References
FE/EIT Review
John A. Camara, Electrical Engineering Reference Manual, 6th edition,
Professional Publications, Inc, 2002.
John A. Camara, Practice Problems for the Electrical and Computer
Engineering PE Exam, 6th edition, Professional Publications, Inc, 2002.
Circuits
National Council of Examiners for Engineering & Surveying,
Principles and Practice of Engineering, Electrical and Computer
Engineering, Sample Questions and Solutions, NCEES, 2001.
Instructor: Russ Tatro
National Council of Examiners for Engineering & Surveying,
Fundamental of Engineering, Supplied-Reference Handbook, NCEES,
2008.
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Introduction
References
The morning FE examination will have 120 questions in a 4 hour
period. The questions cover 12 topic areas:
Michael A. Lindeburg, PE, FE Review Manual, Rapid Preparation for
the General Fundamentals of Engineering Exam, 2nd Edition,
Professional Publications, 2006.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Michael A. Lindeburg, PE, FE/EIT Sample Examinations, 2nd Edition,
Professional Publications, 2006.
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3
Mathematics
Engineering Probability and Statistics
Chemistry
Computers
Ethics and Business Practices
Engineering Economics
Engineering Mechanics (Statics and Dynamics)
Strength of Materials
Material Properties
Fluid Mechanics
Electricity and Magnetism
Thermodynamics
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Introduction
Introduction
Section XI. Electricity and Magnetism includes:
Section XI. Electricity and Magnetism includes:
A. Charge, Energy, current, voltage, power
F. Capacitance and Inductance
B. Work done in moving a charge in an electric field
relationship between voltage and work
G. Reactance and impedance, susceptance and admittance
4
H. AC Circuits
C. Force between charges
I. Basic complex algebra
D. Current and voltage laws
Kirchhoff's voltage law, Kirchhoff's current law, Ohm’s law
E. Equivalent circuit
Series, Parallel, Thévenin/Norton equivalent
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1
Scope of the Systems
Units
You will be expected to analyze Linear, Lumped parameter, time
invariant systems.
Volt
The Potential difference is the energy required to move a unit
charge (the electron) through an element (such as a resistor).
Linear – response is proportional to V or I
(no higher order terms needed)
Amp
Lumped Parameter - Electrical effects happen instantaneously in the
system. Low frequency or small size (about 1/10 of the wavelength).
Electric current is the time rate of change of the charge, measured
in amperes (A).
Ohm
The resistance R of an element denotes its ability to resist the flow
of electric current; it is measured in ohms (Ω).
Time Invariant - The response of the circuit does NOT depend on
when the input was applied.
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Units
Algebra of Complex Numbers
Coulomb
The amount of charge that crosses a surface in one second when a
steady current of one ampere flows.
A complex number z, consists of the sum of real and imaginary
numbers.
z = a ± jb
Farad
Capacitance is the ratio of the charge on one plate of a capacitor to
the voltage difference between the two plates - measured in farads
(F).
The rectangular form can be found from a phasor with polar
magnitude c and an angle θ.
a = c cos θ
b = c sin θ
z = a + jb = c cos θ + jc sin θ
Henry
Inductance is the property whereby an inductor exhibits opposition
to the charge of current flowing through it - measured in henrys
(H).
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Algebra of Complex Numbers
Algebra of Complex Numbers
A complex number z, consists of the sum of real and imaginary
numbers.
Add or subtract complex numbers in the rectangular form.
z = a + jb
y = c − jd
z = a ± jb
z − y = (a − c) + j (b − d )
The phasor form (polar) can be found from the rectangular form as
follows:
Multiply or divide complex numbers in polar form.
c = a2 + b2
m=
⎛b⎞
θ = tan ⎜ ⎟
⎝a⎠
−1
z ⟨θ
y ⟨φ
=
z
⟨ (θ − φ )
y
⎛b⎞
z = c⟨θ = a 2 + b 2 ⟨ tan −1 ⎜ ⎟
⎝a⎠
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2
Algebra of Complex Numbers
Example - Algebra of Complex Numbers
Complex numbers can also be expressed in exponential form by use of
Euler’s Identity.
