Title: PEElectrical Review Course Class 2 AC Circuits
 1PE-Electrical Review Course - Class 2 (AC 
Circuits)
Class 2 - AC Circuits Objectives This review 
session is designed to review material and 
provide practical examples such that the student 
will be able to
1) Identify properties of periodic waveforms and 
calculate the DC and RMS values of periodic 
waveforms. 2) Perform calculations using complex 
numbers and analyze AC circuits using phasor 
analysis. 3) Use various analysis techniques to 
analyze AC circuits, including mesh and node 
equations, source transformations, superposition, 
Thevenins and Nortons theorems. 4) Perform 
power calculations on AC circuits and systems in 
order to determine real power, reactive power, 
apparent power, complex power, and power 
factor. 5) Perform power factor correction on AC 
circuits. 6) Analyze 3-phase circuits, including 
wye and delta generators as well as wye and delta 
loads, with an emphasis on unbalanced 
systems. 7) Perform power calculations in 3-phase 
systems, including the use of the 2-wattmeter 
method and the 3-wattmeter method.
Reading material 1) EE Ref. Manual, 6th Ed., 
Camara, Chapter 27  AC Circuit 
Fundamentals 2) Handout Extra Problems for 
Week 2 
 2PE-Electrical Review Course - Class 2 (AC 
Circuits)
Waveforms Periodic Waveforms A periodic 
waveform satisfies the relationship v(t1)  v(t1 
 T) where T  period (in seconds) f  frequency 
(in Hertz, Hz) w  radian frequency (in rad/s)
Sinusoidal Waveforms Sinusoidal waveforms are 
periodic waveforms described by v(t)  Vpcos(wt  
? ) where Vp  peak or maximum voltage 
 and ?  phase angle in degrees where a shift to 
the left is positive and a shift to the right is 
 negative (as with any function) 
 3PE-Electrical Review Course - Class 2 (AC 
Circuits)
Leading and Lagging Waveforms A waveform leads 
another when a specific point on the waveform 
(such as a zero crossing) occurs earlier in time 
than it does for the second waveform. In the 
example shown, V1 leads V2 by angle q 
and V2 lags V1 by angle q
Note The time difference, tD, between V1 and 
V2 can be converted to an angle using
Average (DC) Value of Periodic Waveforms VDC  
VAVG  DC or average voltage, which is defined as 
follows Similarly, 
Two ways to find the average (DC) value 1) By 
inspection (or by using a simple or weighted 
average) 2) By integration (using the integral 
definitions shown above) 
 4PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 1
Find the average value, IAVG , of the following 
periodic waveform
Example 2
Find the average value, VAVG , of the following 
periodic waveform 
 5PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 3
Find the average value, VDC , of the following 
periodic waveform (the output of an ideal 
half-wave rectifier) 
 6PE-Electrical Review Course - Class 2 (AC 
Circuits)
RMS Value of Periodic Waveforms VRMS  VEFF  
Root-Mean-Square (RMS) or effective voltage, 
which is defined as follows Similarly, 
The name Root-Mean-Square essentially gives the 
definition VRMS  the square-root of the average 
value of the function squared Two ways to find 
the RMS value 1) By inspection (square the 
function, find its average, and take the square 
root) 2) By integration (using the integral 
definitions shown above) 
 7PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 4
Find the RMS value of the following periodic 
waveform
Find the RMS value of the following periodic 
sinusoidal waveform
Example 5 
 8PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 6
Find the RMS value of the following periodic 
waveform
Example 7
A household AC outlet is commonly described by 
VRMS  120V and f  60 Hz. a) Describe the 
waveform as a function v(t) b) Sketch the 
waveform showing the peak value and the time for 
each of the first 2 periods. 
 9PE-Electrical Review Course - Class 2 (AC 
Circuits)
AC Circuit Analysis (also called sinusoidal 
steady-state analysis or phasor analysis) Before 
reviewing AC Circuit Analysis, a review of 
complex numbers is useful. Complex Numbers A 
complex number can be expressed in two 
forms 1) Rectangular form 
2) Polar form 
 10PE-Electrical Review Course - Class 2 (AC 
Circuits)
Converting between rectangular form and polar 
form
Convert 10  j20 to polar form.
Example 8
Convert 25/30o to rectangular form.
