5/26 week 14 day 1 RL circuits and Induction Lab with Oscilloscope

Introduction to inductance

At the beginning of the class, we know that any change in the current through an inductor leads to a change in the magnetic field it produces. This is turn leads to a change in the magnetic flux through the inductor which, by Faraday's law, produces an induced emf in the inductor.

We used the equation for inductance using a given resistance and found that the inductor had an inductance of 760 mH. we found the resistance of the copper wire in the given inductor based on calculations below after find the inductance of the given inductor.. Number of coils is 110, length of 0.05m, radius of 5.12 * 10^-4m. First, we calculate the area of the 18 gauge copper coil using the area of circle formula. We know the density of copper to be 1.72 * 10^-8. By plugging in the numbers for density multiplied by length per area, we can find the resistance to be 0.3 ohm

In this photo, we draw the graph of Current and time according to Voltage vs time graph in the first photo. As the time goes by, the Voltage of inductor decreases and the current increases.

In this photo, the professor gives us a 100 ome resistor and let us measure its resistor according to the color chart. and we know that  the brown is 1, black is 0 * brown is 10 so the resistor is 100.

Then we calculate the inductor according the equation L=u*N^2*A/l. and we have that the area of the inductor is 25cm^2 and the resistor is 100ome and the N is 110 so we get that L=760mH

Then we calculate the resistor of the copper wire in the inductor according to the equation R=pL/A and we find that the R is 0.463, which is very samll.
Then we calculate the time constant of the LR circuit by T=L/R and we get that is 5.1*10^-6 s

In this photo, the professor gives us the period is T=5t=2.55*10^-5 s

Experiments and Analysis

we start to do the experiments about the oscilloscope.

First,we set up the oscilloscppe like in this photo. Here was the set up before we tried to solve for the period and then the time constants and then the number of loops our inductor had.  We had to change the set up of the graph just to solve for those three things.

Then we connect the Function generator , a resistor and the inductor in a circle. we find that the graph in the oscilloscope is changed.

Then in this photo, we begin to measure the inductance using the equipment in the last photo and compare it with the calculation one.  First, according to calculation, we know that a N=110 inductor's inductance is 760mH. Then we first find the T_1/2 IS 3.4*10^-6s and use the equation t=t_1/2/ln2 to find the time constant is 4.9*10^-6. and then according to time constant=L/R we find that L is =736mH

 Then we did a lR Circuit Problem in this photo.

A. we know the inductance is 35mH and use the equaiton time constant=L/R to  the time constant is 290us.
B. we know that the inductor and a resistor is parrel with another resistor so we use the equation I=V/R to fin the I_1=0.062A and ues I=I(1-e^(t/L/R)) to find the I_2 =0.17A
C. in this question, we use the equation V=IR to find the change of voltage after 170us.
D. we know that the voltage of inductor is 11V so we know that the voltage of resistor is 34V and we get the current is 0.28A. Then accodring the equation I=I(1-e^(t/L/R)) we find that after 402us the voltage of inductor is 11V
F. according the equation E=1/2LI^2 and we find the energy in the inductor is 480uJ.

Conclusion:

Today in class, we looked more in depth into inductors and put one to the test by using it in a circuit, from which we measured various experimental values to compare against our theoretical ones. we know that any change in the current through an inductor leads to a change in the magnetic field it produces. When an inductor is added to the circuit and the current is seen to increase rapidly and reaches a steady state as seen in the experiments of our oscilloscope and RL Circuit problems.