two unknowns (I2 and I3). Using the elimination method we solved for I3 and then plugged that back into one of the two new equations. After that we solved for I2 and then using the current equation solved for I1. However, the values shown above are incorrect after placing them in a matrix in teh calculator we doscovered that I1=1.14A, I2=0.999A, and I3=0.14A.

We next set up the circuit above onto a breadboard to verify our calculations. We measured the currents across the resistors and found the values to be extremely close to the theoretical. Our values were as follows: I1=1.151A, I2=1.005A, and I3=0.148A

Day 13: Potential Difference and Continuous Charge Distribution

Electric Potential Lab/Activity

In class we set-up an experiment to determine the electric potential between two points due to a power supply (electric potential difference). We set the power supply to 15 V and using a power meter tested two points on the sheet. First we tested points on the two metallic paint marks and got a reading of 15.54 V. When we measured the potential difference between two points on the lower and higher potential conductor we got a value of 0.

These measurements tell us that there is no work required to to move a charge along a conductor. Because W=q∆V since ∆V=0 then the work equals zero.

Next we clipped one end of the power meter to the painted metallic circle and took measurements of electric potential X cm away, in increments of 1 cm. We also calculated the ratio between each pair of points. Our results are listed above. 

Next we calculated the work to move a given charge. The possible scenarios were:

(a) x=0cm to x=3cm

(b) x=4cm to x=6cm

(c) x=5cm to x=2cm

Our work for two of the given charges is shown below. All the answers can be found in the table above. 

Next we took our Potential and Position values and plotted the results in excel. The graph is shown below. 

Based on our data the direction of the electric field is in the -x direction. As the distance between the points continues down the +x direction the electric potential increases. This means it must be moving against the electric field which would thus have to pint in the -x direction. 

Day 12: Electric Potential

Quiz

At the beginning of class we had a group quiz and were asked to create a circuit in which the two bulbs were as dim as possible using 5 wires, 2 light bulbs, and 2 1.5V Batteries. In order to get the bulbs as dim as possible with the given equipment we decided to line the bulbs up in series and the batteries in parallel. The bulbs were so dim only a small hint of orange could be seen in the filament inside the bulb. 

Change in Temperature of Water using Power

In class there was an experiment set-up in which 200 grams of water was placed in a cup along with a coiled wired heater. The length of coiled wire in the heater was 42 cm. It was attached to a 4.5 V battery and placed in the cup of water for a total of 10 mins. A temperature probe placed inside of the cup recorded the temperature as it changed in logger pro. During the 10 mins the "heater" was submerged in water we calculated what the temperature change of the water should be given the length of the wire, voltage of battery, mass of water, and any necessary constants. Our calculations are shown below.

After we calculated the resistivity (R) of the wire (6.5 Ω) we determined the current (I) for two cases. The reason this was done was because depending on which resistivity constant was used the result for R may have been slightly different. Using this difference we found an uncertainty in current. We then proceeded to calculate the power and found its uncertainty according to the difference between the two cases. At this point we had all the necessary values to calculate ∆T. Using Q=mc∆T and Q=P⋅t we found that the theoretical temperature change of the water should be 2.22°C.

In order to find the uncertainty we propagated the equation for ∆T according to the uncertainties we knew. Giving us a final answer of ∆T=2.22±0.57°C.

The same exact experiment was run again, however this time our voltage was doubled to 9.0V and we were asked to follow the procedure to calculate ∆T. After all the calculations were finished it was found that ∆T=8.90±2.38°C.

The above picture shows the temperature change results from logger pro for each experiment. The blue data set was the 9.0 V experiment and the red data set was the initial 4.5 V experiment. According to logger pro for the initial experiment the temperature started at about 23°C and concluded at about 25°C for a 2°C temperature change. The doubled voltage experiment started at 24.6°C and ended at 32.1°C for a 7.5°C temperature change. 

Contrary to what most may have expected the change in temperature of the system did not double, it actually almost quadrupled. The reason for this can be shown mathematically. 

When we look at what occurs to the current, power, and temperature change we see why ∆T did not double as the voltage was doubled. For both experiments R was the same, however for current the voltage doubled resulting in a value for current that also increased by a factor of 2 because current and voltage are proportional. When the new current is substituted into the power equation we see that the voltage here is actually 4 times that of the initial power calculation. The reason for this is that the current was doubled and the voltage was doubled. When these two values where multiplied to find power the power quadrupled. This value then carried into the ∆T equation thus quadrupling the temperature change. 

