E E 2320
Lecture 06 - Thévenin and Norton
E i l t Circuits
Equivalent
Ci it
Thévenin and Norton Equivalents
RTh
a
a
=
VTh
b
IN
RTh
b
1
Combining Voltage Sources
a
Voltage sources are
added algebraically
a
15 V
25 V
=
10 V
b
b
Combining Voltage Sources
a
Voltage sources are
added algebraically
a
15 V
5V
=
10 V
b
b
2
Combining Voltage Sources
a
5V
Don’t do this.
10 V
Why is this illogical?
Whose fundamental circuit
law is violated by this?
b
Combining Current Sources
Current sources are
added algebraically
a
5A
a
10 A
b
15 A
=
b
3
Combining Current Sources
Current sources are
added algebraically
a
5A
a
10 A
5A
=
b
b
Combining Current Sources
a
5A
Don’t do this.
Why is this illogical?
Whose fundamental circuit
law is violated by this?
10 A
b
4
Source Transformations Simplify
Circuits (1/5)
18 
a
20 
36 
144 V
4A
47 A
b
4
100 V
Source Transformations Simplify
Circuits (2/5)
a
8A
18 
20 
36 
4A
100 V
4
188 V
b
5
Source Transformations Simplify
Circuits (3/5)
24 
a
12 A
12 
288 V
b
Source Transformations Simplify
Circuits (4/5)
a
12 A
12 
24 
12 A
b
6
Source Transformations Simplify
Circuits (5/5)
8
a
24 A
8
a
192 V
b
b
Thévenin Circuit without Source
Transformations
3
60 V
5A
6
2
a
8
b
No Source Transformation for Current Source
By circuit rules
7
Obtaining Thévenin Circuit with
Dependent Sources
• Replace all independent voltage sources with
short
h t circuits
i it (0 resistance).
it
)
• Replace all independent current sources with
open circuits ( resistance).
• Apply a 1.0 amp current source to the terminal
pair.
pair
• Resulting terminal voltage numerically equal
to Thévenin resistance
Another Thévenin Circuit (1/4)
8 ix
5
a
I1
20 V
10 
ix I
2
6
b
Find open circuit voltage Vab:
8
Another Thévenin Circuit (2/4)
8 ix
5
a
I1
20 V
10 
6
ix I
2
b
Solve mesh equations for I2
Then Vab can be found:
Another Thévenin Circuit (3/4)
5  V2
10 
8 ix
ix
V1 a
6
1A
b
Now get Thévenin Resistance by node voltage solution:
9
Another Thévenin Circuit (4/4)
8 ix
5  V2
10 
V1 a
6
ix
1A
b
The Result
5
8 ix
a
I1
20 V
10 
ix
I2
6
b
3
a
12 V
b
10
Check on Previous Example (1/2):
• VTh = 12 V and RTh = 3   IN = 4 A
• We will calculate IN directly.
8 ix
5
a
I1
20 V
10 
ix I
2
6
IN
b
Check on Previous Example (2/2):
5
8 ix
a
I1
20 V
10 
Then:
e : 200 = 5I
51
Since I1 = I2 = IN
ix I
2
6
IN
b
And I1 = 4 A
IN = 4 A
Same short circuit current as predicted by
Thévenin theorem
11