Capacitance

We started the day with a quiz about Kirchoff's Law. Given a circuit as in green below, we need to calculate each current in the circuit. There are 3 currents, 1 is the middle battery going towards the first resistor, second is going to second resistor and coming back to the middle battery. Last one is going from to the bottom battery. Each resistence and voltages are given. We know from Kirchoff's current law that current in = current out. Thus, first current is equals to sum of second and third currents. First, we will look at the top circuits, the current from the middle battery is equal to the current going through resistor 1 and 2. So, we can subtract the currents in the resistor by the currents from the battery. Second, we will look at the bottom circuit. The current from bottom battery is equal to the middle battery but subtracted by the current going through the second resistor. From this, we can find substitute for I2 and find the answer gor I1 and I3 as below. Lastly, we can also calculate the power by using P= ...

Understanding Capacitance:

Capacitance stores energy in electric field. The symbol of this is similar to a battery except the middle lines are in equal length. The unit of capacitance is called Farad (F) and is equal to charge divided by voltage. As we know, work in masses is Force times distance. In charges, this will be Voltage times charge, which can be integrated with change of charge. By substituting charge per capacitance into voltage. We then can integrate charge and fine work to be 1/2 Q(square)/C., which is also 1/2 CV(square).

Next experiment, we place 2 aluminum foils in our lab book and separate them by 1 page. We want to calculate how much capacitance do they have. Using multimeter, we find the capacitances to be as below for each time we add more page to separate the foils. Then, we plug in the data into microsoft excel to see the graph of capacitance vs distance. We find the graph to be inverse exponential as shown below.

Capacitance is proportional to its surface area, but inversely proportional to the distance between its positive and negative charges. We learned that Force multiplied by Area is ... Which is also charges per permittivity. Thus, we can substitute this into the equation and find charges to be permittivity of the material, electrical force, and area multiplied together. We then want to find the relationship of this equation to capacitor. We know that capacitance is charges per voltage. By substituting the equation into the charges, and voltage to electric field multiplied by the distance, we find the equation of capacitance to be multiplication of constant, permittivity, and area, divided by distance.

Given the known for permittivity, distance, dielectric constant, and the capacitance, we can find the area of the plate using the equation that we just derived. We gound the area to be really big for a 1 Farad capacitance. There are two ways to reduce the area, we can either decreases the distance between the plates, or we can increase the constant by changing the material.

Understanding the capacitor system:

We are given 2 capacitors, we want to calculate the total of capacitors in series and parallel. Using the multimeter, we want to obtain the capacitance of each capacitor and we find it to be each, 1.03 and 1.06 microFarad. We then set them in series and find the total capacitance to be halved, and the parallel to be double. From this, we can conclude that in parallel, capacitances add, whereas in series, inverse of capacitances add.

Given a circuit in green color below, we were asked to redraw it into a circuit that we're familiar with, which is parallel and series. We then redrew it in blue below with C1, C2 on top, C3 in the middle, and C4, C5 on the bottom. The capacitances are given below. Recall from past lectures, we add the capacitance in series and add 1/Capacitance in parallel. We found the total resistance to be 6.5 microFarad.

We are now given the voltages, currents, and capacitance of each. By adding the capacitance in series and 1/Capacitance in parallel, we found the total capacitance to be 52 nanoFarad. We can also find the total voltage. We know that charge in series are equal and charge is capacitance multiplied by voltage. By using the known capacitance and voltage, we found the first charge and set it equal to second charge to find the voltage of the second charge. Then, we can add the voltage in series together and set it equal to the voltage in the middle circuit because voltages in parallel are equal. Thus, we found the voltage to be 18.3 V.

Using the capacitances we had above, we want to calculate the electrical force in the circuit. We derived the equation for work earlier to be 1/2 CV(square). Frist, we changed the known to SI units. Then, we plugged in the knowns into the equations and add them the works together to find the total work to be 8.7 x 10^-6 J.

