3/3 Thermal Resistor

In the morning, we were given a pop quiz where we had multiple junctions leaving us with many ugly equations and numerous equations and unknowns. After trying to solve it the conventional way, we use the mesh current method to solve for each of the unknown currents and voltages.

We have a resistor that changes in resistance based on the temperature. We calculated how much of a temperature change would cause the voltage going through this resistor to change by .5V. We found that the temperature change required for this was close to .4704C.

From the graph above, we found the resistance of the thermal resistor to be 11.55 kOhms @ 25C and 7.28 kOhms @ 37C. We also found the difference in voltage between the +Vout and -Vout if the voltage going in is 5V at each 25C and 37C.

We are given a circuit where we have the resistor from the problem before and normal resistor. We calculate the resistance of the resistor at the specific temperature using the graph from before. From this resistance, we calculated the resistance of the other resistor if given the voltage going out when the thermal resistor is at 25C and 37C.

We set up a breadboard circuit where we test the thermal resistor for its actual resistance at room temperature (25C) and at body temperature (37C) if the voltage going through the resistor is 5V.

We plug the thermal resistor into the breadboard and use the multimeter to find the actual resistance of the thermal resistor at body temperature (37C). We found it to be very close to our theoretical value 7.38 kOhms.

3/1 KCL and KVL

My group makes a prediction on what will happen to a hot dog if hooked up in a DC circuit. We say that the hot dog will slow cook over time.

We hook up the hotdog into the circuit and it turns out that the hot dog does slowly cook rather than instantly crisp.

We were then given a situation where LEDs are inserted into the hot dog and asked to predict whether the LEDs placed parallel to the circuit or perpendicular to the circuit will light up. We said that the parallel ones will light and the perpendicular ones won't because there is no voltage difference in the LEDs placed perpendicularly.

Mason places the LEDs into the hotdog.

After the LEDs have been inserted into the hot dog, we found that our predictions were correct. The parallel LEDs lit up because those LEDs had a voltage difference and the perpendicular LEDs didn't.

We are given a problem of this circuit and asked to find Io and Va from the givens. We said that once the current from both current sources reached the middle resistor, The sum of the currents would be equal to the current going through the middle resistor. Once we found the current, we can use V = IR to find the voltage across the middle resistor.

We are given another circuit problem except this time, we have multiple loops with a lot of different currents. We used Kirchoff's Current Law to find equations for each junction in the circuit. We used the 3 equations we had to solve for the 3 unknowns.

We are given a circuit except no information is given and we are only asked to derive the equations for the voltage out and the voltage in.

We use the derived equations previously and use it for the conditions that the voltage going out has to be less than 3.3V and the Voltage going in has to be 12V and the current has to be less than 10 mA. From that information, we are asked to find the largest and smallest resistor 1 and resistor 2 can be in order to satisfy the condition.

This is the same problem except our conditions are different and we use the same equations we derived to find the minimum and maximum resistance of each resistor.

We are given a circuit where we use Kirchoff's Voltage Law to solve for the unknown current and unknown resistance.

We are given another circuit problem and asked to solve for the unknowns again. We use KCL and KVL to make equations for each loop and solve for each unknown.

We created the circuit previously on the breadboard with an LED that would light up when the diode was covered.

We are given a circuit with a bunch of resistors and asked to simplify the circuit down to the equivalent resistor where we only have 1 resistor that represents all the resistors in the circuit.

2/25 Circuit Analysis and Breadboards

We are introduced to the idea of branches, nodes, and loops and were asked to identify how many of each were in the given circuit.

We are given a circuit analysis problem and asked to find the multiple ways the problem can be solved, because there are 3 currents and the first current must be positive, we have 4 total possibilities. We found the current for each of the 4 combinations.

We are given another circuit analysis problem where we have a circuit and we are asked what will happen to two light bulbs in the middle of the circuit once the switch closed. Because the voltage potential difference at the points above and below each light bulb was always 1V, we found that the light bulb did not get brighter or dimmer.

We did a lab with the breadboards again except this time we used a special device that allowed us to control the voltage that we send through the breadboard.

We manually adjusted the voltage from 0 to 2 with .2 increments for the voltage so that we could test for the current going through the circuit as well. We also read the voltage with a multi-meter to see what the experimental voltage is. We found the resistance going through our advertised 100 ohm resistor to have an actual resistance of 98.3 ohms from the data collected. 

We graphed the data we produced from our lab and plotted it on a graph. Once we plotted it, we found the relationship between current and voltage to be linear and our r-squared value to be greater than .98 (which in statistic terms indicates that it fits a linear curve very closely). 

We are given power, voltage, and resistance and are asked to find when the resistance is cold (when voltage is initially applied) vs when the resistance is hot (when the circuit has been running for a while). We found that the resistance of the resistor to be 192 ohms in the beginning and less than 192 ohms after.

We do another lab with the breadboards except this time we introduce something called  MOSFET to the circuit. The purpose of the MOSFET is to keep the voltage at a consistent level. The voltage is drained and applied to the circuit to maintain consistency. We found that the current this time around produced a graph much closer to a linear relationship due to the increased consistency provided by the MOSFET.

We set up another lab with the breadboards and multimeter to measure more data except this time we are using a MOSFET.

This is a graph of the along with the table of the data we collected from the lab with the MOSFET.

2/23 Relationship between Current, Power, Voltage, and Energy(Work).

We were given a problem in class where we were given the graph of charge vs time with the time and q max given and we were asked to create and solve the sinusoidal function of q with respect to t. Since we are given the maximum, we can say that the derivative of the derived equation is equal to the 0 when t is equal to when q is at a maximum.

With the same problem, we created a function of current in terms of time. We used the derived current equation to find the total charge between 0 and 2 seconds.

In this problem, we are given the current vs time graph and the voltage is 10V. We were asked to create the power vs time graph and work vs time graph. Because P = VI, We can multiply 10V to every point on the current vs time graph which is why the power vs time graph is so similar to it because only the scale of the y axis changed. We then did the same thing to find the work vs time graph by saying dw/dt is equal to power. We used the properties of a derivative to draw the work vs time graph. We were then asked to calculate the total work done between 0 to 4 seconds. 

In this lab, we used a breadboard to find the resistance of a row in the breadboard. (Randomly chosen row). We found the resistance to be 1.4 ohms.

Using the same breadboard, we found the resistance of row 22. The breadboard does not have a complete circuit if the circuit extends across the entire row. A complete circuit can only be created if it is in the same column. This is why when we tested the entire row of 22, we found the resistance to be infinity or overload because if the circuit isn't complete, the resistance is infinity. When we tried row 22 in the first column and row 34 in the second column, we also could not create a complete circuit hence the infinite resistance.

This is the breadboard mapped out with what is connected to what.

We are given the problem of trying to derive other graphs given the current vs time graph on the top left corner and the voltage vs time graph on the top right corner. Because of the formula P = VI, we can simply multiply the values of both graphs at the same time and find the power for that same time. The power graph we came out with is on the bottom left. We then found the energy vs time graph by saying that the integral of Power with respect to time is equal to work. We used the properties integration to predict what the graph will come out to be. 

Now that we found the power and energy vs time graphs, we are asked to calculate the total work done between 0 and 4 seconds. We integrate power with respect to time between 0 and 4. Since we do not have an equation for the graph, we found the area under the curve algebraically. The curve between 0 and 1 is not linear so we created an equation for that specific part of the graph to find the area underneath 0 and 1. 

We are given a circuit analysis problem where we are asked to find the power absorbed and each resistor using Kirchoff's Voltage Law and we also calculated the power delivered into the circuit.