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.