Capacitor and Time Constant

We started the day with an expertiment

Understanding the Capacitor system:

When we make a circuit with battery and capacitor placed in parallel to each other, we want to see what would happen to the light bulb. When we connect the switch, nothing happens. When we take off the battery from the circuit and connect the switch again, the bulb lights up for a little while, but eventually turn off. This shows that the energy from the battery stored in a capacitor, then the energy in the capacitor is used in the circuit. We then predict the graph of brightness vs. time and we found the graph to be inverse natural log.

Understanding Time Constant:

Time constant is the amount of time for a constant to charge. Time constant is proportional to resistance and capacitor. Thus, time constant is equals to resistance multiplied by capacitor.

We then separate into 2 groups, each group set 3 batteries and 2 batteries in series respectively. Each of these set is connected to 1 capacitor. In the end, we set the 2 capacitors in parallel and predict the total voltage. We found the voltage to be 1.536 V. Using multimeter, we found the total voltage to be ..., which is closed to what we predicted.

To prove if our prediction is correct, we use logger pro. We connect the logger pro sensor to capacitor as shown below.

We found 2 graphs, the first one is when capacitor is connected to the battery, which is shown in blue, and the other one is when the battery is taken off, which is shown in red.

We want to find the unit of time constant. We know that time constant is CR, but we also know that Capacitor is charge per voltage, and voltage is consist of current multiplied by resistor. By subtituting these equations, we can find CR to be charge per current, but charge per second. Thus, we found the time constant's unit to be second.

From this relation, we can find the equation of final charge. By using current as change of Q over time, we can replace I with this and find change of Q over time to be charge/CR. When we derive the equation, we found Q to be initial Q multiplied by exponential of t/CR.

Since we know that voltage is charge divided by capacitor, we can use this equation in terms of voltage.

Next, we know that graph of voltage vs time is exponential and brightness vs time is inverse exponential. We want to predict the graph of current vs time. We know that power is current multiplied by voltage, which means when voltage increases, power lowers, and current has to lower as well. Thus, the graph of current vs time is inverse exponential.

Given the circuit in circle below, we are given the knowns as below. First, we are asked what the time constant of the circuit when the circuit is closed. Changing all to SI unit, we plug in the knowns to the formula of constant, but the resistor effected only the top resistor because the current does not go in the middle. We can also find the charge by plugging in voltage and capacitor and multiply them.

Then, we want to find the time constant when the circuit is closed. We have to first calculate the resistance affecting the circuit because both resistances are affecting since the current is going on the top circuit. Lastly, we can calculate the time it takes for the capacitor to decrease  from 4.5 to 3. By using the equation that we derived earlier, we just plug in the knowns for time constant, initial voltage, and the ending voltage, which is 3V. We found the time to be 1.09s.

From the initial charge and time constant that we found above, we can find how long it takes for the charge to go down to 1 atom. We know that 1 atom is 1.6 x 10^(-19). By using the equation that we found earlier, we found the time to be about 149s, or 2.5 min.

Vpython 2

Creating a 3D Model
We were given an assignment to code for a 3D Model, using VPython, that would give us a reading on the value of electric potential from 20 observation points, from 3 different charges. This is what our final model looked like:

We took a picture of them at different angles so that their orientation can be better understood.

This is what our coding came to be:

Within the coding, we gave the computer formula to calculate for the electric potential from each charge point and to help us also find the total, using the loop function of VPython. This is the resulting value:

Resistance

Understanding a Circuit:

A circuit is set up as below, we were asked to predict what happens when the switch is closed. Before the switch closed, the bulb marked 1 and 3 was lit and the bulb marked 2 was off when the switch was opened. After the switch is closed, nothing changed. We found iut that there is no effect on the middle path when there is a closed circuit around it. This is called short circuit.

We are then given a different setup as shown on the left bottom. 

