PHYS-1500 PHYSICAL MODELING
Class 23: LC Oscillator
FALL 2006
NAME _________________KEY____________
In this exercise, you will examine an electrical oscillator. The oscillator is made up of an
inductor, L, a capacitor, C, and a resistor, R. When they are connected in series, and a driving
voltage, E, is added, as in the diagram, then E must equal the sum of the voltages across the
other three elements.
di
q
d 2 q dq
q
Then, E  L  iR  , or E  L 2 
which can be written
R
dt
C
dt
C
dt
d 2q
dq 1
L 2  R
 q  E . This equation has the same form as the equation for a driven
dt C
dt
d 2x
dx
mechanical oscillator, m 2   kx  b
 Fd , and the behavior of the two is quite similar.
dt
dt
However, in this case, the driving term is written, E = E 1 + E 2 sin 0t, as it was in the other
electrical cases. For this model, you will use C = 2.5 ×10-6 F, and L = 2.5 ×10-4 H.
Consider the reaction of the system as 0 and R are changed. First set t = 1.0 ×10-5 s, q0 = 0, i0
= 0, E1 = 0, and E 2 = 10.0 V. Then set R = 1.0  and find the maximum value of i for 0 =
20000, 30000, 40000, 50000, and 60000 rad/s. Then do the same for R = 3.0  and R = 10.0 .
Enter the results in the table shown, and graph i(max) vs. 0 for all three values of R on the same
graph.
o
R
20000 s-1
30000 s-1
40000 s-1
50000 s-1
60000 s-1
1.0 
0.708
2.49
8.62
3.62
2.19
Sketch the graph on the back.
1
i (max)
3.0 
0.638
1.70
3.30
2.42
1.73
10 
0.530
0.805
0.997
1.01
0.993
10
i
max
(A)
8
1
6
3
4
10
2
0
10000
20000
30000
40000
50000
60000
70000
 0 (rad/s)
Now set E 1 and E 2 = 0, and q0 = 1.0 ×10-4 C. Set R = 1.0 and sketch the graph of q vs. t. Then
do the same for R = 3.0 and 10.
0.00015
0.00012
0.0001
0.00008
0.0001
0.00006
0.00005
q (C)
q (C)
0.00004
0.00002
0
0
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
-0.00002
-0.00005
0
0.0002
0.0004
0.0006
0.0008
-0.00004
-0.00006
-0.0001
-0.00008
t (s)
t (s)
R = 1.0 



0.00012
0.0001
q (C)
0.00008
0.00006
0.00004
0.00002
0
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
-0.00002
t (s)
R = 10 
2

R = 5.0 
0.001
0.0012
Finally, set q0 = 0, E 1 = 10.0 V, and E 2 = 0. Try the three values of R that you have used before.
Which value of R would you use if you just wanted to get the capacitor charged to a constant
charge in the shortest time?
0.00005
0.000045
0.000045
0.00004
0.00004
0.000035
0.000035
0.00003
q (C)
q (C)
0.00003
0.000025
0.00002
0.000025
0.00002
0.000015
0.000015
0.00001
0.00001
0.000005
0.000005
0
0
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0
0.0002
t (s)



0.00003
0.000025
q (C)
0.00002
0.000015
0.00001
0.000005
0
0.0002
0.0004
0.0006
0.0006
0.0008
t (s)
R = 1.0 
0
0.0004
0.0008
0.001
0.0012
t (s)
R = 10 
R = 10 charges in the shortest time.
3

R = 3.0 
0.001
0.0012