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.