Download Lecture 2 Fundamentals of Electrical Engineering

Lecture 2
Fundamentals of Electrical
Engineering
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
1
“uphill: battery”
“downhill: resistor”
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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POWER AND ENERGY
p(t )  v (t )i (t )
Watts
t2
w   p(t )dt
Joules
t1
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Current is flowing in
the passive
configuration
If the current flows opposite to the passive
configuration, the power is given by p = -vi
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Resistors and Ohm’s Law
a
v  iR
vab  iab R
b
The units of resistance are Volts/Amp which are called
“ohms”. The symbol for ohms is omega: 
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Resistance Related to Physical
Parameters
R
L
A
 is the resistivity of the material used to fabricate
the resistor. The units of resitivity are ohm-meters
(-m)
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Power dissipation in a resistor
2
v
p  vi  Ri 
R
2
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
5
Lecture 3
Circuit Laws,
Voltage & Current Dividers
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
KIRCHHOFF’S CURRENT
LAW
• The net current entering a node is zero.
• Alternatively, the sum of the currents
entering a node equals the sum of the
currents leaving a node.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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KIRCHHOFF’S VOLTAGE
LAW
The algebraic sum of the voltages equals
zero for any closed path (loop) in an
electrical circuit.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
7
v  v1  v2  v3  iR1  iR2  iR3  i ( R1  R2  R3 )  iReq
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
i  i1  i2  i3 
 1
1
1 
v
v
v
v



 v 


R1 R2 R3
 R1 R2 R3  Req
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Circuit Analysis using
Series/Parallel Equivalents
1. Begin by locating a combination of
resistances that are in series or parallel.
Often the place to start is farthest from the
source.
2. Redraw the circuit with the equivalent
resistance for the combination found in
step 1.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
3. Repeat steps 1 and 2 until the circuit is
reduced as far as possible. Often (but not
always) we end up with a single source and
a single resistance.
4. Solve for the currents and voltages in the
final equivalent circuit.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Voltage Division
v1  R1i 
R1
v total
R1  R2  R3
v 2  R2 i 
R2
v total
R1  R2  R3
Of the total voltage, the fraction that appears across a
given resistance in a series circuit is the ratio of the given
resistance to the total series resistance.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Current Division
i1 
R2
v
itotal

R1 R1  R2
i2 
R1
v

itotal
R2 R1  R2
For two resistances in parallel, the fraction of the total
current flowing in a resistance is the ratio of the other
resistance to the sum of the two resistances.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Lecture 4
Node/Loop Analysis
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Node Voltage Analysis
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Lecture 5
Node/Loop Analysis
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Mesh Current Analysis
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Choosing the Mesh
Currents
When several mesh currents flow through
one element, we consider the current in
that element to be the algebraic sum of
the mesh currents.
Sometimes it is said that the mesh
currents are defined by “soaping the
window panes.”
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
13
Solve for the mesh
currents:
20i1  10(i1  i2 )  150  0
15i2  100  10(i2  i1 )  0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
20i1  10(i1  i2 )  150  0
15i2  100  10(i2  i1 )  0
Putting the equations into the standard format:
30i1  10i2  150
 10i1  25i2  100
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Super-mesh
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Combine meshes 1 and 2 into a supermesh. In other
words, we write a KVL equation around the periphery of
meshes 1 and 2 combined.
i1  2i 1 i3   4i2  i3   10  0
Mesh 3:
3i3  4i3  i2   2i3  i1   0
i2  i1  5
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Lecture 7
Thévenin Equivalent Circuits
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Thévenin Equivalent Circuits
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Thévenin Equivalent Circuits
Vt  voc
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Thévenin Equivalent Circuits
Vt
isc 
Rt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Thévenin Equivalent
Circuits
voc
Rt 
isc
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Finding the Thévenin
Resistance Directly
We can find the Thévenin resistance by
zeroing the sources in the original network
and then computing the resistance between
the terminals.
When zeroing a voltage source, it becomes a
short circuit. When zeroing a current source,
it becomes an open circuit.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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I n  isc
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Step-by-step
Thévenin/Norton-EquivalentCircuit Analysis
1. Perform two of these:
a. Determine the open-circuit voltage Vt = voc.
b. Determine the short-circuit current In = isc.
c. Zero the sources and find the Thévenin
resistance Rt looking back into the
terminals.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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2. Use the equation Vt = Rt In to compute
the remaining value.
3. The Thévenin equivalent consists of a
voltage source Vt in series with Rt .
4. The Norton equivalent consists of a
current source In in parallel with Rt .
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Source Transformations
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Maximum Power Transfer
The load resistance that absorbs the
maximum power from a two-terminal
circuit is equal to the Thévenin
resistance.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
SUPERPOSITION
PRINCIPLE
The superposition principle states
that the total response is the sum of
the responses to each of the
independent sources acting
individually. In equation form, this is
rT  r1  r2    rn
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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Superposition Principle
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
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