DC CIRCUIT METHODS
Problem 1
Use source transformation to find vo in the circuit in Fig. 4.17.
Solution:
Problem 2
Find io in the circuit of Fig. 4.19 using source transformation.
Answer: 1.78 A.
Thevenin’s Theorem
THEOREM
Thevenin’s theorem states that “a linear two-terminal circuit can be
replaced by an equivalent circuit consisting of a voltage source VTh in
series with a resistor RTh where VTh is the open circuit voltage at the
terminals and RTh is the input or equivalent resistance at the
terminals when the independent source are turn off”.
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DEVELOPED BYIn 1883 the theorem was developed by LÉON CHARLES THÉVENIN
(1857-1926), an electrical engineer with France’s NATIONAL POSTES ET
TÉLÉGRAPHES TELECOMMUNICATIONS organization.
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THEVENIN’S EQUIVALENT CIRCUIT
Fig 1: A linear two-terminal circuit by its Thevenin equivalent: (a) Original circuit
(b) Thevenin equivalent circuit.
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HOW TO FIND THEVENIN’S VOLTAGE
 Remove the component of interest- normally called the
load resistance (𝑅𝐿 )
 Mark the terminals as 𝑉𝑇𝐻 where the load was removed
from
 Calculate the Thevenin voltage 𝑉𝑇𝐻 by finding the open
circuit voltage across the load terminals
Here,
VTh  voc : open circuit voltage at a  b
Fig 2: Finding 𝑉𝑇𝐻
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HOW TO FIND THEVENIN’S RESISTANCE
CASE 1: (No dependent sources)
 Turn off all independent source.
 RTH: can be obtained via simplification of
either parallel or series connection seen
from a-b
Fig 3: Finding 𝑅𝑇𝐻 (No dependent source)
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HOW TO FIND THEVENIN’S RESISTANCE (cont.)
CASE 2: (With dependent sources)
 Turn off all independent sources.
 Apply a voltage source vo at a-b terminal and
determining the resulting current io .Then,
RTh 
vo
io
 Alternatively, apply a current source io at a-b terminal
and find the terminal voltage vo .
RTh 
vo
io
Fig 4: Finding 𝑅𝑇𝐻 when circuit has dependent sources
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SIMPLIFIED CIRCUIT
VTh
IL 
RTh  RL
RL
VL  RL I L 
VTh
RTh  RL
Fig 5: A circuit with a load: (a) Original circuit
(b) Thevenin equivalent circuit
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EXAMPLE:1
 Find the Thevenin’s equivalent circuit of the circuit shown in Fig 6 to the left
of the terminals a-b. Then find the current through RL = 6 , 16 , and 36 .
Fig 6
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FIND RTH
RTh : 32V voltage source  short
2A current source  open
4  12
RTh  4 || 12  1 
 1  4
16
Fig 7: Finding 𝑅𝑇𝐻
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FIND VTH
VTh :
(1) Mesh analysis
 32  4i1  12(i1  i2 )  0 , i2  2A
i1  0.5A
VTh  12(i1  i2 )  12(0.5  2.0)  30V
(2) Alternatively, Nodal Analysis
(32  VTh ) / 4  2  VTh / 12
VTh  30V
Fig 8: Finding 𝑉𝑇𝐻
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EQUIVALENT CIRCUIT
To get iL :
VTh
30
iL 

RTh  RL 4  RL
RL  6  I L  30 / 10  3A
RL  16  I L  30 / 20  1.5A
RL  36  I L  30 / 40  0.75A
Fig 9: Thevenin equivalent circuit
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EXAMPLE:
 Using Thevenin’s theorem, find the equivalent circuit to the left of the
terminals in the circuit in Fig. Then find i.
Answer: VTh = 6 V, RTh = 3 , i = 1.5 A.
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Problem 1
Find the Norton equivalent circuit of the circuit in Fig. 4.39.
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Solution:
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REFERENCES
1. Fundamental of Electric Circuit by Alexander & Sadiku.
2. Introductory Circuit Analysis by Robert L. Boyelstead.
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