26th of May's Class
The class was started off with an introduction to an inductor. This is what it looked like:
As its colour is black, it blends with the tabletop which is black too. So I had it marked in green.
Since there are a lot of information going on in the image above, I will break them down according to the colors of the boxes.
Green
This box shows the two different graphs that reflect [Induced Voltage vs.Time] at the upper half of the box, and [Current vs. Time] at the lower half of the box. One thing to note is that both of them are reaching an equilibrium value as time pass.
Red
This box contains the equation to calculate the value of a resistor's resistance using its Resistivity, Length and Cross Sectional Area.
Yellow
This box contains the formula to find tao, which is the time constant for inductors. Which is calculated by its inductivity divided by the total resistance of its circuit.
Pink
The method that we will be using to calculate the period of the wave generated is by multiplying the time constant tao by 5. This number is derived previously.
Purple
This box shows the formula for us to calculate Inductivity, which is involved the constant 'mu' (4 * pi * 10^-7), number that the wire is coiled up, the cross sectional area of the inductor, and length of the wire within the conductor.
Next is the experiment. We were supposed to use the oscilloscope to obtain the wave emitted when a current passes through the inductor. With the frequency set at 40 kHz, and 3.00 volt supply, we obtained a graph as shown in the image above. The graph is on the screen of the oscilloscope, which is below the bluish box thingy.
The entire set up is as shown below:
The inductor is set up in a series with the resistor, and function generator, and parallel with the oscilloscope.
Using the graph from the oscilloscope, we measured with our eye power that the half time of the graph is about 4 * 10^-7 s.
Knowing the half time could get us the time constant tao, using the formula of half time divided by ln(2).
With the tao, we could calulate the Inductivity L, with the earlier formula of tao multiplied by total resistance of its path.
With the inductivity, we can calculate an estimate for the number of coils our inductor has.
However, despite the label on the inductor mentioning that the inductor is supposed to have 880 turns, our calculation only showed a meager number of 37. This can be seen as an error on the part of the inductor. From the beginning, it had been giving us problems getting the desired graph, despite our efforts in setting it up correctly.
This is the error and uncertainty that we gathered. The resistor's uncertainty was obtained from the color of its band. While the uncertainty for the time is obtained from dividing the smallest division of the time by 2.
The diagram in question is in the picture above, to the left side of the whiteboard.
There are 4 parts to this question.
First is Blue
We are to find the current flowing in each of the resistors in the circuit, after the switch is closed.
Since the time that the switch is closed is more than the half time, we use a special formula for calculating the current flowing in the upper most horizontal line. It is calculated as I1 in the blue box.
Second is Purple
We are to find the potential drop across both of the resistors.
Third is Green
We are to find the time taken for the voltage within the Inductor to drop to 11V after the switch is opened again. Assume that it had been fully charged prior to this.
Fourth is Black (above Purple)
This is when we are to calculate the amount of energy that the inductor release as the switch is opened.
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