Sixteenth Day

Spring 2015
28th of April's Class

We start the day with a surprise quiz on Resistance, Voltage, Power and Current. The diagram for the quiz is shown in the image below, we were supposed to find the Resistance, Voltage through the Resistor, Current and Power produced by the circuit.

One of the easiest ways to tackle this problem is to note that this circuit is parallel, therefore the voltage in each horizontal lane is equal. The bottom lane, with a voltage of 19.0 V, which is supplied by the battery, tells us that the other horizontal lanes have a voltage of 19.0 V.

The middle lane, which consist of only a power source and a resistor, can be solved by equating the voltage to 19.0 V. Since there is a power source that provides a voltage of 12.0 V, Resistor 2 must possess the remaining 7.0 V. Since V = IR, the current that exist in that lane is equal to the Voltage divided by the Resistance, which is (7/320) Amperes. By the same formula, which is I = V/R, we can find that the top lane possesses (19/510) Amperes.

P = VI, using this formula, we can calculate the power that exist within each of the two loops. The power within the first loop is 19 * (19/510) W, which is approximately 0.708 W. While the power in the second loop is 7 * (7/320) W, which is approximately 0.155 W.

The next part of class, we were taught about capacitors. They function as an energy storage in a circuit by utilizing electric field. In the image above, circled in red is a drawn representation of a capacitor. Due to the gap that exist between the two plates, there is no contact, and therefore the circuit is left open. The positive and negative charges are then separated and isolated at two different ends. When they are at two polar ends, they generate an electric field due to their repulsive and attractive forces with each other.

The capacitance of a material is defined by the amount of charge it can hold per voltage, therefore its formula is likewise, q/V. The energy formula can be written as (1/2)CV^2, derived from the integration of work done from Voltage and Charge.

We then experimented with real life capacitor in the form of aluminium foils, and the distance between them is separated by the number of pages of the lab manual. The distance (d) is in mm. We tested the values empirically using a multimeter, and graph out the table using Excel. It turns out that the relationship between the capacitor's capacitance, measured in Farad, is non-linear.

Capacitance is directly proportional to the Area of the plate of the capacitor, the larger the area, the higher is its value of capacitance. This make sense as the larger the area, there will be more space for the charges to be stored in. However, it is inversely proportional to the distance set between the two plates as the closer they are, the stronger the attraction force which would cause a discharge if the charge is larger than its insulation.

By going back to the previous chapters that we have learnt, charge can be defined by the formula of Epsilon multiplied by Electric field and Area. Voltage can be defined by Electric field times Distance. With this understanding, we can generate another formula for the Capacitance, which is drawn in a rectangle in at the bottom of the image above.

Next, we were given a practice question that relates to Capacitance, where we were given the values for all variables except for Area, which we have to find. After the calculation, we found that the area is stupendously large, therefore Professor Mason discussed with us the two variables that can be adjusted so that the area of the capacitor does not have to be as large to achieve the desired capacitance. One of them is the distance between the capacitor, and the other is the material of the capacitor itself, change it to one with a higher dielectric constant.

We were then given 2 small capacitors, and with the multimeter we got earlier in the class, we were supposed to empirically determine their capacitance in parallel and series arrangement. First, we find each of their respective capacitance, then find their total capacitance respective to each arrangement. Astonishingly, we found that when the capacitor is placed into a parallel arrangement the total capacitance is added together, while in series, it is slightly more complicated. Basically, its calculation method is the opposite with resistors for each respective arrangement. The results are shown in the image above.

The image above is the next practice question given to us. First, we simplify the circuit into the one drawn in the blue marker, then we calculate them based on their arrangement separately before adding them all up to find the collective capacitance of this circuit.

The image above is the next practice question given by Professor Mason, it is similar to the previous one, but the capacitor at the center is replaced by a battery. First, we find the Effective Capacitance, by summing up the capacitance of the series and parallel arrangements after doing separate calculations for each of them.

Next, we are supposed to find the unknown voltage that the battery at the center holds. To do that, we have to first find the Voltage that is present within the top horizontal lane. Since this circuit is parallel, that would mean that all the horizontal lanes would posses the same Voltage. To find the voltage on Capacitor 2, we need to understand that the charge of elements of a circuit set up in series are equal. That means that the charge in Capacitor 1 is equal to the charge in Capacitor 2.

Finding the charge in Capacitor 1 is simply using the formula q = CV, multiplying the capacitance and the voltage it holds, which gives us 220 C. With this, the voltage of Capacitor 2 can be found with the formula V = q/C, which gives us (22/3) V. Adding the voltage from Capacitor 1 and 2 gives us 18.3 V. Due to the law that states that voltage in parallel circuits are equal, this gives us the answer that we were looking for, the voltage at the battery located at the center of the circuit is 18.3 V.