Example 9
Converting between rectangular form and polar 
form using calculators See the handout Complex 
Numbers using the Casio fx-115MS that contains 
examples of complex number calculations for AC 
circuits.
Demonstrate complex number calculations using one 
or more of the calculators listed above.
Demonstration 
 11PE-Electrical Review Course - Class 2 (AC 
Circuits)
Definition A phasor is a complex number in 
polar form that represents magnitude and phase 
angle of a sinusoidal voltage or current.
Express each quantity below in the other form.
Example 10
Phasors - relative quantities Discuss each of 
the following a) should phasors use sin or cos 
as a reference? b) should phasors use peak or 
RMS voltages for the magnitudes? 
 12PE-Electrical Review Course - Class 2 (AC 
Circuits)
Complex Impedances Components are represented 
in AC circuits as follows Z  impedance or 
complex impedance (in ?)
Resistors
Capacitors
Inductors
KVL and KCL in AC Circuits KVL and KCL are 
satisfied in AC circuits using phasor voltages 
and currents. They are not satisfied using the 
magnitudes of the voltages and the currents. 
Consider how KVL applies to the RL circuit below 
in both the time-domain and in the phasor domain. 
 13PE-Electrical Review Course - Class 2 (AC 
Circuits)
AC Circuit Analysis Procedure 1) Draw the phasor 
circuit A) Represent voltage and current sources 
as phasors B) Represent components (resistors, 
inductors, and capacitors) as complex 
impedances. 2) Analyze the circuit in the same 
way that you might analyze a DC circuit. (For 
example, you can combine series/parallel 
impedances like you would resistances and you can 
use node equations, mesh equations, source 
transformations, voltage and current division, 
superposition, etc., to analyze the 
circuit.) 3) Convert the final phasor result back 
to the time domain. 
 14PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 11
Analyze the circuit shown below using phasor 
analysis. 
 15PE-Electrical Review Course - Class 2 (AC 
Circuits)
Complex Power In DC circuits we often calculate 
P, real or average power. In AC circuits, 
several terms related to power are often used. 
Also note that the use of phasors with RMS values 
is very convenient for power calculations. 
I
RMS  
Z  
VRMS  
_ 
 16PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 12
Determine the real power, reactive power, complex 
power, apparent power, and power factor for the 
circuit shown below (include units with each 
answer). Note that this is the same circuit used 
in Example 11. Find the power quantities using 
the total current from Example 11 and the 
relationship 
 17PE-Electrical Review Course - Class 2 (AC 
Circuits)
Other Useful Relationships for Calculating Power 
Quantities Power delivered by the source  Power 
dissipated by the circuit so another way to 
calculate power is to find the power dissipated 
by each component. Only resistors dissipate real 
power and only capacitors and inductors dissipate 
reactive power. The power to each component can 
be calculated using the following relationships 
(note that only the magnitudes VRMS and IRMS are 
needed) 
Total power can then be calculated using
Two approaches to Calculating Power Quantities in 
a Circuit 1) 2) Method 1 was used in Example 
12. Method 2 will be used in Example 13. 
 18PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 13
Repeat Example 12 except find the power 
quantities by first using mesh equations to find 
the current through each component and then 
calculate the power for each component. The 
results should match those found in Example 12. 
 19PE-Electrical Review Course - Class 2 (AC 
Circuits)
Calculating Power in Systems Previous examples 
have focused on circuits. Sometimes the power 
requirements are known for components of a system 
and then it is necessary to calculate the total 
power for the overall system. This can be done 
by finding the real and reactive power for each 
component of the system. 
Example 14
Determine the power factor for the system shown 
below. 
 20PE-Electrical Review Course - Class 2 (AC 
Circuits)
- Power Factor Correction ( common PE Exam topic) 
- Power companies charge their customers based on 
 the real power (P) used.
- If the customer has a significant amount of 
 reactive power (or a low power factor) this has
 no effect on P, but results in higher current
 levels and higher line losses (I2 R) .
- Power companies encourage their large customers 
 to correct their power factor. If their power
 factor drops too low, the power company may
 charge higher rates.
- Lower current levels benefit the customer as 
 well.
