Problem 12.5 - Power Triangle as a Function of Frequency
*
* The voltage across the parallel RLC circuit of the problem will vary
* as a function of frequency, with a maximum at the undamped natural
* frequency (the frequency of maximum response) of the circuit. The
* average power delivered to the resistance will be a maximum when the
* voltage is a maximum. The inductive and capacitive reactances are
* equal in magnitude and opposite in sign at the frequency of the
* voltage maximum, however, and the reactive power delivered to the
* circuit is zero.
*
* Use the .AC sweep command to vary the frequency of the given circuit
* from f=1.59155 to f=795775. Use PROBE to obtain a graph of the
* average power expression and the reactive power expression for the
* circuit. On a second graph plot the circuit power factor as a
* function of frequency, using the expression given in the problem
* statement. Why does the power factor go to zero at both ends of the
* frequency range ?
*
.OPT NOPAGE NOBIAS
I1 1 0 AC 1
R1 1 0 100
L1 1 0 159.155M
C1 1 0 159.155N
.AC DEC 10 ??????? ?????? ; Complete the .AC sweep command.
.PROBE
; Enter .PRINT or PLOT commands as
; desired for list or printer-plot.
.END