Then, we were to find the capacitor with the largest Capacitance, and we found it by calculating the amount of charge it can hold. This calculation was done in the blue bracket located at the bottom of the image shown above.

A continuation from the previous question, which was shown in the previous image, we were supposed to find the work done by this circuit. If we assume that at the start of the system, prior to the switch being closed, that the total energy of this system is zero, then when this system is booted up, the work done would be the change in energy within the system, which is basically the total energy produced by the system now.

Although the image above shows one way of doing it, by using the formula that was derived earlier, that equates energy to (1/2)*C*V^2, there is a simpler way of calculation. By using the effective Capacitance obtained from the previous image, we could just use that as the Capacitance for the energy formula once, instead of repeating this calculation three times to find the total energy.

Fifteenth Day

Spring 2015
21st of April's Class

We started the day introduced to the understanding of how current flow in a circuit. In the image right below, the upper half was the first circuit that was shown to us.

Initially, the switch, located at the left side of the middle row, was opened, and both of the light bulbs to the right, marked with 1 and 3 was on. We were supposed to predict what would happen when the switch was closed. Our initial prediction was that all three would light up, but would all be dim. However, that was not correct, and the true result is that the bulb in the center would not light up at all, and the other two light bulbs would experience no changes at all. The current ignored the center portion of the circuit.

The next circuit Professor Mason presented to us was a little bit more complicated. The circuit is shown at the lower half of the previous image. The same question was asked, what would happen if the switch was to be closed. Initially, both the light bulbs were lit up. Our prediction was that the bulb marked with the number 1 would be brighter, due to the presence of an extra source of Voltage in its circuit.

However, upon experimentation, it was found that there was no change at all. Although during the demonstration, the brightness of both the bulbs were switched after the switch was turned on, Professor Mason explained that it was due to the difference in Voltage left within each battery, as the switch was closed, the flow of voltage was reversed, and therefore the battery supplying the energy for the respective bulbs were switched.

The image right below is the real-life set up of the first circuit mentioned earlier:

Voltage Law works similar to the Law of Conservation of Energy. Voltage is conserved throughout the circuit. However, Current behaves differently depending on the set-up of the circuit. In a series circuit, the value of current is the same at all points. While in parallel set up, its value would be distributed, and vary depending on the resistors present in the circuit.

Professor Mason then asked us to experiment with setting up our own circuit to achieve certain desired outcome. The next image depicts the experiment. The table was made rather confusingly, however, what it was asking is: Which set up would produce a bright light from the bulb, and which would produce a dim light from the bulb. We were to test it out by arranging the batteries in series/parallel and the bulbs too would be arranged alternatively in series/parallel.

We found that:
When the bulbs are arranged in series, they produce dim light.
When the bulbs are arranged in parallel, they produce bright light.
When the batteries are arranged in parallel, the bulbs produce dim light.
When the batteries are arranged in series, the bulbs produce bright light.

And we found a relationship between the brightness produced by the bulb with the Power that is present within the circuit; They have a proportional relationship. The larger the Power, the brighter the light produced by the bulb.

As it turns out, resistors can also be arranged in parallel and series. The calculation for the respective arrangement is also different. The following image depicts the simple formula for calculating the resultant resistance for parallel and series:

As the image above had shown, the calculation for the resistance of the resistors in parallel set up is slightly more complicated than its series counterpart, which is just the sum of all the resistor involved. Each of the ohm value of the resistors has to be powered by (-1) and added all up, the resulting value would then be powered again by (-1) to get the total value of resistance of resistors in parallel arrangement.

We were then taught about resistors and how to read their approximate resistance by reading the colors of the band on the resistor. We start reading from the side that has the smallest distance between a colored band and the end of the resistor (not counting the wire extension). The last band is the approximation of error, it gives a gauge on how accurate the resistor is from the value indicated on the resistor. The second last band signifies a multiplier, in multiples of ten, that was to be multiplied to the preceding digits to its left. The other colored bands signifies a digit, respective to its position.

We were then to use a multimeter to test out the resistance of each resistor, as shown in the picture above. We would then make a record of each reading and see if the value falls within the range of approximation of error. After experimenting, the following image shows the result we obtained from this.

We found that all of them fits within the range of value of error, with the exception of one. Located at the bottom left corner of the image, its degree of error is about 33%, which is way too much. However, we could not pinpoint the reason for the error, it might have been the fault of misreading the colored bands. At this point of time, the powerpoint slide that showed the values of each colored band was long gone. Therefore, we had no way to backtrack and see where we went wrong.