Understanding the difference in battery and bulb system:

We make a table of what makes the difference in brightness between the bulb and the battery. When the circuit is arranged in series, the bulb is dimmer, and when it's arranged in parallel, the bulb is brighter, whereas the battery is the opposite. When the circuit is arranged in series, the battery is brighter, and when it's arranged in parallel, the battery is dimmer.

This brightness is affected by both, current and voltage. From the equation we learned before, power is current multiplied by voltage. These two gives what we know as Watt, which is the Power.

Understanding How to Read Resistor:

There are 3 or 4 lines in resistor. These lines are the color coding of how much resistance does it have. Each color has its own value. When we are reading the resistor color, first and second band are put together as the first and second number of the resistance value (sometimes third band too if the resistor has five lines), the second line from the last is the multiplier of the resistance value, so it will be 10 to the power of that color's value. The last line shows the uncertainty of that resistance in percentage.

We are then given 4 resistors and try to read each resistor value by its color coding. After doing so, we tested their resistence with the multimeter.

The first resistor, by color coding is found to be 1200 ohm with uncertainty of  5% and was tested to be 1256, which was in the range of the uncertainty, so this was correct. The second resistor, by color coding is found to be 20 x 10 ohm with uncertainty of  5% and was tested to be 198.6, which was in the range of the uncertainty, so this was correct. The third resistor, by color coding is found to be 30 x 10^2 ohm with uncertainty of  5% and was tested to be 3060, which was in the range of the uncertainty, so this was correct. The forth resistor, by color coding is found to be 33 ohm with uncertainty of  5% and was tested to be 40.8, This was slightly off because it was not in the range of the uncertainty. This may be due to ...

Using multimeter to find resistance:

Given 3 resistors, we want to set it up in different arrangements. First, we want to find the resistance of 1 resistor, which is foun to be 612 ohm. Then, we want to attach another resistor next to it and find the resistance of both to be 1225 ohm. Attaching another resistor, we found the resistance to be 1835 ohm.

Next, we want to take them all apart and attach one of the resistor on top of the first one instead of next to it and find the resistance using multimeter. When we place the multimeter needle in between the two resistors, we found the resistance to be 306 ohm. Then, we add more resistor parallel to the two resistors. We found the resistance to be 204 ohm.

From these data, we can conclude that in series, the resistances are added, whereas in parallel, the resistances are divided depending on how many resistors there are.

Understanding Equation of Resistance:

When in parallel, the equation of resistance will be 1/Req=1/R1+ 1/R2. When in series, the equation of resistance will be the sum of all resistance.

Given the mixture of parallel and series in a circuit as below, we have to separate each of them. Start by adding the series first. The top series resistance will give the sum of both, which results in 320 ohm. Then calculate the parallel of that part, this gives about 76 ohm. Then add it with the series next to it, which results in about 176 ohm. Then we calculate the bottom parallel, which gives the out resistance to be 110 ohm. The last resistance is in parallel. Using the parallel equation, we found the final resistance to be about 70 ohm.

Understanding Kirchoff's Law:

Components in Electricity

Understanding charges in 2-D:

We have learned a lot about voltage in the past. Voltage is electric force multiplied by distance, which is also k multiplied by charge and divided by radius squared when the charge is in a straight line. Today, we are learning about the charges when it has a component on it. To find the radius, we can use pythagorean theorem. The equation of this radius can be substituted into the equation of the voltage when the radius is unknown. First, we are integrating in respect to r, then we found the equation to be k multiplied by charge divided by the radius. The radius is substituted with the sqrt(x(square) + a(square). From that, we find the equation as below.

The electric force itself can have component.

We are now given a ring with 20 point charges acring on a single point. The point is at a distance 20cm away from the center of the ring, and the radius of ring is 30 cm. We can calculate this by hand, which will be using equation of voltage as kq/sqrt(radius(square)+x(square)). By plugging in the known, we find the voltage to be 4.99x 10(5) as below.

