CHAPTERS 4

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CHAPTERS 4 5

• NETWORKS 1 0909201-01
• 4 October 2005 Lecture 5a
• ROWAN UNIVERSITY
• College of Engineering
• Dr Peter Mark Jansson, PP PE
• DEPARTMENT OF ELECTRICAL COMPUTER ENGINEERING
• Autumn Semester 2005 Quarter One

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admin
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networks I

• Todays learning objectives
• define when to best apply new methods for
  analyzing circuits
• introduce chapter 5 key concepts

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when is it best to use node-voltage vs.
using mesh-current

• it depends
• when circuit contains only voltage sources use
  m-c
• when circuit contains only current sources use
  n-v
• when it has both, use either, but look to
  minimize your equations (how many nodes vs.
  meshes?)

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important concepts in ch. 4

• Node Voltage Method is an easy use of KCL and
  Ohms Law.
• Mesh Current Method is an easy use of KVL and
  Ohms Law.
• Excellent methods for handling dependent voltage
  and current sources when adding currents and/or
  voltages.

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new concepts from ch. 5

• electric power for cities
• source transformations
• superposition principle
• Thevenins theorem
• Nortons theorem
• maximum power transfer

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electric power to the cities

• generation ? transmission ? distribution
• the network of electric power

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Basic Components of Electric Power
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Electric Power Delivery Efficiency
Source PJM Website
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Why increase voltage to transmit electricity?

• power loss on conductors
• are i2R
• increase efficiency of delivery by decreasing
  losses (example 23 Megawatts)
• 230kV 100 Amps vs. 23kV 1000Amps
• i2R 10,000R vs. 1,000,000R 100x

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Electric Power Production Technologies
Source EPRI Website
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Learning checks 5 6

• LC 1
• Put these in the appropriate order as to how
  electric power gets to your home (HINT Last
  number is 2)
• 1 Substation (to step voltage down)
• 2 Electric meter on your home
• 3 Transmission line
• 4 Distribution line
• 5 Electric power generator
• 6 Substation (to step voltage up)
• Answer should look like -----

LC 2 What is the equation for power
losses on an electric power line?
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source transformations

• procedure for transforming one source into
  another while retaining the terminal
  characteristics of the original source
• producing an equivalent circuit

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why transform?

• One more technique to add to your tool kit of
  reducing and solving circuits
• it may be easier to solve a circuit when the
  sources are all the same type (i.e., current or
  voltage)

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lets transform this circuit
Rs
a
?
_
vs
b
?
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to this circuit

a
?
is
Rp

b
?
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for any applied load R

• both circuits must have the same characteristics
• lets apply the extreme values of R

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When R 0

• we essentially have a short circuit
• therefore the short circuit current of each
  circuit must be equal
• for first circuit
• i vs/Rs
• for second circuit
• i is , so
• To be equivalent circuits is vs/Rs

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When

• we essentially have an open circuit
• therefore the open circuit voltage of each
  circuit must be equal
• for second circuit
• v is Rp
• from the first circuit v vs
• so
• To be equivalent circuits vs is Rp

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combining what we know

• when R 0
• is vs/Rs
• when
• vs is Rp
• so from R 0 to
• vs (vs/Rs) Rp
• Therefore Rs Rp

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dual circuits

• circuits are said to be duals when the
  characterizing equations of one network can be
  obtained by simple interchange of v and i and G
  and R
• Rp 1/Gp
• is vs Gp and vs is Rs

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examples this circuit is equivalent to
Rs 12O
a
?
_
36V
b
?
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this one..
Rp Rs 12?
vs is Rs or is vs/Rp
a
?
is ?A
Rp ??
3 A
12 ?
b
?
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examples make these circuits equivalent
Rs
a
?
_
12V
b
?
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how..
Rp Rs 10?
So vs 12V is Rs or is vs/Rp
a
?
is
Rp 10?
1.2 A
b
?
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examples make these circuits equivalent
Rs
a
?
_
12V
b
?
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how..
Rp Rs 10?
So vs -12V is Rs or is vs/Rp
a
?
is
Rp 10?
-1.2 A
b
?
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examples make these circuits equivalent
Rs 8?
a
?
_
vs
b
?
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how..
Rp Rs 8 ?
So vs is Rs or 24 V
a
?
is 3 A
Rp
b
?
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examples make these circuits equivalent
Rs 8?
a
?
_
vs
b
?
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how..
Rp Rs 8 ?
So vs is Rs .. Show vs as LC 3
a
?
is 3 A
Rp
b
?
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example 5.3-2