The final part of our day was spent learning about Kirchoff's Law. It is one of the fundamental laws that would help us gain an understanding as to how the elements within a circuit interact and behave. One of the most important understanding gained from this law has to do with the nodes of a circuit. The nodes of a circuit is basically a junction, where three or more elements in the circuit would meet. At this point, the input and output of the current has to equal zero. With this understanding, we can also predict the direction of the current and voltage.

VPython Project

VPython Project

For this assignment, we were tasked to code for a model that would  place 3 point charges in a 3D space, and it is supposed to be encircled with observation points. Our model turned out something like this:

The Red, Green, Blue spheres represented the 3 point charges. The encircling Cyan spheres are observation points. We gave each of the point charges simple coordinates, and made a loop defined by a circle using Trigonometrical functions to create the encircling observation points. The coding would look like this:

First, we defined the constant k and the coordinates and charge of each spheres. Then coding the loop using n, that describes the changing theta as the observation points are distributed equidistantly from the origin of (0, 0, 0). Using a combination of Triginometrical function and Pythagoras Theorem, we derivate a formula to define the x and y coordinate of each respective point charges.

Once that is done, we coded the formula for electric potential, and requested the program to aid us in the calculation of electric potential for each point charges. Finally we sum them up to get the total electric potential from this system, which resulted in this:

Our model's total electric potential turns out to be [3.40 * 10^12 V]

Fourteenth Day

Spring 2015
16th of April's Class

Class started with on of the variants of the concept of Electric's Potential Energy, whose base formula is (kq/r). In the picture above, we were given a ring that possess charge and a point of observation in hovering in the z-axis, away from the center of the ring.

The radius required was modified to (a^2 + x^2) ^0.5. Since the Q, which is the total charge of the ring is 20 microcoulomb, we have to divide it by 20 as there are 20 point charges in the ring. This cause us to have 1 microcoulomb of charge per point.

Calculating the total potential for this system, using the formula of V = [(kq)/(a^2 + x^2) ^0.5)], we also have to multiply it by 20 as there are 20 point charges.

This picture is taken from the Excel spreadsheet, that was made to calculate the same calculation that was done by hand in the previous picture. The spreadsheet calculated the potential from each point of charge, then added them together.

Next, the point of observation was shifted away from the center of the ring. Now it has the x, y and z components of a vector. Using a variation of Pythagoras Theorem, the radius of P2 can be defined by [(x^2) + (y^2) + (z^2)]^0.5. We would then formulate a definition for the radius such that no matter which of the point charges is affecting the observation point, we can still obtain the electric potential, using angle theta and x.

We did the calculation for part (c) of the questions above, and it yielded the results shown in the picture right above this text, which is the same answer as what we obtained previously. In this method, we obtained the formula for electric potential using integration of the formula in the image right below:

In the image below, we derivated the formula for electric potential using a mixture of integration and Pythagoras Theorem's definition for hypothenuse. With this, we have used 3 different ways to derivate for the formula of electric potential.

Next, we deal with electric fields, using the same model for the charged points on a charged ring with the same observation point, as the previous images and questions.

We derivate a way to calculate the x-component of an electric field without relying on angles. We substitute the definition of cos(theta) using Pythagoras Theorem into the original formula.

The area circled in pink is the next part that we were working on. We moved away from the charged ring into a charged rod model. With the given observation point located at (0.1, 0.15), we have to define its electric potential.

With 'a' being the x-component of the observation point, and 'x' is the x-component of the charged point, and 'b' is the y-component of the observation point. We integrate it from 0 (which is the left end of the rod that was treated as an origin) to L (which is the right end of the rod, that describes its length in full).

The last part of class was an experiment we conducted using an electric power supply, multimeter, thumbtacks and a sheet of conducting paper. We were supposed to gauge the electric potential between the two thumbtacks that were pinned to the conductive paper as shown:

Two clips from the electrical power supply would be attached to the thumbtacks. Then we would use the probes from the multimeter to test the potential at both of the thumbacks, and the distance (in 1cm increment) between the two thumbtacks.

Thirteenth Day

Spring 2015
14th of April's Class

The first experiment we conducted was done with two batteries, and two light bulbs. We were tasked to design a circuit that would create the brightest glow from these bulbs, using all the items given. We designed the brightest glow by having the batteries set up in a series. Image is shown below.

The next part of the experiment assigned us to design a circuit that would create the dimmest glow. To do that, we gave the batteries a parallel set up.

The reasoning for the two different results from the series and parallel, is that they were designed to calibrate the set up differently. The parallel set up causes a higher rate of Current while series' set up creates a higher Voltage. A higher voltage means that there is a higher potential energy in the system, therefore a larger amount of work will be done.

