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Electrical Engineering
Principles & Applications
Chapter 2
Resistive Circuits
Slide 1

Objectives
1
...
e
...
Apply the voltage-division and current-division principles
3
...
Solve circuits by the mesh-current technique
5
...
Apply the superposition principle
7
...
3

Ans
...
1

Slide 7

Ans
...
1
...


Begin by locating a combination of resistances that are in series
or parallel
...


2
...


3
...
Often (but not always) we end up with a single source
and a single resistance
...


Solve for the currents and voltages in the final equivalent
circuit
...


5
...


Slide 8

Example

Slide 9

Example (cont
...


Ans
...
04 A, i2=0
...
32, i4=0
...
5 A, i3=0
...
5V
1000+1000+ 2000+ 6000
v4 =
Slide 13

6000
× 15 = 9V
1000 + 1000 + 2000 + 6000

Current Division
• The total current flowing into a
parallel combination of resistors
will be divided among them
• For two resistances in parallel, the
current passing into one resistance
is related to the ratio of the other
resistance to the sum of the two
resistances
• If more than two resistances, you
need to combine them until you are
left with two and then use the
above step

Slide 14

R1 R2
v=
itotal
R1 + R2

v
R2
i total
i1 =
=
R1 R1 + R 2
R1
v
i2 =
=
i total
R2
R1 + R 2

Example
Find vx using voltage division and then find is and use it to find i3 using
current division

R 2 R3
30 × 60
Req =
=
= 20 Ω
R2 + R3 30 + 60
is =
Slide 15

vs
100
=
= 1
...
25 = 0
...
05 V, v2=5
...
07 V

Note
Although series/parallel equivalents and
the current/voltage division
principles are very
important concepts,
yet they are not sufficient to
solve all circuits !!

Slide 18

Node-Voltage Analysis
•
•

Techniques learnt so far are not applicable to more complex networks
...
Also, voltage division and current division
cannot be used
...
7v1 − 0
...
2v1 + 0
...
32 V
v2 = 6
...
35v1 − 0
...
05v3 = 0
− 0
...
3v2 − 0
...
05v1 − 0
...
35v3 = 0
Solving
v1 = 45
...
73 V
v3 = 27
...
909 A
20
Slide 27

Another Example
With a different choice of the reference node, find ix using nodal
analysis

v1 = −27
...
73 V
v3 = −45
...
909 A
ix =
20
Note: Different values of node voltages but final
answer for current is the same
...
Assume
the is is given
...
)
Next, we find an expression for the controlling variable
ix in terms of the node voltages

v3 − v 2
ix =
R3

Slide 35

Example (cont
...
Assume the is is given
...
5v x

Node 3
v3 v3 − v2 v3 − v1
+
+
=0
R4
R3
R1
supernode (nodes 1 and 2)
v1 v1 − v3 v2 − v3
+
+
= is
R2
R1
R3

Contolling variable
v x = v3 − v1
Slide 37

Summary: Node-Voltage Analysis
1
...
Write network equations
•
•
•

First, use KCL to write current equations for nodes and super nodes
Write as many current equations as you can without using all of the nodes
Then, if you do not have enough equations because of voltage sources
connected between nodes, use KVL to write additional equations

3
...
Put the equations into standard form and solve for the node
voltages
5
...
33 A
Slide 39

ib = −0
...
5 A
Slide 40

i y = 2
...
This circuit has three loops

Not every loop current is required to compute all
the currents through components
a

fabef

ebcde

fabcdef

A mesh is a loop that does not enclose
any other loop
...
i1, i2 are mesh currents
Slide 41

1

b

2
−

I1

3

c

Using two loop currents
−

7

4

Ia

f

−

= − I1− I3
−

e
d
f
6
5
A BASIC CIRCUIT −

I3

Ibe = I 1
−

Ibc = I

3

For every circuit there is a minimum number of loop
currents that are necessary to compute every current
in the circuit
...


Loops, Meshes and Loop Currents
For a given circuit let
B: number of branches
n: number of nodes
The minimum required number of
loop currents is

L = B − ( N − 1)

Mesh currents are always independent

Determination of loop currents
kvl on left mesh

KVL on the right mesh
+ vs 2 + v4 + v5 − v3 = 0
Using

An example:

Ohm's

law

v 1 = i1 R 1 , v 2 = i1 R 2 , v 3 = ( i1 − i 2 ) R 3
v 4 = i2 R 4 , v 5 = i2 R 5
Replacing and rearranging

B=7
N =6
L = 7 − (6 − 1) = 2
Slide 42

Two loop currents are
required
...
Hence
they are independent and
form a minimal set

These are loop equations for the
circuit

Writing the Mesh Equations
+ v R1 − = i1R1
Branches = 8
Nodes = 7
Loop currents needed = 2

