RC Circuits and Capacitors (4/30/2015)

Charge Buildup and Decay in Capacitors

We started by creating a circuit consisting of a 4.5 V battery, a #14 bulb, and a 0.47 F capacitor. When we connected this circuit as shown above on the left, the bulb lit up. What we then had to see, was if the bulb would still light if no battery was included in the circuit. The answer is yes, and can be seen in the picture on the top left. The only issue was that the brightness died down over time. What is going on is that when the whole circuit is connected, the capacitor is storing energy, and after the battery is disconnected there is still energy in the battery that is used to light the bulb until the energy is used up.

This sketch depicts the brightness over time of the bulb when it is connected to only the capacitor (no battery involved).

We also created a circuit with parallel capacitors to observe what happens in this case. What we realized is that the bulb took nearly twice as long to discharge the energy stored in the capacitors than in the circuit with one capacitor. This makes sense considering our previously established notion that when capacitors are in parallel, capacitance adds. We had twice as much capacitance in this circuit based on the time it took for the energy to discharged.

A Capacitance Puzzle

For this activity, we hooked up two identical .47 capacitors to two separate circuits with different batteries (3V and 4.5V). The actual charges were 2.62V and 3.87V in our batteries. After charging the capacitors up, we were asked to predict what would happen if the capacitors were then hooked directly to each other. As shown in the quick calculation below, we predicted that our voltages would add u to 6.49 V.

Shown below are the capacitors connected to each other after being charged. What we discovered was that our voltages did not add up. Our final voltage ended up being smaller than our theoretical. This is most likely due to the fact that capacitors in series inversely add up. So the actual numbers would have looked like 1/2.62 + 1/3.87 = 1/0.64 = 1.56 V.

Quantitative Measurements on an RC System

Next, we worked on conducting a more quantitative study of our "RC" system. We did this by charging and discharging a capacitor while measuring the voltage over time using Logger Pro. The graph below shows our results, with the blue line representing the charging of the capacitor, and the green being the discharge. In the photos above you see the best fit equation up close, as well as the meaning of each variable. A and B are volts, while C is 1/time. This result has given us a mathematical relationship between voltage across a capacitor and the time that best describes the voltage change during capacitor discharge, which is V = Ae^(-C*t)+B. It should be noted that C was nearly identical in both readings, which means that the time a capacitor charges signifies how long it'll take for it to discharge as well.

The Exponential Nature of Charging or Discharging a Capacitor

From the sketched graphs above, we can describe a relationship between Voltage, Brightness, and Current. What we see is that brightness is proportional to power. We also see that current is the derivative of voltage.

Blowing Up a Capacitor

Here we can see the dangerous results of overloading a capacitor. This was a very small capacitor, so one can imagine the immense amount of damage that a significantly larger capacitor can cause. It's important to be very careful when working with such things.

RC Circuit Problem

Here we have a 100 mF capacitor is connected in series with a 10.0 W resistor. This combination is connected in parallel with a 25.0 W resistor. Both branches are then connected in parallel to a 4.50 V battery that can be switched on and off. The capacitor starts off fully discharged. 

We were tasked with solving various questions about this system, the calculations for which can be seen below. For simplicity I will give the questions and answers here:
A) The time constant in both branches when the switch is closed is 1 second.

B) The maximum charge that the capacitor can attain after the switch is closed is 0.45 C.  

C) The voltage drop across the 10.0 W resistor will be equal to 1.50 V after the switch is closed when t = 1.1 seconds.

D) If the switched is opened, the value of the new time constant is 3.5 seconds.

E) If the switch is opened after the capacitor is fully charged, it will take 156.74 seconds for there to be only one electron on the capacitor.

Summary:

In summation, we have done extensive testing and analysis on capacitors in order to understand how they work, and how they affect a circuit. We know how they behave when connected in series or parallel, and also developed a relationship between their charging and discharging. We also learned how dangerous capacitors can be when used incorrectly by watching a small one explode. At the end of the lab we tested our new knowledge of capacitors by calculating various characteristics of a circuit, which included capacitance, time constants, and maximum charge.

Capacitors and Capacitance (4/28/2015)

Quiz

We were hit with a surprise quiz today! The question is pictured above, and we had to solve for every unknown in the problem, which were the currents and power.

Shown above are Professor Mason's and our table's calculations for the problem. We solved for the currents I1, I2, and I3, and power P1 and P2 being used in the circuit. In the photo on the right, the value of I2 is a negative number. We cannot have negative current, so the negative sign is actually direction, telling us that the way we placed I2 is incorrect and it should be flowing in the opposite direction.