The rectangular form of a given complex number is
e
± jθ
= cos θ ± j sin θ
What is the number when using trigonometric functions?
The trigonometric functions then become:
cos θ =
e jθ + e − jθ
2
sin θ =
e jθ − e − jθ
2j
z = 3 + j4
D
( A) 5e j 36.86
( B) (5)(cos 36.86D + j sin 36.86D )
(C ) cos 0.64 + j sin 0.64
( D) (5)(cos 0.93 + j sin 0.93)
The correct answer is D.
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Electrostatics
Electrostatics
Electric charge is a fundamental property of subatomic particles.
An electric field E with units of volts/meter is generated in the vicinity
of an electric charge.
A Coulomb equals a very large number of charged particles. This
results in a Farad also being a very large number of charged particles.
Thus we usually deal with very small amount of charge.
The force applied by the electric field E is defined as the electric flux of
a positive charged particle introduced into the electric field.
F = QE
The charge of one electron is -1.602 x 10-19 C.
The work W performed on a moving charge QB a certain distance in a
field created by charge QA is given by
The charge of one proton is +1.602 x 10-19 C.
r2
r2
r1
r1
r2
QQ
QAQB
dr = A B
2
4πε
4
πε
r
r1
W = −Q ∫ E dL = − ∫ F dr = − ∫
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Electrostatics
Electrostatics
Work is performed only if the charges are moved closer or farther apart.
Current is the movement of charge.
For a uniform electric field (such as inside a capacitor), the work done
in moving a charge parallel to the E field is
W = −QΔV
The last equation says the work done is the charge times the change in
voltage the charge experienced by the movement.
The electric field strength between two parallel plates with a potential
difference V and separated by a distance d is
E=
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V
d
⎛1 1⎞
⎜ − ⎟
⎝ r2 r1 ⎠
By convention the current moves in a direction opposite to the flow of
electrons.
Current is measured in Amperes and is the time rate of change of
charge.
i (t ) =
dq(t )
dt
If the rate of change in the charge is constant, the current can be
written as
I=
dQ
dt
The above equations largely describe the behavior of a capacitor.
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3
Magnetic Fields
Practice Problem - Electrostatics
A magnetic field can exist only with two opposite and equal poles.
Determine the magnitude of the electric field necessary to place a 1 N force on an
electron.
While scientists have searched for a magnetic monopole it has not yet
been found.
A magnetic field induces a force on a stationary charge.
Conversely a moving charge induces a magnetic field.
F = −Q E
Thus
E=
The magnetic field density is a vector quantity and given by B
F
Q
=
1N
= 6.24 ×1018
1.602 ×10−19 C
N
C
= 6.24 ×1018 Vm
φ
Magnetic Flux
B= =
A
Area
An inductor (or transformer) relies on the magnetic field interaction
with moving charges to alter the behavior of a circuit.
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Practice Problem - Electrostatics
DC Circuits
A current of 10 A flows through a 1 mm diameter wire. What is the average number
of electrons per second that pass through a cross section of the wire?
DC Circuits include the following topics:
( A) 1.6 ×1018
electrons
sec
( B ) 6.2 ×1018
electrons
sec
(C ) 1.6 ×1019
electrons
sec
( D ) 6.3 ×1019
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DC Voltage
Resistivity
Resistors in Series and Parallel
Power in a Resistive Element
Capacitors
Inductors
electrons
sec
Capacitors in Series and Parallel
The closest answer is D.
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DC Circuits
Circuit Symbols
Electrical Circuits contain active and passive elements.
Active elements can generate electric energy – voltage sources, current sources, opamps
Passive elements absorb or store electric energy – capacitor, inductor, resistor.
An ideal voltage source supplies power at a constant voltage regardless of the current
the external circuit demands.