In simple drawings, the image above is what the circuit looked like.

Below, we were taught how to draw circuits in simple sketches with symbols. The upper half of the image below are the legends to what each symbol meant. While the lower half is the representation of the series and parallel circuits we designed in simple sketch form.

In the image below, the words written in green marker is the variables we need to find the change in the temperature of water. Words written in blue is the demonstration that Professor Mason is doing to show us how the Voltage affects the temperature of water.

First he would use a water heater to heat water constantly for 2 minutes, and graph that. Then, he would double the voltage, and graph the next two minutes.

This is the resultant graph of the first two minutes (before the doubled voltage):

This next one is the resultant graph of the next two minutes (after the doubled voltage)

In the graph above, there are two distinct graphs. The red graph is from before the doubled voltage, while the blue is the graph from after the doubled voltage. We can clearly observe an increase in the slope of the blue graph, comparing to the red. We were then asked to estimate the factor that the slope would be increased due to the doubled voltage. After working out some equations, the image below shows how we derived that the slope would be increased by a factor of about 4.

Professor Mason then gave us a refresher on Kinematics. Relative to Gravitational field, work done in perpendicular motion from the direction of the field is always 0.

In the image below, we were asked to rank the magnitude of work done in increasing manner, from the different angles presented to us. From this, we learnt that the work done relative to the field, ranges from a multiplication factor of 0 (perpendicular) to 1 (parallel).

We derived an equation to find another definition for Voltage in terms of charge and radius. We start of by the definition of potential energy of an electrical system is the negative of its Work Done (circled in pink). It can also be defined by the equation written below (circled in yellow). Potential energy of electrical system can be defined with the integration of charge multiplied by Energy of electrical system. Voltage is then derived into the integral of Electrical field over distance, which would ultimately lead us into the definition of voltage being (kQ)/r.

For the last part of class, we had to complete a VPython assignment. We were tasked to predict the resulting graph from the code given above. Our predicted resultant graph is this image below, along with the hand calculated Voltage total from the given respective observation points.

As the last task for the day, we were asked to code a similar coding with the image above, by adding another charge and also 1 more observation point. The image below is part of the coding we made for the program. We would then calculate its total voltage from the 3 different observation points by requesting VPython to do the calculation for us.

This picture below is our calculation by hand:

Twelfth Day

Spring 2015
9th of April's Class

The class started with the basic concepts of electricity, demonstrated through wire, battery and a light bulb. We were to experiment and draw out two possible arrangements of the set up that would allow the bulb to light up, and also two that does not.

The reason for the failure of a bulb to light up would be that the charge does not flow through the filament of the bulb. Since it received no energy, it did not heat up, and therefore unable to generate light.

We were introduced to an apparatus named electroscope. The image below shows the apparatus up close:

Professor Mason rubbed an iron rod with a pelt so that it may build up charges. He then placed it in contact with the conductor of the electroscope, circled in green. The two films of conductor within the electroscope then repelled each other.

Professor Mason then placed batteries to be in contact with the conductor, but nothing happened. This is as predicted as there is not circuit in the system, where there is an inlet and an outlet for the charges to flow through.

The way charges move around to produce electricity is very much like the way water flows. The potential energy stored within water when it is at a certain height is analogous to the voltage of electricity, and the rate at which it flows is analogous to current. These two systems also need a circular path which allows a one-directional motion for the charge or water to flow, to produce work.

We were then taught the concept of Drift Velocity, which is the flow velocity of a particle.

Professor Mason, then experimentally determined a graph of Voltage vs. Time and Current vs. Time. With them, he formulated the Voltage vs. Current graph, which was found to be linear. Although we were given a warning that in real life, materials tend to not conform to that linear relationship.

After a material is heated up beyond a certain temperature, the heat will cause the chemical properties of the material to change, causing the graph to become erratic and oddly shaped.

Professor Mason then taught us the relationship between Resistance and Length. He charged a copper wire of length 160 cm and 200 cm. The 160 cm yield 17.4 Ohm, while the 200 cm yield 21.4 Ohm.

Calculating their ratio of resistance to length, both of them yield nearly the same ratio, therefore we can conclude that the length of material do affect the resistance.

We were then asked the question: What if diameter was increased? How would it affect resistance?

Using the formula circled in pink, we can see that the area affects the Current. As the radius gets larger, the area gets larger too; and as the area increases, the current increases too. Using that in conjunction with Ohm's Law, depicted in the image below, as the current increases, resistance would decrease. Therefore, larger cross-sectional area actually leads to smaller resistance.

As Resistivity rises, Temperature also rise as the particles would collide more frequently.