Day 11: Current, Resistance, and Voltage

Light Bulbs and Batteries

For this small experiment we were asked to find a way to make the light bulb light up with one piece of wire and a battery.  The top picture shows us successfully accomplishing this. Next we wanted to make the bulb shine twice as bright. To do this we just doubled the power supply and got the result shown in the bottom picture. 

Coils and Voltage/Current

For our first experiment we were asked to select a resistor and connect it to an ammeter and power supply, in which the voltage applied was known. We also used a power meter to determine the voltage at the resistor. For each trial we increased the voltage into the circuit using the power supply. We recorded our data below as well as that of the group across from us. 

The data above shows the results from the experiment. The group across from us had a much smaller resistor with more wire wrapped around it. But for the most part the voltage was consistent in both experiments. The current however was very different for each group.

We took the above data and plotted it in a Current vs. Voltage graph. The results were clear. Our data clearly fit a linear relationship, meaning that Voltage and Current are directly proportional  to each other. We also learned that the slope of this line was in fact the resistance of the resistor. For our group our resistance was about 28.272 while group #2's resistance was 17.745. This means that our groups resistor had greater resistance. The reason our groups resistor had much greater resistance is because it was a longer, therefore the electron had a much more difficult time traveling across it. 

 Copper vs. Nickel-Silver

For our next experiment we were given several different coils of varying diameters and lengths of wire wrapped around them. One coil was made of copper and the rest were made of nickel-silver. The goal for our experiment was to determine the resistance of each coil. Using a power meter we tested each coil to see what the resistance was and got the following data. After measuring the resistance we needed to correct it by determining the resistance in the power meter. When we did this we found that the power meter had a 1.9 Ω error.

 This data shows that as the length of wire became larger the resistance of the coil also increased. In order to show this relationship we plotted Length vs. Resistance.

This graph clearly shows a proportional relationship between the Length of the Wire and Resistance. As the length increased the resistance also increased. 

When we looked at the effect the Area had on the resistance we found out that the relationship was actually inversely proportional. This can clearly be seen when examining the two final NS 200 coils. They had the same length but different diameters. As the diameter of the coil was increased (same as increase in area) the resistance actually decreased. This results in an inverse relationship.

Day 10: Gauss's Law

ActivPhysics: 11.8 Gauss's Law

Question 1: Electric Field Outside a Charged Sphere
A solid sphere of 10-mm radius has +10 x 10^-10 C of electric charge distributed uniformly throughout the sphere. Use Gauss's law to determine the electric field caused by this charge at a distance of 15 mm from the center of the sphere. The answer is shown in the meter. If you have difficulties, the Advisor will help you in this first application of Gauss' law.

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Question 2: Electric Field Outside a Charged Spherical Shell:
A 10-mm radius spherical shell, like a basketball, has +10 x 10^-10 C of electric charge distributed uniformly on its surface. Use Gauss' law to determine the electric field caused by this charge at a distance of 15 mm from the center of the sphere. After your prediction, you can click the "Shell" Object type and set R = 15 mm and compare the answer with your prediction. Consult the Advisor if you have difficulties.

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Question 3: Electric field inside the charged spherical shell
Leave the Object type on "Shell." Use Gauss' law to predict the value of the electric field inside the spherical shell. If you want, you can calculate the magnitude of the electric field at R = 5 mm from the center of the 10-mm radius spherical shell. After your prediction, move the R slider to 5 mm to check your prediction. If you have difficulty understanding the result, consult the Advisor.

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Question 4: Electric Field Inside a Charged Solid Sphere
Predict the value of the electric field 5.0 mm from the center of a 10-mm radius solid sphere. The sphere is uniformly charged with +10 x 10^-10 C of electric charge. (a) First, determine the electric charge inside a 5.0 mm radius Gaussian surface. (b) Then, use Gauss' law to determine the electric field 5.0 mm from the center of the charged sphere. When finished, you can check your result with the simulation. Consult the Advisor if you have difficulties.

Calculated Answers for 1 to 4:

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Question 5: Electric Field Inside a Charged Solid Sphere
Show that the electric field INSIDE the solid uniformly charged sphere varies as:

E = [k Q/(R_{charged sphere})²](R/R_{charged sphere})

where R_{charged sphere} is the radius of the charged sphere (10 mm in the simulation), Q is the total charge on the sphere (adjustable with the Q slider), and R is the distance from the center of the sphere at the position where the electric field is being calculated.