• a little more complex transformation

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superposition principle (SP)

• In a single element
• if the application of
• i1 yields v1 and i2 yields v2 then
• i1 i2 will yield v1 v2
• the total effect of several causes acting
  simultaneously is equal to the sum of the effects
  of the individual causes acting one at a time

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SP can help

• how to apply SP to simplify analysis
• disable all but one source
• find partial response to that source
• disable all but the next source
• find partial response to that source
• iterate
• sum all the partial effects to get total

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How to continued

• set current sources to 0
• Note zero current sources are open circuits
• Memory Aid no current ? no connection
• solve for partial effect
• set voltage sources to 0
• Note zero voltage sources are short circuits
• Memory Aid no voltage rise ? across a short
• solve for partial effect
• sum effects together

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examples

• p. 156
• see example 5.4-1
• Now you try it - see 5.4-1 on p. 183
• Show v as LC 4

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Thévenins theorem

• GOAL reduce some complex part of a circuit to
  an equivalent source and a single element (for
  analysis)
• THEOREM for any circuit of resistive elements
  and energy sources with a terminal pair, the
  circuit is replaceable by a series combination of
  vt and Rt

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examples

• see example 5.5-1

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Thévenin method

• If circuit contains resistors and ind. sources
• Connect open circuit between a and b. Find voc
• Deactivate source(s), calc. Rt by circuit
  reduction
• If circuit has resistors and ind. dep. sources
• Connect open circuit between a and b. Find voc
• Connect short circuit across a and b. Find isc
• Connect 1-A current source from b to a. Find vab
• NOTE Rt vab / 1 or Rt voc / isc
• If circuit has resistors and only dep. sources
• Note that voc 0
• Connect 1-A current source from b to a. Find vab
• NOTE Rt vab / 1

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HW example

• see HW problem 5.5-1

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Nortons theorem

• GOAL reduce some complex part of a circuit to
  an equivalent source and a single element (for
  analysis)
• THEOREM for any circuit of resistive elements
  and energy sources with a terminal pair, the
  circuit is replaceable by a parallel combination
  of isc and Rn (this is a source transformation of
  the Thevenin)

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Norton equivalent circuit

a
?
isc
Rn Rt

b
?
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Norton method

• If circuit contains resistors and ind. sources
• Connect short circuit between a and b. Find isc
• Deactivate ind. source(s), calc. Rn Rt by
  circuit reduction
• If circuit has resistors and ind. dep. sources
• Connect open circuit between a and b. Find voc
  vab
• Connect short circuit across a and b. Find isc
• Connect 1-A current source from b to a. Find vab
• NOTE Rn Rt vab / 1 or Rn Rt voc / isc
• If circuit has resistors and only dep. sources
• Note that isc 0
• Connect 1-A current source from b to a. Find vab
• NOTE Rn Rt vab / 1

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HW example

• see HW problem 5.6-2

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maximum power transfer

• what is it?
• often it is desired to gain maximum power
  transfer for an energy source to a load
• examples include
• electric utility grid
• signal transmission (FM radio receiver)
• source ? load

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maximum power transfer

• how do we achieve it?

a
?
Rt
_
vt or vsc
RLOAD
b
?
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maximum power transfer

• how do we calculate it?

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maximum power transfer theorem

• So
• maximum power delivered by a source represented
  by its Thevenin equivalent circuit is attained
  when the load RL is equal to the Thevenin
  resistance Rt

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efficiency of power transfer

• how do we calculate it for a circuit?

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Norton equivalent circuits

• using the calculus on pi2R in a Norton
  equivalent circuit we find that it, too, has a
  maximum when the load RL is equal to the Norton
  resistance Rn Rt

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HW example

• see HW problem 5.7-5

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