Energy Storage

Above is the derived and modified formula of the energy stored in a capacitor. The work it takes to move a differential charge dq through a voltage V is V dq. Integrating from when there is no charge on the capacitor until the capacitor is fully charged with a charge Q we get:

deltaU = W = (integral from 0 to Q) V dq. 

If you continue to integrate that out, and replace Q with CV, you end up with the equation above.

Predicting Dependence on Area and Separation

Depicted above is the fact that capacitance (a measure of the ability of a device to store charge per unit of voltage applied across the device) is proportional to the area of the conductors in a capacitor, and inversely proportional to the distance between the plates. A capacitor itself is any two conductors separated by an insulator which can be electrically charged so that one conductor has a positive charge and the other conductor has an equal amount of negative charge.

An example of a charging capacitor, the formula for capacitance, and the units of capacitance (F).

Measuring How Capacitance Depends on Area or Separation

In order to verify the prediction above, we were asked to create our own capacitors from which we would gather data and plot a graph so that we can see the relationship between capacitance and area/distance.

What we ended up doing was separating two foil sheets with varying sheets of paper. We placed the foil in a notebook, separated it by paper pages, and closed the book in order to keep the book as closed as possible. Above you can see the results of capacitance, which was inversely proportional to the number of pages used (the distance apart). We did not conduct a test to see if capacitance was proportional to area, but logically we can expect that since a larger surface can hold more charges.

Our graph depicting the inverse relationship between capacitance and distance between conductors.

Deriving a Mathematical Expression for Capacitance

Given that Q = eEA = ke₀EA and that in a uniform electric field the voltage drop over a distance d is given by V = Ed, we can use what we know about charges in conductors and the relationship between potential difference, V, and electric field to derive an expression for the capacitance, C, of a parallel plate capacitor in terms of the area, A, and separation, d, of the aluminum plates. Above, you can see the results of plugging in the known values into the equation C = Q/V.

To test this newly derived equation, we were asked that if we have two square foil sheets, separated by paper with a dielectric constant of 3.5 that is 1 mm thick, how long (in miles) would each side of the sheets have to be in order to have C = 1 F? The answer shown above is incorrect. The answer for the area is 32284100 meters-squared, which square-rooted becomes 5681.9 m. After converting to miles you get 3.53 miles. In order to make a circuit element smaller, while still having this same capacitance would be to use a material with a better (larger) dielectric constant.

Capacitors in Series and Parallel

For this test, we took the capacitance of two capacitors individually, and in two different configurations. What we see from the data above, is that the capacitance of capacitors in series add inversely as 1/C1 + 1/C2 = 1/F. Alternatively, capacitors in parallel add directly as C1 + C2 = F.

Equivalent or Effective Capacitance

To test this capacitor addition, we solved the problem above by first adding the capacitors in series, and then adding all the resulting capacitance values in parallel, ending up with 6.5 uF.

Capacitors in a Circuit Problem 

For the final activity, we were given the oddly shaped circuit shown above, where three capacitors are connected to a battery of unknown voltage. Professor Mason's interpretation of the circuit is shown above, while ours is shown below where we proceeded to calculate the unknown voltage Vb.

We started by solving for the Q of each capacitor since Q = CV, and we were given the voltage drop at V₁ = 11 V. Since the top capacitors, C₁ & C₂, are in series we were able to calculate V₂, since they must both have the same charge. We now have the Q of the top capacitors, and know that their added capacitance is 12nF. We can now determine the voltage drop across the effective capacitance C₁₂ which is also equal to the voltage of the battery using the equation V_b = V₁₂ = Q₁₂/C₁₂= 220nC/12nF = 18.3 V.

We then go on to calculate the work done by the battery to charge up the system of capacitors.

The work needed to charge up the capacitors is equal to the energy stored in each of the capacitors, so we use the equation Ut = U1 + U2 + U3, where U = 1/2*C*V^2. As seen above, we did this in one fell swoop by using the total C and total V of the circuit, and ended up with a rounded answer of 9x10^-6 J.

Summary: 

In summation, we reviewed the basics of how to calculate current and power in a circuit using Kirchhoff's Laws. We also learned about capacitance and how it functions in a circuit, by seeing how changing different variables affects capacitance, and how to compute total capacitance of circuits. Our final task was a culmination of what we learned about capacitance, as well as recalling various rules about potential difference through a circuit.

DC Circuits and Current Distribution (4/21/2015)

Circuit Theory

We have a circuit with two batteries in series, two lit bulbs on the opposite end, and in the middle is an open switch with an off bulb next to it. We were asked to predict what will happen to the bulbs once the switch is closed. We predicted that the upper and lower bulbs will remain the same, and that the center bulb will stay off.

