Linear Circuit Analysis:
Charge and Current
Work, Power, Voltage, and
Resistance
Kirchoff's Laws
Y Delta, Node, and Loop
Circuits with Operational
Amplifiers
Network Theorems
Analysis of Diode Circuits
Capacitance and Inductance
First-Order Transient Circuits
AC Steady State Analysis
Steady State Power
The Power Factor
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Introduction
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circuits.92g
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Linear Circuit Analysis:
The Power Factor





For the network shown, determine the real and reactive power losses in the line and the real and reactive power required at the input of the transmission line. How much power would be saved if the power factor was increased to .90. The frequency of the system is 60 Hertz.
Irwin. page 483 drill 10.12






The complex power is defined to be


S = Vrms I*rms

Where I*rms is the complex conjugate of Irms. The real part of S is referred to as real or average power and the imaginary part of S is called the reactive or quadrature power. Therefore S can also be expressed as:

S = P + jQ




Given in terms of the power used and the power factor, the complex power for the load can be written as:



Finally the power factor is defined to be the ratio of the average power to the apparent power:


The power factor also refers to the phase of the current with respect to the voltage. All of these formulas are shown as being stored in the Text Editor. The last line has a command line next to it for evaluating on the Home Screen. Also, a function was written for finding S in terms of the power and the power factor.
Since we know the power consumed by the load and the power factor we can use the SLoad(power,pf) function to find the complex power of the load. For easier access this function has been placed in the custom toolbar with the complex functions.










With the custom menu in place call the SLoad function by pressing [F3] [4]. Enter the parameters as shown from the figure of the network.
Press [F3] [ENTER] ENTER] to view the complex power as a magnitude and an angle.
Since we now know the power and the voltage across the load, if we rearrange the formula S = V I* to solve for I, we get:



Store the value found for the current as the variable named i for later calculations.
Press [F3] [ENTER] ENTER] to view the complex current as a magnitude and an angle.
Recognizing that this is a simple series circuit, we can say that the current through the load is the same as the current through the wires.



Multiply the current times the impedance of the wires to find the voltage dropped by the wires. Store this value to the variable vline for later calculations.
Press [F3] [ENTER] [ENTER] to view the complex voltage as a magnitude and an angle.
By KVL we can find the voltage supplied by the source.
Add the voltage dropped along the wires to the voltage across the load to find the value of the voltage supplied by the source. Store this value to the variable v for later calculations.
Press [F3] [ENTER] ENTER] to view the complex voltage of the source as a magnitude and an angle.



The real part of the complex power of the line is the real or average power and the imaginary part of complex power of the line is the reactive or quadrature power.
The real part of the complex power of the source is the real or average power and the imaginary part of complex power of the source is the reactive or quadrature power.
The power factor of the source was found by using the Execute feature from the text editor.









Following the formula relating the power factor to the power of the load and the rms values for current and voltage, we can solve for the rms value of the current.
Substituting in the value for the power consumed, and the new increased power factor, along with the voltage across the load we find the value for the current for the network with the increased power factor.
Following the formula P = I2R we multiply the current squared times the real part of the impedance of the wire to find the power used by the wire.