- Large companies often have lagging power factors 
 due to large amounts of machinery (inductive
 loads) and they can correct their power factor by
 adding in large parallel capacitors (or
 synchronous motors that act like capacitive
 loads).
- This generates negative reactive power which 
 cancels the positive reactive power due to the
 inductive loads resulting in a power factor which
 is near or close to unity.
- The corrected power factor results in lower 
 current levels and thus lower line
 losses (I2 R) as the power company delivers the
 power to the customer.
21PE-Electrical Review Course - Class 2 (AC 
Circuits)
Illustration of correcting power factor to unity 
 22PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 15
Correct the power factor to unity in the circuit 
below by adding a capacitor in parallel with the 
source. (This circuit was used in Examples 11 - 
13.) A) Determine the value of C to be added B) 
Determine the source current before and after 
correction 
 23PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 16
Correct the power factor to 0.95, lagging in the 
circuit below by adding a capacitor in parallel 
with the source. (This circuit was used in 
Examples 11 - 13.) A) Determine the value of C 
to be added B) Determine the source current 
before and after correction 
 24PE-Electrical Review Course - Class 2 (AC 
Circuits)
Complete the table shown below using the results 
from the last two examples 
 25PE-Electrical Review Course - Class 2 (AC 
Circuits)
- Three Phase Circuits and Systems ( common PE 
 Exam topic)
- 3-phase circuits have the following advantages 
- more efficient (smaller I2R losses) 
- less vibration in machinery 
- smaller conductors 
- Single-phase and 3-phase generators
26PE-Electrical Review Course - Class 2 (AC 
Circuits)
Phase sequences The three phases may be arranged 
in two possible phase sequences
1) abc (or positive) phase sequence
2) acb (or negative) phase sequence 
 27PE-Electrical Review Course - Class 2 (AC 
Circuits)
Balanced versus unbalanced generators Balanced 
generators have the same magnitude for each phase 
and exactly 120o of phase shift between each 
phase. Systems with balanced generators are easy 
to analyze . (Note unbalanced systems are 
commonly seen on the PE Exam.)
Example 17
For each case shown below, is the generator 
balanced? What is the phase sequence? Circle 
the correct responses.
Generator connections There are two common ways 
to connect the three phase generators 1) Wye 
(Y) connection 2) Delta (D) connection
Load connections There are two common ways to 
connect the three phase loads 1) Wye (Y) 
connection 2) Delta (D) connection 
 28PE-Electrical Review Course - Class 2 (AC 
Circuits) 
 29PE-Electrical Review Course - Class 2 (AC 
Circuits) 
 30PE-Electrical Review Course - Class 2 (AC 
Circuits)
Generator-load configurations (discuss each) 1) 
Y-Y a) 4-wire b) 3-wire 2) Y-D 3) D -D 4) 
D -Y 
 31PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 18 
 32PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 19 
 33PE-Electrical Review Course - Class 2 (AC 
Circuits)
Power calculations in 3-phase circuits ( common 
PE Exam topic) Power calculations can be made in 
several ways, including 1) finding the power 
dissipated by each component in the load 2) 
using the 3-wattmeter method 3) using the 
2-wattmeter method If a wattmeter is shown 
connected in a circuit, determine the phasor 
voltage across the wattmeter and the phasor 
current through the wattmeter and then use the 
fact that the wattmeter will read For example, 
consider the wattmeter shown below 
 34PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 20
Calculate the power dissipated by the load by 
calculating the power dissipated by each 
resistor. (Same circuit as in Example 19.) 
 35PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 21
Use the 3-wattmeter method to calculate the power 
dissipated by the load by calculating the power 
read by each of the three wattmeters. (Same 
circuit as in Example 19.) 
 36PE-Electrical Review Course - Class 2 (AC 
Circuits)
Example 22
Use the 2-wattmeter method to calculate the power 
dissipated by the load by calculating the power 
read by each of the two wattmeters. (Same 
circuit as in Example 19.) 
 37PE-Electrical Review Course - Class 2 (AC 
Circuits)
Line impedances For Y loads the line impedances 
can be added to the load impedances for 
calculating line currents. For D loads it is 
sometimes convenient to convert the D load to a Y 
load such that the line impedances can be added 
to the Y load. Y-D and D-Y Conversions
Note that in the special case where the load is 
balanced, the Y-D and D-Y equations reduce to