We can also use spreadsheet to calculate the total voltage. We set the known in for radius, x, charge, and constant as below. Then we set the equation to be ... For 20 points. Then we add all of the voltages and found the answer to be the same as done by hand.

In the spreadsheet, we recorded the x and y-axis. Y is the same for every point of x, but x changes every 0.01 since we will split it up into 16 parts. K is always the same and charge is also given. We wanna calculate r first which equation would be ... And also set equation of voltage to be. Finally, we will sum all the voltages to find the V total as below.

A rod with length of 0.16m with charges that is located at (0.1,0.15) is given. We want to find the total voltage at the rod. If we split the rod into 16 parts, doing the calculation by hand would be tidious; thus, we use spreadsheet to calculate them.

We also find the relationship between k, lamda, q, a, x, and b. We know that the volte is integration of k

Voltage, temperature, potential energy, and work in electricity

Understanding battery system:

We started the day doing an experiment on lightbulb and battery again. First, we connect 1 battery to two bulbs and see how the bulbs light. Then, we connect two battery side by side to 1 lightbulb and see how the bulb lights.

The brightness of both trials are equal.

The bulb thats connected to batteries that are side by side is dimmer than when the batteries are stacked up because when the batteries are stacked up, the voltage of both batteries are added, whereas when the batteries are side by side, is the same as using 1 battery voltage. Because of this, using 2 batteries or 1 battery earlier give the same brightness result.

When drawn in a circuit form, there are parts on the circuit that can be drawn to indicate each, which are battery, resistor, bulb, and switch as shown below. 

Understanding relationship between voltage and temperature:

The next experiment is water heater. The Water heater is powered by battery.

The graph below shows the 2 minutes that it is heated.

Then, the voltage is doubled and graphed for the next 2 minutes as shown below. When the voltage is doubled, the scalar factor is multiplied by 4, theoretically because the current is squared in temperature. Thus, the slope is steeper.

Understanding work in electricity:

There are 3 paths that can be taken to move on a slope. Going parallel to the charge then perpendicular, going perpendicular to the charge then parallel, or just going parallel to the slope. We wanted to see the difference in work on each path. Work on a path that's perpendicular to the current is 0. Thus, we only count the work thats parallel to the current. After doing the calculation, the total work for all path turn out to be equal.

When given a current and 3 different ... On it, we want to rank them from the least work to the biggest work. The work perpendicular to the current is zero as we learned because the angle between the current and the .. Is cos 90*, which is zero. This makes the work also becomes zero. The next one would be the .. Slanted to the current, then the ... Parallel to the current as calculated below.

Recall from past lecture, that Electric force is equal to charge multiplied by electric field; then, we can substitute force with them as well.

Understanding potential energy and work:

We now learn that potential energy in electric is equal opposite to work. But we know that work is force times distance, which force is also charge times electric field; therefore, work is integration of charge multiplied by electric field and change in distance. We also know that voltage is work per charge. By equating these equations, we can find the voltage to be integration of electric field multiplied by distance.

Vpython:

Given program as below, we have to predict how it would look in the whiteboard.

We predicted the 2 charges to be on the x-axis, one one (-2,0,0), and the other on (2,0,0). We also predicted the observation point on (-1,0,0).

After making the prediction, we calculate the V total using the given charges in the program, which are (-1 x 10(-9)) and (1 x 10(-9)). We also use the distance from each charges to the observation point. Using this known, we plug it in to the formula of voltage that we have known from earlier, which is kq/r, and add the voltage from both charges. We found the total voltage to be -6V. When the observation point is moved to (1,0,0), we can use the same method and found the voltage to be 6V.

When we run the program in vpython, the screen shows as below, which followed what we predicted.

The next blog will show vpython that we are assigned to do and we are allowed to work together, but we have to remember, "collaborating, not plagiarizing"! :)

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