As seen in the photo above, our prediction was correct! The lit bulbs stayed on with the same amount of light, and the bulb in the middle remained off. The reason for this is because each battery produces 1.5 V, and as the charge travels upward starting from the bottom battery, the charge builds up to 3V. As the charge travels across the circuit and reaches the top bulb on the right, it uses 1.5 V to light it. Afterwards, the rest of the charge travels through the bottom bulb to complete the circuit.

In the second circuit we needed to analyze, we have two 1.5 V batteries on the left lighting two bulbs in the center, and an open switch keeping a third battery from interaction with our circuit. We predicted that once the switch is closed, the upper bulb will be brighter, and the lower bulb will remain the same.

We ended up being incorrect in regards to this circuit. The upper bulb and lower bulb remained at the same level of brightness. The reason for this is because the power flowing through the 2 bulb/2 battery circuit remains the same. After closing the switch, the power flowing in from the third battery on the right stays continues to travel around the rectangular path on the right. One bulb takes 1.5 V, another 1.5 V is lost at the junction to the left of the switch, and the remaining 1.5 V goes into the lower bulb.

DC Circuit Analysis

For this activity, we used two bulbs and two batteries to make the bulbs as dim as possible, and as bright as possible. The table above are our results, which indicate that the bulbs dim when the bulbs themselves are in series, or if their batteries are parallel. Likewise, when the bulbs were parallel or the batteries were in series, it resulted in brighter bulbs.

We also needed to explain power in terms of voltage and current. Basically, the brightness on the bulb depends on the amount of power flowing through a circuit. Therefore, voltage and current affects the brightness of the bulbs.

Using a Multimeter

The first test we ran on how to use a multimeter involved creating a circuit like the one depicted above, and testing the amount of voltage flowing through the resistors. As you can see from the data above, the voltage of the battery power source doubled when we went from one to two batteries. The voltages at resistors 1 and 2 were equal when using 1 battery, and equal when using 2. What we realized is that the voltages in each resistor could be added up to get the value of the voltage in the power source.

Next, we created three more circuits as depicted above, and tested the voltages at various points (which are also shown in the diagrams) using 1 or 2 batteries as well. In the first circuit above (which is the same as the circuit from the first part of this activity), we realized that the current is equal throughout the entire circuit. The final two circuits shown above depict parallel resistors this time, instead of in series. What we realized in the final circuits are that the voltages flowing through the resistors are still equal, however the magnitude of voltages matches that of the power supply. Also, it turns out that the current decreases by half after passing through each resistor.

Voltage Law

The photos above summarize what was just learned, where in a parallel circuit, the voltage stays the same and the current divides: "Components connected in Parallel with each other have the same voltage drop across. If one of the components is a battery it is not a voltage drop but it is still the same voltage only now it is a gain."

Alternatively, in a series circuit the voltage divides and the current stays the same: "Components connected in Series with each other have the same current flowing through them (or the same charge if they are capacitors)." 

The theory behind why the circuits behave this way, is because it must obey the Voltage Law, which states that the next voltage in a circuit must be 0.

Resistor Color Code

We learned that resistors are color coded in a way that tells us their resistance in Ohms, as well as the amount of error in their values. Under the Theoretical table we have our interpretation of the codes of 4 different resistors, and under Actual we see the actual resistance of each piece after we manually tested them using a multimeter.

Below that data, we used three identical resistors of 620 Ohm +/- 5% (actual values are located at the very bottom of the photo above) and testing them in various circuits. What we saw is that when in series, the amount of resistance added up according to how many resistors were used. When in parallel, the amount of resistance divided by the amount of resistors used.

Now that we know and understand how resistors work in series and parallel, we were tasked with finding the total resistance of the system above. As you can see, we gradually combined the various R1 and R2 resistors in parallel and series until we ended up with a final resistance of 68 Ohms.

Kirchhoff's Laws

Depicted above is an example of Kirchhoff's Laws, which include the Loop Rule and Junction Rule. The Junction Rule states that the sum of all the currents entering any node or branch point of a circuit must equal the sum of all currents leaving the node. The Loop Rule states that around any closed loop in a circuit, the sum of all emfs, voltage gains provided by batteries or other power sources, (e = emf) and all the potential drops across resistors and other circuit elements must equal zero.

The way we implement this is to first assign currents to a circuit, and the apply the loop rule to (in this case) come up with three equations which we can use to solve for the current. We would have to solve for I2 and I3, then use the current rule to find I1, which equals I2+I3.

Summary:

In summary, we went over DC circuits and tested them in various layouts using a multimeter. We also anaylzed Voltage Law, and deciphered the color code on resistors. Finally, we went over Kirchhoff's Laws, which include the junction and loop rules. Using all we learned today, we are able to solve for various issues in circuits such as calculating the amount of current flowing through a circuit, and the total resistance that a circuit has, as well as what are the best circuit layouts for brightness and dimness.

