Problem 10.6 - Equivalent Impedance as a Function of Frequency
*
* The circuit for this problem is from textbook Chap. 10, Problem 26.
*
* Use the .AC command to sweep the source frequency from f=1 Hz to
* f=100 kHz. Print the amplitude and angle (magnitude and phase) of
* the complex current through resistance R1.
*
* Use PROBE to obtain a graph of the current angle versus frequency.
* Use the printed table and the PROBE graph to determine the range of
* frequencies over which the circuit is capacitive (RC equivalent
* circuit), inductive (RL equivalent circuit), or resistive.
* Mark the areas of equivalent capacitive, resistive, and inductive
* (two-element) equivalent circuit on the graph.
*
.OPT NOPAGE NUMDGT=5 NOBIAS ; Use this statement with Versions 4.03 or
; later. For earlier versions delete NOBIAS.
V1 1 0 AC 1
R1 1 2 1
L1 2 3 ??? ; Enter inductance and capacitance values.
C1 3 0 ???
.AC DEC 10 ??? ??? ; Complete the .AC sweep command.
.PROBE V(1) I(R1)
.PRINT AC VM(1) VP(1) IM(R1) IP(R1)
.END