*Disclaimer: I was absent this day due to a severe stomach flu which had me admitted to the hospital. I'll try describing the following as best I can from what I've been able to figure out myself through pictures.*
Introduction to Inductance
For this experiment, we had a 110 turn inductor for which we had to calculate the inductance L of. We did this using the equation L=(u(0)*N^2*A)/l, where u(0) is a constant, N is the number of turns, A is the cross sectional area of a solenoid, and l is the length of the solenoid. The value came out to 760 microHenry, and we will use this expected value to compare with the value we get when we experimentally test the inductor. We also calculated the resistance of the inductor, which came out to be 0.27846 Ohms.
After calculating these values, we created a circuit consisting of the inductor and a function generator connected in series to an oscilloscope. By running a current through, we were able to calculate a time constant for the inductor by looking at the reading shown above, and performing the proper calculations shown below.
After that, we added a resistor to the circuit as shown below, along with black grounding leads:
After running a current using the function generator once again, we received the reading shown below on the oscilloscope:
Using all the data we calculated before, the readings achieved from the latest circuit, and certain assumptions about the system (such as the internal resistance of the function generator), we calculated a half-time of the decay of the induced emf in the solenoid. Afterwards, we used the half-time measurement to calculate the actual inductance of the coil which came out to 7.36 microHenry. This results in a 3.16% error based on our initial theoretical calculations. We then used that inductance to calculate the number of turns N in the solenoid, which came out to 108.3 turns, giving us a 1.55% error. All this work can be see below:
LC Circuit Problem
For this next experiment, we imagine a circuit containing a 35.0 mH inductor in series with a 120 W resistor, both of which are in parallel with a 730 W resistor and a 45.0 V battery. The switch is closed for 170 ms and then opened. Given that, we were asked to answer the follow questions (for simplicity, the answers are placed immediately under the questions):
What is the time constant when the switch is first closed?
290ms |
|
What is the value of the current in both resistors at the
moment the switch is opened at 170 ms?
0.061A
|
|
What is the voltage drop across both resistors at 170 ms?
20V
|
|
How long will it take for the voltage drop across the inductor
to be equal to 11.0 V after the switch is closed?
105ms
|
|
(E)
|
What is the time constant of the circuit after the switch
is open at 170 ms?
41ms
|
(F)
|
How much energy is dissipated in the two resistors after
the switch is open at 170 ms?
480mJ
All the work for the previous questions can be seen in the slides below: |
Summary:
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