An ideal current source supplies power at a constant current regardless of the voltage
the external circuit demands.
Dependent sources deliver voltage and current at levels determined by voltages or
currents elsewhere in the circuit.
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4
DC Voltage
DC current
Symbol: V or E (electromotive force)
Symbol: A
Circuit usage: V
or v(t) when voltage may vary.
Circuit usage: I
or i(t) when current may vary with time.
Voltage is a measure of the DIFFERENCE in electrical potential
between two points.
Amperage is a measure of the current flow past a point.
Current THROUGH a circuit element.
Voltage ACROSS two points.
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(Coulomb per second)
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Resistivity
Definitions
A Direct Current (dc) is a current whose polarity remains constant with
time. The amplitude is usually considered to remain constant.
Symbol: R measured in ohms, Ω
Circuit usage: R
Resistance is the property of a circuit or circuit element to oppose
current flow.
The amplitude is usually considered to remain constant.
R=ρ
An Alternating Current (ac) is a current that varies with time.
A common form of AC is the sinusoidal power delivered by the power
company.
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Resistors in Series and Parallel
L
Length
= (resistivity )
A
Area
A circuit with zero resistance is a short circuit.
A circuit with an infinite resistance is a open circuit.
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Example - Calculating equivalent resistance
Series Resistors:
Req = R1 + R2 + ……
Req = ?
= 5 + 6 + 7 = 18Ω
The equivalent resistance REQ is larger than the largest resistor.
Parallel Resistors:
1
1
1
= +
+ ......
Req R1 R2
Req = ?
For Two Resistors in Parallel:
Req =
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R1 R2
R1 + R2
=
10(20) 200
=
30
10 + 20
= 6.67Ω
The equivalent resistance REQ is smaller than the smallest resistor.
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5
Power in a Resistive Element
Passive Sign Convention
The power dissipated across two terminals is
Use a positive sign for the power when:
Current is the direction of voltage drop.
P = ( ± ) VI
= (±)
V2
R
= (±) I 2R
P = the power in watts
V = the voltage in volts
I = the current in amperes
I.A.W. with the Passive Sign Convention
+ (positive) – element is absorbing power.
- (negative) – element is delivering power.
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Power - Example
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Capacitors
Symbol: F for capacitance
Power delivered/absorbed.
Circuit usage: C for capacitor
C=
εA
Q = CV
d
Capacitor resists CHANGE in voltage across it.
Passive charge storage by separation of charge - Electric field energy.
P = (+)
2
V
R
= (+)
dv
1 t
v(t ) = ∫ i dτ + v(t0 )
dt
C t0
The total energy (in Joules) stored in a capacitor is
i (t ) = c
( 5V ) = + 25V = + 1 Watt
( )
( )
4
100Ω
100Ω
2
energy =
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CV 2 VQ Q 2
=
=
2
2
2C
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Inductors
Capacitors and Inductors in Series and Parallel
Symbol: H for Henries
Capacitors add in parallel (CAP)
Circuit usage: L for inductor
L=
Nφ
I
CEQ = C1 + C2 + C3 + ...
Where N is the number of turns through a magnetic
flux φ which results from the current I.
Use the following form for series capacitance.
CEQ =
Inductor resists CHANGE in current thru it.
LEQ = L1 + L2 + L3 + ...
1 t
i (t ) = ∫ v dτ + i (t0 )
L t0
The total energy (in Joules) stored in an inductor is
Use the following form for parallel inductors
1
LEQ = 1 1
L1 + L2 + ...
LI 2
energy =
2
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1
+ C12 + ...
Inductors add in series (just like resistors)
Passive energy storage by creation of magnetic field.
di
v(t ) = L
dt
1
C1
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6
Example - Calculating equivalent capacitance
Example - Calculating equivalent inductance
= (1 + 2 + 3) H = 6 H
Leq = ?
Ceq = ?
= (1 + 2 + 3) μ F = 6 μ F
The equivalent capacitance Ceq is larger than the largest capacitor.
