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PHYS 192 - Physics II Labs

Online Experiment - LR Circuit on AC Power

Line 1: Name of the experiment? (LR Circuit on AC Power)

Line 2: Your full name: (last, first)

Line 3: Lab course? (Phys 192)

Line 4: Section? (two digits code)

Line 5: Purpose of the experiment? (find the purpose after reading the theory and procedure)

Theory

This circuit consists of an inductor with inductance , a resistor with resistance and an AC L R power supply with the source voltage where is the the maximum (ωt ) V s = V m cos +ϕ0 V m value of the source voltage, is the angular frequency that relates to the frequency as ω f πfω = 2 and is the phase shift of the source voltage. The switch was closed a long time ago so the ϕ0 system is at its “steady” cyclic state.

Due to changing current, the inductor self induces an opposing magnetic field to the current such that the absolute value of the voltage across the terminals of the inductor always given by

. Also remember that, according to Ohm’s law, the absolute of the voltage across a (dI dt) V L = L / resistor is given by . IV R = R

Let us now apply the Kirchhoff’s voltage rule onto this single loop circuit . V V VΔ s + Δ R + Δ c = 0 The inductor and resistor voltages drop in the direction of current. If we do the summation of the voltage changes in the direction of the current, we get

Eq1: (ωt ) I (dI dt) V m cos +ϕ0 − R − L / = 0

Prepared by the lab supervisor, Dr. A�lgan (ea�lgan01@manha�an.edu) Online 192 - LR Circuit on AC Power PAGE 1 / 10

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Let’s reorganize Eq1: Eq2: . (dI dt) I (ωt ) L / + R = V m cos +ϕ0

The solution to this inhomogeneous 1st order ordinary differential equation is in the form of a cosine function that has the same frequency with the source voltage: where is (ωt ) I = Im cos +ϕ Im the maximum value of the current and is the phase shift of the ϕ current. We, however, need to find and to complete the Im ϕ solution.

The cosine function of the current can be written as the real part of the complex function . Similarly the cosine function on the ei(ωt+ϕ) right hand side of Eq2 can be written as the real part of . ei(ωt+ϕ )0 Remember from the complex algebra that . (x) (x)eix = cos + i sin Inserting these two expressions into Eq2 gives us

Eq3: ωLI e I e ei mi(ωt+ϕ) + R m

i(ωt+ϕ) = V mi(ωt+ϕ )0

which simplifies to

Eq4: . e (R ωL) eImiϕ + i = V m

iϕ0

Since , the last equation can also be written as i = e+iπ 2/

Eq5: . e (R e ) eImiϕ + XL

+iπ 2/ = V miϕ0

Here, we assigned to which is called “the reactance” of the Lω XL inductor. Reactance is the effective resistance of a device on an AC power supply circuit and it is in units of ohms.

is a basic addition operation in complex numbers R e )( + XL

+iπ 2/

algebra. It is equal to . Inserting this last e √R2 + XL2 +itan (X R)1−

L/ expression into Eq6 results in

Eq7: . e e Im√R2 + XL2 i[ϕ+tan (X R)]1−

C/ = V miϕ0

Comparison of the both sides gives the following two final results:

Im = V m/√R2 + XL2 and (X R) ϕ = ϕ0 − tan 1−

L/

Note that the maximum value of the current can also be written as

Im = V m/√R 2πfL)2 + ( 2

In this experiment we will compare this last result to the experimentally measured value of the current (let’s assign it to I*m to distinguish it from ) for varying values of , in order to justy Im f the theory.


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Procedure

Step 1: Download and launch. Go to https://phet.colorado.edu/en/simulation/legacy/circuit-construction-kit-ac-virtual-lab download the java file. Your computer should have a java compiler to run such files. If not, google it, download and install java onto your computer. Also google and learn how to run java files. Run the program you downloaded. You will see this window >>>


https://phet.colorado.edu/en/simulation/legacy/circuit-construction-kit-ac-virtual-lab

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Step 2: Components. Pick up 4 wires, one AC power supply, one resistor and one inductor. Also pick up an ammeter; for that you need first click the “Ammeter(s)” option located inside the “Tools” section which will make an ammeter appear in the components panel of the simulation. Ammeter is a device to measure the current. Arrange the components as shown. >>>

Set the resistance to R=30 Ω. Right click on the resistor and choose “Change Resistance”. Type 3, hit enter and click “Done”. In the same way, set the value of the capacitance to C=25 H and the voltage of the AC power supply to 100 volts i.e. . Similarly you 00 VV m = 1 can set the frequency of the AC power supply to any value between 0 and 2 Hz.

Also make sure that the internal resistance of the AC power supply is set to 0.01 Ω (the default setting).


