Chapter 36: Alternating Currents

Assignment:  Your assignment is in parts. I have entered all the necessary Mathematica  commands you will need. All you have to do is adjust the values for r, c, and L, and plot the functions.  Hand in the print out.

1) First save this Notebook to your disk so you can make changes. 
2) Execute the entire notebook, and look at the figures. 

Now, your job as a radio engineer is to design a radio that will pick up KPLX 105.9 MHz on the FM radio band. 
3) Convert MHz to radians/sec. 
4) Adjust the values for c and L to make the resonant frequency equal to the above frequency. (Try to pick the values to make the curve more or less symmetric.) 
5) Pick an value for r so that the current from the next station  105.7 MHz is about 1/10th of I(max).  (I only want this to be approximate. Do it by trial and error, and eyeball it.)
6) Plot the current vs. frequency. 
7) Plot the current vs. frequency, BUT make r three times larger than the above value. 
8) Comment on the difference you get using the 2 values for r.

LRC Circuit

First, Clear the variables. Always a good idea when you get strange results.

Clear["Global`*"];

Now define the effective resistances.

xr=r;
xc=1/( w c);
xL=w L;

Add them together in the proper way.

z=Sqrt[ xr^2 + (xL-xc)^2]

The resonant frequency is defined as:

resonance=1/Sqrt[ L c ]

Using a modified Ohm's Law, the current will be:

current=voltage/z

And the phase will be:

phase=ArcTan[xr,(xL-xc)]

Now let's pick some values to plot.

r=50;
c=1 * 10^-6;
L=1;
voltage=1;

See what the resonant frequency is:

resonance

Examine the expression for the current.

current

Examine the expression for the phase.

phase

Now plot the current as a function of the frequency, w.  Note, I plot w from zero to twice the resonant frequency value.  We get a divide by zero error, but Mathematica is easy going about such problems.  Call this plot p1.

p1=Plot[current, {w,0, 2 resonance}]

Now plot the phase in units of Pi as a function of the frequency, w.  Note, I plot w from zero to twice the resonant frequency value.  We get a divide by zero error, but Mathematica is easy going about such problems. The phase varies from -Pi/2 to +Pi/2.

Plot[phase/Pi, {w,0, 2 resonance}]

Now plot the current as a function of the frequency, w, BUT now change r to 200 ohms, and see what changes.  Note, I plot w from zero to twice the resonant frequency value.  Call this plot p2.

r=200;
current=voltage/z;
p2=Plot[current, {w,0, 2 resonance}]

Compare plots p1 and p2. Force the PlotRange to show All points.

Show[p1,p2, PlotRange->All]