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ELECTRICAL TECHNOLOGY
EET 103/4
 Explain and analyze series and
parallel circuits
 Explain, derive and analyze Ohm’s
Law, Kirchhoff Current Law,
Kirchhoff Voltage Law, Source
Transformation, Thevenin theorem.
1
NETWORK THEOREM
(CHAPTER 9)
2
9.2 Superposition Theorem
 Used to find the solution to networks with two
or more sources that are not in series or
parallel.
 The current through, or voltage across, an
element in a network is equal to the algebraic
sum of the currents or voltages produced
independently by each source.
 Since the effect of each source will be
determined independently, the number of
networks to be analyzed will equal the
number of sources.
3
9.2 Superposition Theorem
 When removing a voltage source from a
network schematic, replace it with a direct
connection (short circuit) of zero ohm. Any
internal resistance associated with the source
must remain in the network.
4
9.2 Superposition Theorem
 When removing a current source from a
network schematic, replace it with an open
circuit of infinite ohms. Any internal
resistance associated with the source must
remain in the network.
5
9.2 Superposition Theorem
The total power delivered to a resistive
element must be determined using the
total current through or the total voltage
across the element and cannot be
determined by a simple sum of the
power levels established by each
source.
6
9.2 Superposition Theorem
Example 9.1
Using superposition theorem, find I1:
7
9.2 Superposition Theorem
Example 9.1 - solution
Open the current source:
E 30
I '1 

5A
R1 6
8
9.2 Superposition Theorem
Example 9.1 – solution (cont’d)
Replace the current source and short the
voltage source:
 Rsc 

I "1  I 
 Rsc  R1 
 0 
 3
0A
06
9
9.2 Superposition Theorem
Example 9.1 – solution (cont’d)
With both sources in the circuit, the total
current is therefore;
I1  I '1  I "1
 50  5 A
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9.2 Superposition Theorem
Example 9.2
Use superposition theorem to find I2:
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9.2 Superposition Theorem
Example 9.2 – solution
Short the voltage source E2:
12
9.2 Superposition Theorem
Example 9.2 – solution (cont’d)
Redraw the circuit:
R2 R3
R2 R3 
R2  R3
12  4

12  4
 3
13
9.2 Superposition Theorem
Example 9.2 – solution (cont’d)
RT  R1  R2 R3
I 's
 24  3  27 
E1 54
I 's 

2A
RT 27
 R3   4 
  2
I '2  I 's 
  0.5 A
 R2  R3   12  4 
14
9.2 Superposition Theorem
Example 9.2 – solution (cont’d)
Replace E2 and short the voltage source E1:
I "s
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9.2 Superposition Theorem
Example 9.2 – solution (cont’d)
Redraw the circuit:
R1 R2
R1 R2 
R1  R2
24  12

24  12
I "s
8
16
9.2 Superposition Theorem
Example 9.2 – solution (cont’d)
RT  R3  R1 R2
 4  8  12 
I "s 
E2
RT
I "s
48

4A
12
 R1   24 
  4
I "2  I "s 
  2.67 A
 R1  R2   24  12 
17
9.2 Superposition Theorem
Example 9.2 – solution (cont’d)
With both sources in the circuit, the total
current is therefore;
I 2  I "2 I '2
 2.67  0.5
 2.17 A
18
9.2 Superposition Theorem
Example 9.3
(a) Use superposition theorem to find I2
(b) Demonstrate that the superposition theorem is not
applicable to power level
19
9.2 Superposition Theorem
Example 9.3 – solution
(a)
Replace the current source
with an open circuit:
E
I '2 
R1  R2
36

2A
12  6
P'2  I '2  R2  2 2  6  24 W
2
20
9.2 Superposition Theorem
Example 9.3 – solution (cont’d)
Reconnect the current source and replace the
voltage source with a short circuit:
 R1 

I "2  I 
 R1  R2 
 12 
 9
6A
 12  6 
P"2  I "2  R2  6 2  6  216 W
2
21
9.2 Superposition Theorem
Example 9.3 – solution (cont’d)
With both sources in the circuit, the current
through R2 is;
I 2  I '2  I "2
 62 8A
P2  I 22 R2
 82  6  384 W
22
9.2 Superposition Theorem
Example 9.3 – solution (cont’d)
P'2 P"2  24  216  240 W
which is not equal to:
P2  384 W
Hence, the superposition is not applicable to power level
23
9.3 Thevenin’s Theorem
Any two-terminal dc network can be
replaced by an equivalent circuit consisting
of a voltage source and a series resistor.
24
9.3 Thevenin’s Theorem
 Thévenin’s theorem can be used to:
Analyze networks with sources that are not
in series or parallel.
Reduce the number of components required
to establish the same characteristics at the
output terminals.
Investigate the effect of changing a particular
component on the behavior of a network
without having to analyze the entire network
after each change.
25
9.3 Thevenin’s Theorem
 Procedure to determine the proper values of
RTh and ETh
26
9.3 Thevenin’s Theorem
 Preliminary
1. Remove that portion of the network across which
the Thévenin equation circuit is to be found. In the
figure below, this requires that the load resistor RL
be temporarily removed from the network.
27
9.3 Thevenin’s Theorem
2. Mark the terminals of the remaining two-terminal
network. (The importance of this step will become
obvious as we progress through some complex
networks.)
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9.3 Thevenin’s Theorem
3. Calculate RTh by first setting all sources to zero (voltage sources
are replaced by short circuits, and current sources by open circuits)
and then finding the resultant resistance between the two marked
terminals. (If the internal resistance of the voltage and/or current
sources is included in the original network, it must remain when the
sources are set to zero.)
RTh  R1 R2
R1 R2