And we are told to use mesh currents!
This defines the loop currents to be used

+
v R3 = − i2 R3

− v R 2 + = (i1 − i2 ) R2+

v R5 = i2 R5
−

−

+ v R 4 − = − i2 R4

Identify all voltage drops
Write kvl on each mesh

Top mesh : − vS 1 + vR1 − vS 2 + vR 2 = 0

Bottom : − vR 2 + vR 5 − vR 4 + vS 3 − vR 3 = 0
Use Ohm’s law

Slide 43

Example
Find Io using loop analysis

An alternative selection of loop currents

Shortcut: polarities are not needed
...
5mA
Rearrange 12kI1 + 6kI 2 = 12 multiply by 3

12 kI 1 = 12 + 6 kI 2 ⇒ I1 =

5
mA
4

express variable of interest as function
of loop currents
IO = I1 − I 2
Slide 44

6kI1 + 9kI 2 = 9

multiply by 2 and substract

3
24kI1 = 18 ⇒ I1 = mA
4

Example
Using this pattern for mesh 1 of the figure shown, we
have:

R2 (i1 − i3 ) + R3 (i1 − i2 ) − v A = 0

For mesh 2:

R3 (i2 − i1 ) + R4 i2 + v B = 0

For mesh 3:

R2 (i3 − i1 ) + R1i3 − v B = 0

Slide 45

Example
Similarly, using this pattern for mesh 1 of the figure
shown, we have:

R1i1 + R2 (i1 − i4 ) + R4 (i1 − i2 ) − v A = 0
For mesh 2:

R5i2 + R4 (i2 − i1 ) + R6 (i2 − i3 ) = 0

For mesh 3:

R7 i3 + R6 (i3 − i2 ) + R8 (i3 − i4 ) = 0
For mesh 4:

R3i4 + R2 (i4 − i1 ) + R8 (i4 − i3 ) = 0

Slide 46

Example
Solve for the currents in each element in the circuit

mesh 1 : 20 i1 + 10 (i1 − i2 ) − 150 = 0
mesh 2 : 10 (i2 − i1 ) + 15i2 + 100 = 0
30 i1 − 10 i 2 = 150

i1 = 4
...
308 A
The current in the 10 - Ω is i1 -i 2 = 6
...
In other
words, we write a KVL equation around the periphery of
meshes 1 and 2 combined
...
Define the mesh currents flowing around each mesh
...
Write KVL equations, stopping after the number of equations is
equal to the number of mesh currents
•
•
•

First, use KVL to write voltage equations for meshes that do not
contain current sources
Next, if any current sources are present, write expressions for their
currents in terms of the mesh currents
Finally, if a current source is common to two meshes, write a KVL
equation for the supermesh

3
...
Substitute into the
network equations, and obtain equations having only the mesh currents as
unknowns
4
...
Solve for the mesh currents by use
of determinants or other means
5
...
Turn off independent sources in the original network:
- A voltage source becomes a short circuit
- A current source becomes an open circuit

2
...
15

Slide 60

Note: voltage across R2 = 0 (short circuit)
i2=0
i1=20/5 = 4 A
isc = 4 + 2 = 6 A
Vt=Rtisc= 4 x 6 = 24 V

Example
Find the Thévenin resistance for each circuit

Rt=10+(5||20)=14

Slide 61

Rt=10+20 = 30

Rt = ((20||5)+6)||10) = 5

Thévenin Equivalent Circuits

Slide 62

Circuits with Dependent Sources
1
...
Compute Rt = Voc/Isc and Vt = Voc

Slide 63

Circuits with Dependent Sources
ix + 2ix =
ix =

voc
10

10 − voc
5

10 − voc voc
3
=
5
10

voc = 8
...
57 V
=
= 1
...
Perform two of these:
• Determine the open-circuit voltage Vt = voc
• Determine the short-circuit current In = isc
• Zero the sources and find the Thévenin resistance Rt looking
back into the terminals

2
...
The Thévenin equivalent consists of a voltage source Vt
in series with Rt
4
...
25voc
R2 + R 3
0
...
62V
and we got: isc =

Rt =
Slide 68

vs 15V
=
= 0
...
62V
=
= 6
...
75A

voc

Circuits with Dependent Sources
voc = 4
...
75A
Slide 69

Exercises
I n = 1
...
375Ω

Slide 70

I n = 2A, Rt = 15Ω

Source Transformations

Slide 71

Source Transformations

R1i1 + R2i2 + 10 − 20 = 0

10
i1 =
= 0
...
667 A
Another approach
Slide 72

Source Transformations

Slide 73

Maximum Power Transfer
The load resistance that absorbs the maximum power
from a two-terminal circuit is equal to the Thévenin
resistance

V t2
2
PL =
= iL R t
4Rt
Slide 74

Example
Find the load resistance for maximum power transfer
...
25 W
Slide 75

Superposition Principle
The superposition principle states that the total response
is the sum of the responses to each of the independent
sources acting individually

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