1. DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING
CHAPTERS 4 & 5
NETWORKS 1: /03
18 November 2002 – Lecture 5a
ROWAN UNIVERSITY
College of Engineering
Professor Peter Mark Jansson, PP PE
DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING
Autumn Semester 2002

2. networks I Today’s learning objectives – describe a supernode
define and apply new methods for analyzing circuits: mesh current analysis
w/ current sources
w/ current and voltage sources
w/ dependent sources
describe a supermesh
introduce chapter 5 key concepts

3. key concepts in ch 4 node voltage analysis of circuits w/
current sources - done
voltage sources - done
dependent sources – done
supernode concept
mesh current analysis of circuits w/
current sources
voltage sources
dependent sources
supermesh

4. Supernode
consists of two nodes connected by an independent or dependent voltage source.
Using KCL we can say that the algebraic sum of all currents into a supernode is zero at all times

5. illustration

6. 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)

7. But what is the current in R2?
MESH-CURRENT METHOD + – V R1 R2 R3 R4 i1 i2
Current in R1 is i1
Current in R3 is i2
But what is the current in R2?

8. MESH-CURRENT METHOD + – V R1 R2 R3 R4 i1 i2
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?

9. MESH-CURRENT METHOD + – V R1 R2 R3 R4 i1 i2
It depends on which mesh you are adding voltages in.
In mesh 1 i1 is positive, in mesh 2 i2 is positive.

10. 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

11. MESH-CURRENT METHOD + – V R1 R2 R3 R4 i1 i2 + + + + +
Mesh 1:
Mesh 2:

12. Examples Example 4.6-1 page 126 Dorf & Svoboda Problem 4.6-3

13. 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

14. MESH-CURRENT METHOD WITH A CURRENT SOURCE+ I
Mesh 1:
2 eq. / 3 unk.?
No, i1 = I
Mesh 2:

15. 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

16. Example
Problem 4.7-7
page 148 Dorf & Svoboda

17. MESH-CURRENT METHOD WITH A DEPENDENT SOURCE+ + + + i1 i2 ia R4
2ia R2 +
2 eq. / 3 unk.?
Mesh 1:
No: i1 = 2ia
Mesh 2:
ia = -i2
i1 = -2i2

18. Example(s)
Problem
Problem
page 149 Dorf & Svoboda

19. MESH-CURRENT METHOD WITH A DEPENDENT V-SOURCE+ + + + i1 i2 ia + R4
2ia _ R2 +
2 eq. / 3 unk.?
Mesh 1:
No: ia = -i2
Mesh 2:
2ia = -2i2

20. 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?)

21. supermesh
one larger mesh created from two or more meshes that have an independent or dependent current source in common
reduces the number of KVL loops and allows KVL to be applied around the periphery of the supermesh

22. example 2.8-1 we’ll review just the mesh and supermesh portion…
see page 150

23. 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.

24. new concepts from ch. 5 electric power for cities
source transformations
superposition principle
Thevenin’s theorem
Norton’s theorem
maximum power transfer

25. electric power to the cities
generation  transmission  distribution
the network of electric power

26. Basic Components of Electric Power:

27. Electric Power Delivery Efficiency
Source: PJM Website

28. Electric Power Production Technologies
Source: EPRI Website

29. source transformations
procedure for transforming one source into another while retaining the terminal characteristics of the original source
producing an equivalent circuit
identical effect at terminals, not within the circuits themselves

30. why transform?
it may be easier to solve a circuit when the sources are all the same type (i.e., current or voltage)

31. let’s transform this circuit…Rs О a + _ vs О b

32. to this circuit… • a О is Rp • О b

33. for any applied load R
both circuits must have the same characteristics
let’s apply the extreme values of R
R = 0
R = 

34. 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…
is = vs/Rs

35. 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

36. 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

37. 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

38. examples: this circuit is equivalent to…
Rs= 12Ω О a + _
36V b О

39. this one….. vs= is Rs or is = vs/Rp Rp = Rs = 12 is = ?A Rp = ? 3 AО
is = ?A
Rp = ?
3 A
12  О b

40. examples: make these circuits equivalent…Rs О a + _
12V b О

41. how….. So vs= 12V =is Rs or is = vs/Rp Rp = Rs = 10 is Rp = 10 1.2 AО is
Rp = 10
1.2 A О b

42. examples: make these circuits equivalent…Rs О a _ +
12V b О

43. how….. So vs= -12V =is Rs or is = vs/Rp Rp = Rs = 10 is Rp = 10a О is
Rp = 10
-1.2 A О b

44. examples: make these circuits equivalent…
Rs = 8 О a + _ vs b О

45. how…..
Rp = Rs = 8 
So vs= is Rs or 24 V a О
is =3 A Rp О b

46. examples: make these circuits equivalent…
Rs = 8 О a + _ vs b О

47. how…..
Rp = Rs = 8 
So vs= is Rs or -24 V a О
is =3 A Rp О b

48. example 5.3-2
a little more complex transformation