E E 2415 Lecture 03 - Thévenin and Norton Equivalent Circuits
E E 2415
Feb 22, 2016
E E 2415. Lecture 03 - Thévenin and Norton Equivalent Circuits. Th é venin and Norton Equivalents. Combining Voltage Sources. Voltage sources are added algebraically. Combining Voltage Sources. Voltage sources are added algebraically. Combining Voltage Sources. Don’t do this. - PowerPoint PPT Presentation
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E E 2415
Lecture 03 - Thévenin and Norton Equivalent Circuits
Thévenin and Norton Equivalents
VTh
RTh a
b
IN RTh
a
b
=
Th Th NV R I ThN
Th
VIR
Combining Voltage Sources
10 V
15 V
25 V=
a
a
b
b
Voltage sources areadded algebraically
Combining Voltage Sources
10 V
15 V
5 V=
a
a
b
b
Voltage sources areadded algebraically
Combining Voltage Sources
5 V 10 V
b
a
Don’t do this.
Why is this illogical?Whose fundamental circuitlaw is violated by this?
Combining Current Sources
5 A 10 A
b
a
15 A
a
b
=
Current sources areadded algebraically
Combining Current Sources
5 A 10 A
b
a
5 A
a
b
=
Current sources areadded algebraically
Combining Current Sources
5 A
10 A
b
aDon’t do this.
Why is this illogical?Whose fundamental circuitlaw is violated by this?
Source Transformations can Simplify Circuits (1/5)
144 V
18
4 A36
20
100 V
47 A 4
b
a
Source Transformations Simplify Circuits (2/5)
8A 18 4 A36
20
100 V
4 188 V
a
b
Source Transformations can Simplify Circuits (3/5)
12 A 12
24
288 V
b
a
Source Transformations can Simplify Circuits (4/5)
12 A 12 24 12 A
a
b
Source Transformations can Simplify Circuits (5/5)
24 A 8
a
b
192 V
8 a
b
Obtaining Thévenin Circuit with Dependent Sources
• Replace all independent voltage sources with short circuits (0 resistance).
• Replace all independent current sources with open circuits ( resistance).
• Apply a 1.0 amp current source to the terminal pair.
• Resulting terminal voltage numerically equal to Thévenin resistance
Another Thévenin Circuit (1/4)
Find open circuit voltage Vab:
20 V
5
10
8 ix
6
a
b
ix
I1
I2
1 2 1 2
1 2 1 2
20 15 108 10 16 8( )
x
x
I I i I Ii I I I I
Another Thévenin Circuit (2/4)
Solve mesh equations for I2
Then Vab can be found:
20 V
5
10
8 ix
6
a
b
ix
I1
I2
2 2 12abI A V V
Another Thévenin Circuit (3/4)
Now get Thévenin Resistance by node voltage solution:
5
10
8 ix
6
a
bix
V1V2
1 A
2 2 1 15 10 6V V V
22 1 28 1.8
10VV V V
InjectOneAmp:
Another Thévenin Circuit (4/4)
5
10
8 ix
6
a
b
ix
V1V2
1 A
1 3 3abV Volts R
The Result
20 V
5
10
8 ix
6
a
b
ix
I1
I2
12 V
3 a
b
Check on Previous Example (1/2):
• VTh = 12 V and RTh = 3 IN = 4 A• We will calculate IN directly.
KVL 1: 20 = 15I1 - 10I2
KVL 2: 0 = 10ix + 8ix ix = 0 I1 = I2
20 V
5
10 6
a
bix
I1
I2IN
8 ix
Check on Previous Example (2/2):• VTh = 12 V and RTh = 3 IN = 4 A• We will calculate IN directly.
KVL 1: 20 = 15I1 - 10I2
KVL 2: 0 = 10ix + 8ix ix = 0 I1 = I2
20 V
5
10 6
a
bix
I1
I2IN
8 ix
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