Download Example 4.10

Example 4.10 Irwin, Page 127
Find the Thevenin equivalent circuit at points A-B for the circuit shown.
 We need to find two quantities: VTH = Voc and RTH = Voc/ISC
 If we use nodal analysis and/or mesh analysis or any other appropriate
techniques, we find VTH = 0V and ISC = 0A (because there are no
independent sources of power in the circuit)
 RTH = Voc / ISC = 0V/0A = ?????
 VTH = 0V is equivalent to a short circuit
 ISC = 0A is equivalent to an open circuit
 The Thevenin (Norton) equivalent consists of a single resistor.
Fall 2001
ENGR-201 Example 4.8
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Example 4.10 – Find RTH
To make our work easier, we will
find RTH to the at points C-D first.
To find RTH apply a known voltage
source (1V) and determine the
amount of current demanded. .
1V
R TH =
IS
Fall 2001
ENGR-201 Example 4.8
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Example 4.10 – Find RTH at C-D
1V
 1mA
1k
VX =2000I X =2V
IX =
1V-2V
IS =I X +
 1mA-0.5mA =0.5mA
2k
1V
 2k
 R TH C D 
0.5mA
We can now replace the circuit to
the left of points C-D with its
Thevenin equivalent.
Fall 2001
ENGR-201 Example 4.8
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Example 4.10 – Find RTH
 R TH A-B =2k||  3k  2k 
 R TH A-B  1.43k
The Thevenin (and Norton)
equivalent circuit, then, is
Fall 2001
ENGR-201 Example 4.8
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