R1  R2

R1 R2
R1  R2
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9.3 Thevenin’s Theorem
In a laboratory, RTh may be measured:
30
9.3 Thevenin’s Theorem
4. Calculate ETh by first returning all sources to their original
position and finding the open-circuit voltage between the marked
terminals. (This step is invariably the one that will lead to the most
confusion and errors. In all cases, keep in mind that it is the opencircuit potential between the two terminals marked in step 2.)
 R2 

ETh  E
 R1  R2 
 6 
 9
6V
 3 6
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9.3 Thevenin’s Theorem
In a laboratory, ETh may be measured:
32
9.3 Thevenin’s Theorem
5. Draw the Thévenin equivalent circuit:
33
9.3 Thevenin’s Theorem
6. If required, reconnect the portion which had been
removed previously:
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9.3 Thevenin’s Theorem
Example 9.7
Find the Thevenin equivalent circuit for the
network in the shaded area:
35
9.3 Thevenin’s Theorem
Example 9.7 – solution
Remove the component external to the relevant
network i.e. R3.
36
9.3 Thevenin’s Theorem
Example 9.7 – solution (cont’d)
Replace the current source with an open circuit
and calculate RTh:
RTh  R1  R2
 24  6
37
9.3 Thevenin’s Theorem
Example 9.7 – solution (cont’d)
Reconnect the current source and calculate ETh:
ETh  IR1
 12  4  48 V
38
9.3 Thevenin’s Theorem
Example 9.7 – solution (cont’d)
Draw the Thevenin equivalent circuit:
39
9.3 Thevenin’s Theorem
Example 9.7 – solution (cont’d)
The original circuit
The corresponding
Thevenin equivalent
circuit
40
9.3 Thevenin’s Theorem
Example 9.7 – solution (cont’d)
If required, reconnect the external component
which had been removed previously:
41
9.3 Thevenin’s Theorem
Example 9.8
Find the Thevenin equivalent circuit for the
network in the shaded area:
42
9.3 Thevenin’s Theorem
Example 9.8 – solution
Remove the component external to the relevant
network i.e. R3.
43
9.3 Thevenin’s Theorem
Example 9.8 – solution (cont’d)
Replace the voltage source with a short circuit:
44
9.3 Thevenin’s Theorem
Example 9.8 – solution (cont’d)
Redraw the circuit and
calculate RTh:
R1 R2
RTh  R1 R2 
R1  R2
6 4

 2.4 
64
45
9.3 Thevenin’s Theorem
Example 9.8 – solution (cont’d)
Reconnect the voltage source:
46
9.3 Thevenin’s Theorem
Example 9.8 – solution (cont’d)
Redraw the circuit:
47
9.3 Thevenin’s Theorem
Example 9.8 – solution (cont’d)
Redraw the circuit further and calculate ETh:
 R1 

RTh  E
 R1  R2 
 6 
 8

 63
 4.8 V
48
9.3 Thevenin’s Theorem
Example 9.8 – solution (cont’d)
Draw the corresponding Thevenin equivalent
circuit:
49
9.3 Thevenin’s Theorem
Example 9.8 – solution (cont’d)
Reconnect R3 which had been removed
previously:
50
9.3 Thevenin’s Theorem
Example 9.10
Find the Thevenin
circuit for network
within the shaded
area.
51
9.3 Thevenin’s Theorem
Example 9.10 – solution
Redraw the circuit:
RL
52
9.3 Thevenin’s Theorem
Example 9.10 – solution (cont’d)
Disconnect the component(s) external to the
network:
53
9.3 Thevenin’s Theorem
Example 9.10 – solution (cont’d)
Replace the voltage sources with short
circuits and find RTh:
R2 R3
Ra 
R2  R3
46

46
 2.4 k
54
9.3 Thevenin’s Theorem
Example 9.10 – solution (cont’d)
Replace R2 and R3 with Ra:
R1 Ra
Rb 
R1  Ra
0.8  2.4

0.8  2.4
Ra
2.4 k
 0.6 k
55
9.3 Thevenin’s Theorem
Example 9.10 – solution (cont’d)
Replace R1 and Ra with Rb:
RTh  Rb  R4
 0.6 1.4
 2 k
Rb
0.6 k
56
9.3 Thevenin’s Theorem
Example 9.10 – solution (cont’d)
Use superposition theorem to find ETh:
E Th
57
9.3 Thevenin’s Theorem
Example 9.10 – solution (cont’d)
Replace E2 with a short circuit and find E’Th:
E 'Th  V3
 R2 R3 

 E1 

R

R
R
2
3 
 1
 46 
  4 .5 V
 6

0
.
8

4
6


58
9.3 Thevenin’s Theorem
Example 9.10 – solution (cont’d)
Reconnect E2 and replace E1 with a short
circuit and find E”Th:
E"Th  V3
 R1 R3 

 E2 

R

R
R
1
3 
 2
 0 .8 6 
  1 .5 V
 10

4

0
.
8
6


59
9.3 Thevenin’s Theorem
Example 9.10 – solution (cont’d)
Since E’Th and E”Th are of opposite polarities:
ETh  E 'Th  E"Th  4.5  1.5  3 V
60
9.3 Thevenin’s Theorem
Example 9.10 – solution (cont’d)
Draw the corresponding Thevenin equivalent
circuit and reconnect RL.
61