The equivalent inductance Leq is larger than the largest inductor.
Leq = ?
Ceq = ?
=
=
1
1
1μ F
+ 2μ F +
1
1
6 + 3+ 2 1
μF
6
=
1
( + + )
μF
6
= μ F = 0.55μ F
11
1
1
1
11 1
6 μF
=
1
=
1
3μ F
1
2
1
3
The equivalent capacitance Ceq is smaller than the smallest capacitor.
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DC Circuits
=
1
1H
1
=
+ 21H + 31H
6
H
11
1
6 + 3+ 2 1
6
H
=
1
11 1
6 H
= 0.55 H
The equivalent inductance Leq is smaller than the smallest inductor.
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DC Circuit Analysis
DC Circuit Analysis include the following topics:
Ohm’s Law
Kirchhoff's Laws
Short Break – 5 minutes
Rules for Simple Resistive Circuits
Superposition Theorem
Superposition Method
Loop-Current Method
Node-Voltage Method
Source Equivalents
Maximum Power Transfer
We will also briefly look at RC and RL Transients
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Ohm’s Law
Ohms Law:
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Definitions
V=IR
Circuit Connections:
This version of Ohm’s Law assumes a linear circuit.
R=
V
I
Ω=
V
A
Branch – a connection between two elements
Nodes – point of connection of two or more branches.
Loops – any closed path (start/end same point) in a circuit.
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Kirchhoff’s Laws
Rules for Resistive Circuits
Kirchhoff’s Current Law – sum of all current equals zero.
The current through a simple series circuit is the same in all circuit
elements.
sum of all currents in = sum of all currents out.
I = I R1 = I R2 = I R3
This is a restatement of conservation of charge.
∑I
in
The sum of all voltage drops across all elements is equal to the
equivalent applied voltage.
= ∑ I out
VEQ = V1 + V2 + ... = IREQ
Kirchhoff’s Voltage Law – sum of all voltages around a closed path is
zero.
∑V
=0
∑V
= ∑ Vdrop
closed path
rise
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Rules for Simple Resistive Circuits
Rules for Simple Resistive Circuits
Voltage Divider for Series Resistors
Current Divider for Parallel Resistors:
v1 =
R1
R1
v=
v
REQ
R1 + R2
v2 =
R2
R2
v=
v
REQ
R1 + R2
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i1 =
R2
i
R1 + R2
i2 =
45
R1
i
R1 + R2
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Example – Current Divider
Example - Voltage Divider
Find i2
Find voltage V2
v2 = vs
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R2
R1 + R2
= (10V )
2
4+2
= (10V )
2
6
=
10V
3
= 3.33V
i2 = i
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R1
R1 + R2
= (30mA)
9k
9k + 18k
= (30mA)
9
27
=
30mA
3
= 10mA
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8
Superposition Theorem
Superposition Method
1. Deactivate all independent sources except one.
Voltage source = zero when shorted.
Determine the contribution (responses) of each independent source to
the variable in question.
Then sum these responses to find the total response.
Current source = zero when opened.
2. Solve the simplified circuit.
3. Repeat until all independent sources are handled.
The circuit must be linear to use superposition.
4. Sum the individual responses to find the total response.
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Methods of Analysis
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Loop-Current Method
The loop-current is also known as the mesh current method.
Loop-Current Method – assign currents in a loop and then write the
voltages around a closed path (KVL).
A mesh is a loop which does not contain any other loops within it.
Assign mesh currents to all the (n) meshes.
Node-Voltage Method – assign a reference voltage point and write
current as voltages and resistances at the node. (KCL).
Apply KVL to each mesh. Express the voltages in terms of Ohm’s law
– i.e. currents and resistances.
Solve the resulting (n) simultaneous equations.
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Node-Voltage Method
Example – Loop-current Method
The node-voltage method is also known as nodal analysis.
1. Convert all current sources to voltage sources.
2. Chose a node as the voltage reference node. Usually this is the
circuit’s signal ground.
