# Class 3 Thevenin_ Norton

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```					    EC 101

Electrical Sciences

1
Welcome to Lecture 3

CIRCUIT
ANALYSIS
TECHNIQUES

2
Thevenin’s Theorem

M.L.Thevenin published this theorem in 1883.
Theorem : Any two-node linear network may be replaced by a voltage
source equal to the open circuit voltage between the nodes in series with
the resistance as seen by a load at this port

Ex Find voltage Vab across terminal ab

5Ω                       4Ω
a
25V                                 3A
20Ω                          24Ω

b
3
5Ω                      4Ω
a
25V                                3A
20Ω                          24Ω

b

Replace by
Req                         Thevenin’s Equivalent Circuit
                            a

VTh                           24Ω

b
Thevenin’s Equivalent Circuit
4
VTh – Open circuit voltage across ab

Req - Equivalent Resistance as seen by the load after deactivating
all independent sources

1. Open Circuit ab

5Ω                        4Ω
a
25V                                 3A
20Ω

b

5
VTh         VTh = v1
5Ω   v1         4Ω
a
+
25V                    3A     VTh
20Ω
-
b

Applying nodal analysis,
v1  25 v1
    3
5     20             v Th  32V
v1  32V
6
Req

Deactivate all independent sources

Voltage sources – short circuit

Current sources – Open circuit

5Ω                 4Ω
a

20Ω                       Req

b

R eq  4  20 5  4  4  8Ω
7
Req = 8Ω
                      a
+

VTh                    24Ω    Vab
=32V
b       -

24
Vab  32          24V
24  8

8
Norton’s Theorem

E.L.Norton published this theorem

Theorem : Any two -node linear network may be replaced by a current
source equal to the short circuited current between the nodes in parallel
with the resistance as seen by a load at this port

Ex Find voltage Vab across terminal ab

5Ω                       4Ω
a
25V                                3A
20Ω                          24Ω

b
9
Norton’s Equivalent Circuit


a
ISC           Req   24Ω

b
1. Short Circuit ab

5Ω                      4Ω
a
25V                             3A       ISC
20Ω

b
10
5Ω   v1         4Ω
a
25V                      3A    ISC
20Ω

b
Applying nodal analysis,

v1  25 v1 v1
     3
5      20 4
v1  16V
v1 16
I SC          4A
4   4                       11
Req

Deactivate all independent sources

Voltage sources – short circuit

Current sources – Open circuit

5Ω                 4Ω
a

20Ω                       Req

b

R eq  4  20 5  4  4  8Ω
12

a       +
Req
ISC = 4A                           Vab
=8Ω           24Ω

b       -

Norton’s Equivalent Circuit

8
Vab  4          24  24V
8  24

13
Thevenin’s Equivalent Circuit

By Source Transformation

Norton’s Equivalent Circuit

Req = 8Ω              
a                                        a
Req
ISC = 4A        =8Ω           24Ω
VTh                           24Ω
=32V
Source                          b
b        Transformation
Thevenin’s Equivalent Circuit                Norton’s Equivalent Circuit

VTh                        VTh
I SC                      R eq   
R eq                       I SC
14
Req     When the Circuit contains dependent sources also

Deactivate all independent sources only

Apply a test voltage source or a test current source to the Thevenin
terminals a,b

VTest
R eq   
I Test

15
Ex Find I applying Thevenin’s Theorem

2kΩ               3kΩ
+
I
4V                        VX            4kΩ   VX
4000
-

16
Solution    VTh

1. Open Circuit the 4kΩ resistance
2kΩ        VTh    3kΩ
+
4V
VTh   VTh
4000
-
Applying nodal analysis,

VTh  4 VTh

2000    4000
or, 2VTh  8  VTh
VTh  8V                            17
Req   1. Deactivate the independent sources

2kΩ               3kΩ
VX
+
VX         VX
4000
-

18
1. Apply a test voltage VTest

2kΩ                3kΩ
I Test
VTest      VTest
4000

Applying nodal analysis,

VTest VTest  I Test  3000
I Test         
4000          2000

19
4000I Test  VTest  2VTest  6000I Test

VTest  10000I   Test

VTest
 10000  10k 
I Test

Req = 10kΩ

20
Thevenin’s Equivalent Circuit
Req = 10kΩ

I
VTh                           4kΩ
=8V

8
I         0.571mA
10  4

21
Alternative Way

2kΩ     3kΩ
+
I
4V                VX         4kΩ   VX
4000
-

VTh  8V

22
2kΩ             3kΩ
+

4V                        VX
0   ISC VX  0
4000
-

2kΩ            3kΩ

4V                            ISC

4
I SC         0.8mA
23
23
VTh      8V
R eq                 10k Ω
I SC   0.8mA

Req = 10kΩ

I                    8
I         0.571mA
VTh                             4kΩ
10  4
=8V

Thevenin’s Equivalent Circuit

24

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