Potential of Continuous Charge Distributions (4/16/2015)

Potential Difference Due to Continuous Charge Distributions

To start, we have a ring where Q = 20 µC and the radius is a = 30 cm. These numbers come into play later, because first we were asked to formulate an equation which would solve for the potential difference V at point P, where x = 20 cm. In the first image, our calculations to solve for this equation are shown. The final correct answer is actually written in black on the side, where V = kDq/square-root(x^2+a^2). The purpose here is to see one of the ways to solve for finding potential difference, where we can use an integral method in which the potential dV from each element of charge dq is integrated mathematically to give a total potential at the location P. But first we will discretely calculate V in Excel.

Estimate the Potential from a Charged Ring

Shown here is the second method of solving for the same ring. We can approximate the value of the potential V by summing up several finite elements of charge Dq by using a computer spreadsheet. We divided the ring into 20 elements of charge Dq and calculated the total V at a distance of x = 20 cm from the center. The final answer we found was V = 4.99*10^5.

Calculation of the Potential from a Charged Ring

To further analyze how to approach this problem, we were tasked with figuring out the equations needed to solve for V at points P, when P is at the top of the ring, the middle, and the bottom. The correct equations are shown in the picture on the right.

What we realized when solving for P at the top of the ring, was that we ended up with an angle to solve for. Depicted above-right is the proper equation for the angle. Two other equations for the angle are possible, written as theta = arccos(x/sqrt(x^2+4a^2)) and theta = arcsin(2a/sqrt(x^2+4a^2)).

The final step in this long process, after calculating all possible equations and variables to be used, is to finally calculate for the actual potential difference V and compare it to what we found through the spreadsheet. The final answer is shown above, but without rounding it comes to 4.99*10^5 V, which is the exact same value as the one presented on Excel.

DV from a Ring Using the E-Field Method (Gauss' Law)

The third, and final, way to find potential difference of the charged ring is to use Gauss' Law in a way that will allow us to solve for V. Shown above is the derivation for converting Gauss' Law into a form that applies to potential difference. 

(r=x^2+a^2 should be r^2=x^2+a^2)

After plugging in the appropriate values, we again end up with the correct value for potential difference. We now have three different approaches which we can use to solve for the same unknown.

Electric Potential Due to a Finite-Length Line-Charge

For the next activity, we are tasked with finding the total charge and potential difference of a 16.0 cm long, horizontal rod has a uniform charge density of 2.70 mC/cm, while taking the left end of the rod to be the origin. We also need to determine the magnitude of the potential at (10cm, 15cm).

We have calculated the total charge discretely using Excel as shown above. By separating the rod into 16 segments, and integrating through the length of the bar from 0.055cm to 0.095cm, we can calculate the charge at each segment and add them together for the total charge of 2.46*10^6 V. This is the value of electric potential at any point on the bar.

In order to solve for the magnitude of potential at (10cm,15cm), the above equation for V needs to be used, which takes into account the distance r from the origin of the bar to the point. The equation for V with variables for a and b was actually derived using Wolfram Alpha, and afterwards we simply plugged in our appropriate values. Not displayed above is the answer for the potential, which is 2.43*10^4 V.

Sketches of Electrical Field Lines and Equipotentials

If an electric charge can travel along a surface without doing any work, the surface is called an equipotential surface. Pictured above are three equipotential surfaces depicted as the line going through the center of the first drawing (going clockwise from left, to right), around the center of the second drawing, and along the curvature of the third drawing. The premise is that the path of the charge and the electric field lines are perpendicular of one another, resulting in the equation E(dot)ds= 0.

Electric Potential Lab/Activity

Power Source

Voltmeter

Conductive Paper

Pictured above are the various pieces of equipment used to run this activity. We had a power source connected to two points on the conductive paper, which we then turned on to run a voltage through. We then used a voltmeter to measure the magnitude of the voltage at various distances. This was done by placing both of the voltmeter needles are precise lengths from each other. The Position values shown below depict the distance and direction that one of the needles moved in relation to the other stationary needle, which was left at the origin (one of the metal circles on the paper). It should be noted that the potential difference between the two metal circles is 15.08 V. 15 V was the amount of power we chose to produce from the power supply.

After gathering this data, we were asked to answer several questions in the lab notebook shown below:

Summary:

In summary, we have established three different methods of finding potential difference V between two points: discretely using a spreadsheet, algebraically/trigonometrically by solving for V, and solving for V using Gauss' Law. We have also looked into what it means for a surface to be equipotential and how a charge can do zero work by traveling along such a surface. We also got some hands on experience by manually testing potentially differences between two points on conductive paper. This allowed us to gather some data and calculate how much work a charge would have to do to move from one point on the paper to another using the equation W=QDV.