Write the mesh equation for loop I1.
−10V + I1 (10Ω) + ( I1 − I 2 ) (30Ω) + I1 (20Ω) = 0
3. Write the KCL equations at all unknown nodes
I1 (60Ω) − I 2 (30Ω) = 10V
Remember you are writing the sum of all currents entering a node are
equal to the sum of all current leaving a node.
Write the mesh equation for loop I2.
( I 2 − I1 ) (30Ω) + I 2 (10Ω) + I 2 (20Ω) = 0
A convention is to assume all currents are leaving the node - the
direction of the voltage drop is away from the node.
− I1 (30Ω) + I 2 (60Ω) = 0
Solve for the currents.
− I1 (30Ω) + I 2 (60Ω) = 0 ⇒ I 2 = 12 I1
10V
(60Ω) − 12 I1 (30Ω) = 10V ⇒ I1 =
= 0.22 Amps
45Ω
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I 2 = 12 I1 = 0.11Amps
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9
Example – Node-Voltage Method
Example – Node-Voltage Method
1
1 ⎞ V2
⎛ 1
V1 ⎜
+
+
= 10 Amp
⎟−
⎝ 1Ω 5Ω 2Ω ⎠ 2Ω
⎛ 17 ⎞ V2
V1 ⎜
= 10 Amp
⎟−
⎝ 10Ω ⎠ 2Ω
1
1 ⎞ V2
⎛ 1
+
+
= 10 Amp
V1 ⎜
⎟−
⎝ 1Ω 5Ω 2Ω ⎠ 2Ω
17V1 − 5V2 = 100V ⇒ V1 =
−
V1
⎛ 6 ⎞
+ V2 ⎜
⎟ = 2A
2Ω
⎝ 10Ω ⎠
100V + 5V2
17
Clear the fractions for node 2 equation and Sub above result in the equation for node 2.
Write the node equation at node 2.
V2 − V1 V2
+
− 2A = 0
2Ω
10Ω
V1
1 ⎞
⎛ 1
+ V2 ⎜
+
⎟ = 2A
2Ω
⎝ 2Ω 10Ω ⎠
One way to simplify is the clear the fractions. From node 1, we then have
Write the node equation for at node 1.
V1 − 10V V1
V −V
+
++ 1 2 =0
1Ω
5Ω
2Ω
−
−
−5V1 + 6V2 = 20V
V1
1 ⎞
⎛ 1
+ V2 ⎜
+
⎟ = 2A
2Ω
⎝ 2Ω 10Ω ⎠
Solve the simultaneous equations for V1 and V2.
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⎛ 100V + 5V2 ⎞
⇒ −5 ⎜
⎟ + 6V2 = 20V
17
⎝
⎠
⇒ 77V2 = 340V + 500V
⇒ V2 = 10.91V
500
⎛ −25
⎞
⇒⎜
+ 6 ⎟ V2 = 20V +
V
17
⎝ 17
⎠
⇒ V1 = 9.09V
55
Node-Voltage Method
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Source Equivalents
As you see from the last example, even relatively simple equations can
require significant time to solve.
Watch your time and tackle time intensive problems only if you have
time to spare.
A source transformation (equivalent) exchanges a voltage source with a
series resistance with a current source with a parallel resistor.
v s =i s R
is =
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vs
R
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Example – Source Transformation
Example – Source Transformation
The above equivalent circuit will behave exactly as the original circuit
would.
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10
Thevenin’s Theorem
Example – Thevenin’s Theorem
Thevenin’s Theorem: a linear two-terminal network can be replace with
an equivalent circuit of a single voltage source and a series resistor.
VTH is the open circuit voltage.
RTH is the equivalent resistance of the circuit.
Use node analysis to find voltage V1. Note that V1 = VTH!
