Problem 12.3 - Power Factor Improvement with Added Capacitance
*
* This problem is based on textbook Examples 12.3 and 12.4.
* The reactive power delivered to a capacitance is opposite in sign to
* the reactive power delivered to an inductance. Because of this, a
* capacitance placed in either series or parallel with an inductive
* load will reduce the total reactive power delivered to the circuit.
*
* For the circuit of the problem, use a capacitance model with a
* variable parameter C. Use the .STEP command to step the capacitance
* value from 0 to 12 microfarads in 3-microfarad steps. Print the
* amplitude and angle (magnitude and phase) of the current through
* the voltage source and the amplitudes of the resistance current and
* the capacitance current.
* Using the output data, hand-calculate the apparent power delivered
* by the source, the real power delivered to the load, and the total
* reactive power to the circuit. Draw a set of nested power triangles
* (triangles inside each other, with the same origin for P and Q)
* showing the power factor improvement or decrease in power factor
* angle as the capacitance is increased.
.OPT NOPAGE NOBIAS
V1 1 0 AC 120
V2 1 3 AC 0 ; The voltage source is added to
; measure the current through the
; top circuit branch (the source
; current) to avoid the double
; negative sign confusion of using
; I(V1).
R1 3 2 90
L1 2 0 .3183
C1 3 0 CLOAD 1
.MODEL CLOAD CAP(C=3E-6)
.STEP ; Complete the .STEP command.
.AC LIN 1 60 60
.PRINT AC ; Complete the .PRINT command using
; I(V2) as the source current.
.END