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Step 3: Build the circuit. Now first click the PAUSE button and then build the circuit. >>>


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Step 4: Current chart. Click the “Current Chart” button and connect it to the circuit as shown.


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Step 5: Test the system. Set the frequency of the AC power supply to 1.0 Hz. Leave the frequency window open. In the next step, we will change the frequency systematically. Hit the play button. In the Current Chart you will see that the current goes up and down sinusoidally. Let the circuit run at least 5 full periods and observe that the Ammeter and the Current Chart work synchronously. Since reading the values on the Current Chart is difficult, we will use the ammeter together with the Chart to acquire the value of the maximum current with better accuracy.

Pause the simulation.


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Step 6: Acquire the data. (Before you start applying the instructions of this step read the whole paragraph first). You are now ready to run the simulation for different values of the frequency. First, set the frequency to 0.2 Hz (type 0.2 and enter. Do not hit the “Done” button since this will close the window. If you close it by accident that is alright too, you can always reopen it by right-clicking on AC power supply and choosing “Change Frequency”). Now start the simulation by simply hitting the play button (see the figure >>>). Let the system cycle at least 5 times and meanwhile watch the current plot in the Chart. Pause the simulation near a maximum before the maximum arrives and take advantage of the “step forward button” (see the figure >>>). Forward the simulation step by step, by looking at the current value on the Ammeter. The value will increase, increase, … and will start to decrease after the maximum value. Take a note of the numbers and find this maximum value. The program does not have a step backward button. If you are confused or made a mistake go forward: hit the play, pause before another maximum and apply this procedure to find the maximum value of the current. Record the values (the frequency and the max current) somewhere, and change the frequency to 0.4 Hz, similarly find the max current for that frequency as well. Repeat this procedure to find the max currents for Hz. .2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.8 and 2.0f = 0

Step 6: Process the data. Open a google spreadsheets program and enter the frequencies in units of Hz into the first column, the maximum currents in units of A into the second column. Title the columns as “f [Hz]” and “I_m [A]”, respectively. First, plot I_m versus f.

Line 6: The current relates the frequency __________ . I f [ linearly | logarithmically | exponentially | quadratically | none ]

Although the data looks like an exponential saturation curve, in this plot you will not do any fits to this data because our theoretical expectation is a kind of function that is not in a simple form and not included in the fit library either.

Our theoretical expectation is such that . By using Im = V m/√R 2πfL)2 + ( 2 google online graph you can plot this function very easily: go to google search bar and type 100/sqrt(30^2+(2*3.14*x*25)^2) >>> Note that the current is respectively higher at the lower frequencies. For this reason this circuit is called “Low Pass Filter”.


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Now on the third column of the spreadsheet program calculate the theoretical values of the current by implementing the function =100/SQRT(30^2+(2*3.14* [address of the frequency] *25)^2). Title this column as “I*_m [A]”. On the fourth column find the percent discrepancy between the experimental and theoretical value of the current, i.e. between I_m and I*_m. Title the column as “%disc”.

Line 7: Are the absolute value of all the percent discrepancies less than 5 ? _________ [ Yes | No ]

Now let’s justify the behaviour of the current as a function of frequency by converting it to a form that we can apply a fit from the spreadsheet

program library. If we take the natural logarithm of the current function , this gives us Im = V m/√R 2πfL)2 + ( 2

. (I ) (V ) .5 (R 2πfL) ) ln m = ln m − 0 ln 2 + ( 2

If we take and the relationship becomes linear. On the fifth column of the spreadsheet compute x and title it as (R 2πfL) )x = ln 2 + ( 2 (I )y = ln m “x”. On the sixth column compute y and title it as “y” . Plot y versus x and do the line fit. As always show the equation of the fit on the graph .

Line 8: What is the theoretical expectation of the slope? ________ (it is a certain decimal number and do not neglect its sign)

Line 9: What is the experimental value of the slope? (read it from the equation of the fit on y vs x graph) ________

Line 12: Compare the theoretical value of the slope to the experimental one. Give your answer in terms of percent discrepancy: ______%

Line 13: What is the theoretical value of the intercept? __________ (the answer is a decimal number. Use 3 digits after the decimal point and round it properly)

Line 14: What is the experimental value of the intercept? (read it from the equation of the fit on y vs x graph) ________

Line 15: Compare the theoretical vs experimental values of the intercept. Give your answer in terms of percent discrepancy: ______%

Line 16: State your conclusion.


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Your spreadsheet work should look like this >>>


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O n l i n e E xp e r i m e n t - L R C i r cu i t o n A C P o w e r · 2021. 2. 13. · O n l i n e E xp e r i m e n t ... St e p 3: B ui l d t he c i rc ui t . Now first click the