V1 − 25V
V
+ 1 − 3A = 0
5Ω
20Ω
4V1 − 100V + V1 − 60V = 0
5V1 = 160V
V1 = VTH = 32V
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Example – Thevenin’s Theorem
Norton’s Theorem
Now deactivate all independent sources and find the equivalent resistance.
Norton’s Theorem: a linear two-terminal network can be
replaced with an equivalent circuit of a single current source
and a parallel resistor.
IN =
Now we can write the Thevenin equivalent circuit.
VTH
RTH
IN is the short circuit current.
RTH is the equivalent resistance of the circuit.
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Example – Norton’s Theorem
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Example – Norton’s Theorem
Now simplify the circuit by combining resistances and the current sources
5 / /20 =
5(20)
Ω = 4Ω
5 + 20
The current isc must be half the 8A input (resistive current divider with equal
resistances in each leg.)
If you already have the Thevenin equivalent circuit (previous example) - do not start
from scratch. However in this example, we will solve the circuit again.
Use a source transformation to put the circuit in terms of current sources.
Does the previous Thevenin equivalent circuit yield the same answer?
i=
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25V
= 5A
5Ω
ieq = 5 A + 3 A = 8 A
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11
Maximum Power Transfer
Maximum Power Transfer
The maximum power delivered to a load is when the load resistance
equals the Thevenin resistance as seen looking into the source.
The maximum power delivered to a load is when the load resistance
equals the Thevenin resistance as seen looking into the source.
RL = RTH
RL = RTH
The voltage across an arbitrary load is
VL = VTH
When the load resistance equals the Thevenin resistance, the maximum
power delivered to the load is given by
RL
RL + RTH
pmax =
The maximum power delivered to an arbitrary load is given by
2
⎛
RL ⎞
⎜ VTH
⎟
RL + RTH ⎠
V
RL
⎝
=
= VTH2
PL =
2
RL
RL
( RTH + RL )
2
VTH
4 RTH
2
L
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RC and RL Transients
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RC and RL Transients
The capacitor and inductor store energy.
An inductor stores energy in the form of a magnetic field.
A capacitor stores this energy in the form of an electric field.
If a charged inductor (with a steady current flowing) is connected to a
resistor, it will give up its energy over a short period of time as follows.
If a charged capacitor is connected to a resistor, it will give up its
energy over a short period of time as follows.
vc (t ) = vc (t = 0)e
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iL (t ) = iL (t = 0)e
− τt
Where τ = L/R.
Where τ = RC.
This leads to a short hand rule – a capacitor acts as an open circuit in a
DC circuit.
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− τt
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AC Circuits
This leads to a short hand rule – an inductor acts as a short circuit in a
DC circuit.
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Alternating Waveforms
AC Circuits include the following topics:
Alternating Waveforms
Sine-Cosine Relations
Phasor Transforms of Sinusoids
Average Value
The term alternating waveform describes any symmetrical waveform
including:
square, sawtooth, triangular, and sinusoidal waves
Effective or rms Values
Sinusoidal waveforms may be given by
Phase Angles
v (t ) = Vmax sin(ωt + θ )
Impedance
The phase angle θ describes the value of the sine function at t = 0.
Admittance
Ohm’s Law for AC Circuits
Complex Power
Resonance
Transformers
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Sine-Cosine Relations
Phasor Transforms of Sinusoids
We often need to write sine in terms of cosine and vice versa.
A convention is to express the sinusoidal in terms of cosine. Thus the
phasor is written as a magnitude and phase under the assumption the
underlying sinusoid is a cosine function.
cos(ωt ) = sin(ωt + π2 ) = − sin(ωt − π2 )
sin(ωt ) = cos(ωt − π2 ) = − cos(ωt + π2 )
The period of the waveform is T in seconds.
The angular frequency is ω in radians/second.
The frequency f in hertz is given by
f =
Trigonometric:
Vmax cos(ωt + φ )
Phasor:
Veff ⟨ φ
Rectangular:
Vreal + jVimag = Vmax (cos θ + j sin θ )
Exponential:
Vmax e jφ
1 ω
=
T 2π
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Average Value
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Effective or rms Value
The root mean squared (rms) of any periodic variable is given by
The average (or mean) value of any periodic variable is given by
T
X ave =
1
x(t ) dt
T ∫0
T
X rms =
The average value of a sinusoid is zero.
