Download R - s3.amazonaws.com

CHAPTERS 4 & 5
NETWORKS 1:
0909201-03/04
18 November 2003 – Lecture 5a
ROWAN UNIVERSITY
College of Engineering
Dr Peter Mark Jansson, PP PE
DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING
Autumn Semester 2003
networks I
Today’s learning objectives –

define and apply new methods for
analyzing circuits: mesh current analysis
 w/ current sources
 w/ current and voltage sources
 w/ dependent sources

introduce chapter 5 key concepts
key concepts in ch 4
node voltage analysis of circuits w/



current sources - done
voltage sources - done
dependent sources - done
mesh current analysis of circuits w/



current sources
voltage sources
dependent sources
Mesh-Current Method
A mesh is a special case of a loop


contains no other loops within it
planar networks only
 can be drawn in a plane
 contains no crossovers

Mesh current is the current that flows
through all the elements in a loop (and the
convention is clockwise)
MESH-CURRENT
METHOD
R
R
3
1
V
+
–
i1
R2
i2
Current in R1 is i1
Current in R3 is i2
But what is the current in R2?
R4
MESH-CURRENT
METHOD
R
R
3
1
V
+
–
i1
R2
i2
R4
The two currents, i1 and i2, are flowing
through R2 in opposite directions.
The current in R2 is the difference between i1 and i2.
But which one is positive and which one is negative?
MESH-CURRENT
METHOD
R
R
3
1
V
+
–
i1
R2
i2
R4
It depends on which mesh you are adding voltages in.
In mesh 1 i1 is positive, in mesh 2 i2 is positive.
MESH-CURRENT METHOD (v)
we use KVL and Ohm’s law around mesh
move in the mesh in a clockwise direction
if a ‘+’ sign is encountered first ‘add’ the
mesh current
is a ‘–’ sign is encountered first ‘subtract’
put the current of mesh you are in first for
elements where two currents are flowing
MESH-CURRENT
METHOD
R
R
3
1
+
+
+
+
V
+
–
i1
R2
+
i2
Mesh 1:
 V  R1i1  R2 i1  i2   0
Mesh 2:
 R2 i2  i1   R3 i2  R4 i2  0
R4
MESH-CURRENT METHOD (i )
we use KVL and Ohm’s law around mesh
move in the mesh in a clockwise direction
if a current source is encountered ‘add’ the
voltage associated with it
follow passive sign convention for other
resistive elements encountered
put the current of mesh you are in first for
elements where two currents are flowing
Examples
Example 4.6-1
page 121 Dorf & Svoboda
Problem 4.6-3
page 141 Dorf & Svoboda
MESH-CURRENT METHOD
WITH A CURRENT SOURCE
R
R1
3
+
+
+
+
I
i1
R2
i2
R4
+
Mesh 1:
 v I  R1i1  R2 i1  i2   0
Mesh 2:
 R2 i2  i1   R3 i2  R4 i2  0
2 eq. / 3 unk.?
No, i1 = I
NOTE:
Either mesh-current method is useful
and can be used depending upon your
preference
One is merely a mirror image of the
other:


voltage +, others negative
voltage -, others all positive
Example
Problem 4.7-7
page 142 Dorf & Svoboda
MESH-CURRENT METHOD
WITH A DEPENDENT SOURCE
R
R1
3
+
+
+
+
2ia
i1
R2
i2
ia
R4
+
Mesh 1:
Mesh 2:
 v 2 ia  R1i1  R2 i1  i2   0
 R2 i2  i1   R3 i2  R4 i2  0
2 eq. / 3 unk.?
No: i1 = 2ia
ia = -i2
i1 = -2i2
Example(s)
Problem 4.7-10
Problem 4.7-13
page 142-143 Dorf & Svoboda
MESH-CURRENT METHOD
WITH A DEPENDENT V-SOURCE
R
R1
3
+
+
+
+
2ia
+
_
i1
R2
i2
ia
R4
+
Mesh 1:
Mesh 2:
 2 ia  R1i1  R2 i1  i2   0
 R2 i2  i1   R3 i2  R4 i2  0
2 eq. / 3 unk.?
No: ia = -i2
2ia = -2i2
when to use node-voltage
vs. mesh-current
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?)
important concepts in ch. 4
Node Voltage Method is an easy
combination of KCL and Ohm’s Law.
Mesh Current Method is an easy
combination of KVL and Ohm’s Law.
Excellent methods for handling
dependent voltage and current sources
when adding currents and/or voltages.
new concepts from ch. 5
electric power for cities
source transformations
superposition principle
Thevenin’s theorem
Norton’s theorem
maximum power transfer
electric power to the cities
generation  transmission  distribution
the network of electric power
Basic Components of Electric Power:
How electricity gets to you
When electricity leaves a po wer
plan t (1), its voltage is increased
at a “ste p-up” substation (2).
Next, the energ y tr ave ls along a
transmi ssion line to the area
where the p ower is neede d (3 ).
Once th ere, the vo ltag e is
decreased, or “ stepped -down,”
at a noth er substati on ( 4), and a
distribution power line (5) carr ies
the electricity until it reaches a
home or business (6).
– EEI, Getti ng Ele ctri city Whe re It’s Ne ede d,
May 20 00
Electric Power Delivery Efficiency
Source: PJM Website
Electric Power Production Technologies
Source: EPRI Website
source transformations
procedure for transforming one source
into another while retaining the terminal
characteristics of the original source
producing an equivalent circuit
why transform?
it may be easier to solve a circuit
when the sources are all the same type
(i.e., current or voltage)
let’s transform this circuit…
Rs
+
О
a
О
b
vs
_
to this circuit…
•
is
О
a
О
b
Rp
•
for any applied load R
both circuits must have the same
characteristics
let’s apply the extreme values of R

R=0

R=
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…
i = vs/Rs
 s
When R =

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…

vs= is Rp
combining what we know…
when R= 0

is = vs/Rs
when R =

vs= is Rp
so from R= 0 to

vs = (vs/Rs) Rp
Therefore Rs = Rp
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
examples:
this circuit is equivalent to…
Rs= 12Ω
О
+
a
36V
_
О b
this one…..
Rp = Rs = 12
vs= is Rs or is = vs/Rp
О
a
О
b
is = ?A
3A
Rp = ?
12 
examples:
make these circuits equivalent…
Rs
О
+
a
12V
_
О b
how…..
Rp = Rs = 10
So vs= 12V =is Rs or is = vs/Rp
О
a
О
b
is
1.2 A
Rp = 10
examples:
make these circuits equivalent…
Rs
О
_
a
12V
+
О b
how…..
Rp = Rs = 10
So vs= -12V =is Rs or is = vs/Rp
О
a
О
b
is
-1.2 A
Rp = 10
examples:
make these circuits equivalent…
Rs = 8
О
+
a
vs
_
О b
how…..
Rp = Rs = 8 
So vs= is Rs or 24 V
О
a
О
b
is =3 A
Rp
examples:
make these circuits equivalent…
Rs = 8
О
+
a
vs
_
О b
how…..
Rp = Rs = 8 
So vs= is Rs or -24 V
О
a
О
b
is =3 A
Rp
example 5.3-2
a little more complex transformation
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
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
How to… continued
set current sources to 0 (open circuits)
solve for partial effect
set voltage sources to 0 (short circuits)
solve for partial effect
sum effects
examples
see examples 5.4-1 and 5.4-2
Thévenin’s 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
examples
see example 5.5-1
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
HW example
see HW problem 5.5-1
Norton’s 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)
Norton equivalent circuit
•
isc
О
a
О
b
Rn = Rt
•
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
HW example
see HW problem 5.6-2
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
maximum power transfer
how do we achieve it?
О
a
Rt
+
vt or vsc
RLOAD
_
О
b
maximum power transfer
how do we calculate it?
p  i 2 RL
vs
i
RL  Rt
vs
p(
) 2 RL
RL  Rt
dp

0
dRL
RL  Rt
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
efficiency of power transfer
how do we calculate it for a circuit?
  pout / pin
2
vs
vs
pin  vs i  vs (
)
RL  Rt
2 RL
2
pout
vs
vs
2
 pmax (
) 
RL  RL
4 RL
  pout / pin  50% max
Norton equivalent circuits
using the calculus on p=i2R 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
HW example
see HW problem 5.7-6