The rms value of a sinusoid is
X rms =
A waveform may be rectified which results in a different average
value.
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x (t ) dt
T ∫0
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X max
2
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Phase Angles
Impedance
It is common to examine the timing of the peak of the volt versus the
timing of the peak of the current. This is usually expressed as a phase
shift.
The term impedance Z in units of ohms describes the effect circuit
elements have on magnitude and phase.
A capacitor’s behavior may be described by its’ phasor as
ic(t) = vc(t)∠90º.
Z = R ± jX = Y ⟨θ
The resistor has only a real value. Z = R
The current in a capacitor leads the voltage by 90º.
An inductor’s behavior may be described by its’ phasor as
vL(t) = iL(t)∠90º.
1
−j
=
jωC ωC
The capacitor has a negative imaginary value.
Zc =
The inductor has a postive imaginary value.
Z L = jω L
The current in an inductor lags the voltage by 90º.
R = Z cos θ
X = Z sin θ
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Admittance
Ohm’s Law for AC Circuits
The reciprocal of impedance is the complex quantity admittance Y.
Ohm’s Law for AC circuits is valid when the voltages and currents are
expressed in similar manner. Either peak or effective but not a mixture
of the two.
Y=
1
Z
V = IZ
The reciprocal of resistive part of the impedance is conductance G.
G=
1
R
B=
1
X
Y = G + jB
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Complex Power
Complex Power
The complex power vector S is also called the apparent power in units
of volts-amps (VA).
The power factor angle is defined as
S is the vector sum of the real (true, active) power vector P and the
imaginary reactive power vector Q.
p. f . = cos θ
Since the cosine is positive for both positive and negative angles – we
must add a description to the power factor angle.
S = I * V = P + jQ
Lagging p.f. is an inductive circuit.
The real power P in units of watts (W) is
Leading p.f. is a capacitive circuit.
P = 12 Vmax I max cos θ = Vrms I rms cos θ = S cos θ
Power factor correction is the proces of adding inductance or
capacitance to a circuit in order to achieve a p.f. = cos (0) = 1.
The reactive power Q in units of volts-amps reactive (VAR) is
Q = 12 Vmax I max sin θ = Vrms I rms sin θ = S sin θ
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Resonance
The average power of a purely resistive circuit is
Pave = Vrms I rms =
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In a resonant circuit at the resonant frequency, the input voltage and
current are in phase and the phase angle is zero.
2
Vrms
2
= I rms
R
R
Thus the circuit appear to be purely resistive in its response to this AC
voltage (again at the resonant frequency).
For a purely reative load p.f. = 0 and the real aveage power = 0.
The circuit is characterized in terms of resonant frequency ω0, the
bandwidth B, and the quality factor Q.
Pave = Vrms I rms cos 90D = 0
Each of these parameters can be found from the circuit element values.
ω0 =
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= 2π f 0
LC
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Transformers
Good Luck on the FE/EIT Exam!
Transformers are used to change voltage levels, match impedances,
and electrical isolate circuits.
It is a time exam. Answer what you know. Mark what you might
know and come back later. Do not get bogged down on a few long
questions. Move along!
The turns ratio a indicates how the magnetic flux links to the mutual
inductances of the transformer.
It is a multiple choice exam. Look for hints in the answers.
V
I
N
a = 1 = primary = sec ondary
N 2 Vsec ondary
I primary
If totally in doubt – Guess. Use your intuition and science to guess.
A lossless transformer is called an ideal transformer. The secondary
impedance can be expressed as a reflected impedance given by
Z EQ =
V primary
I primary
= Z primary + a 2 Z sec ondary
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FE/EIT Review
Circuits
Instructor: Russ Tatro
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