This document discusses Thévenin's and Norton's theorems, which state that a linear two-terminal circuit can be reduced to an equivalent circuit with either a voltage source and resistor or current source and resistor. It provides definitions and steps for determining the equivalent components. Examples are given to demonstrate simplifying circuits using source transformations between the Thévenin and Norton models.

2. Objective of Lecture
 State Thévenin’s and Norton Theorems.
 Chapter 4.5 and 4.6 Fundamentals of Electric Circuits
 Demonstrate how Thévenin’s and Norton theorems
can be used to simplify a circuit to one that contains
three components: a power source, equivalent
resistor, and load.

3. Thévenin’s Theorem
 A linear two-terminal circuit can be replaced with an
equivalent circuit of an ideal voltage source, V Th, in
series with a resistor, RTh.
 V Th is equal to the open-circuit voltage at the terminals.
 RTh is the equivalent or input resistance when the
independent sources in the linear circuit are turned off.

4. Circuit Schematic:
Thévenin’s Theorem

5. Definitions for Thévenin’s Theorem
Linear circuit is a circuit where the
voltage is directly proportional to the
current (i.e., Ohm’s Law is followed).
Two terminals are the 2 nodes/2
wires that can make a connection
between the circuit to the load.

6. Definitions for Thévenin’s Theorem
+
Voc
_
Open-circuit voltage Voc is the voltage, V, when the load is an open
circuit (i.e., RL = ∞ ).
V OC
V Th

7. Definitions for Thévenin’s Theorem
 Input resistance is the resistance seen by the load
when V Th = 0V.
 It is also the resistance of the linear circuit when
the load is a short circuit (RL = 0 ).
R in
R Th
V Th i SC

8. Steps to Determine VTh and RTh
Identify the load, which may be a resistor or a part of
the circuit.
Replace the load with an open circuit .
Calculate VOC. This is V Th.
Turn off all independent voltage and currents sources
in the linear 2-terminal circuit.
Calculate the equivalent resistance of the circuit.
This is RTh.
1.
2.
3.
4.
5.

The current through and voltage across the load in
series with V Th and RTh is the load’s actual current and
voltage in the original circuit.

9. Norton’s Theorem
 A linear two-terminal circuit can be replaced with an
equivalent circuit of an ideal current source, IN, in
parallel with a resistor, RN.
 IN is equal to the short-circuit current at the terminals.
 RN is the equivalent or input resistance when the
independent sources in the linear circuit are turned off.

10. Circuit Schematic:
Norton’s Theorem

11. Definitions for Norton’s Theorem
Short-circuit current Isc is the current, i, when the load is a short circuit
(i.e., RL = 0 ).
I SC
IN

12. Definitions for Norton’s Theorem
 Input resistance is the resistance seen by the load
when IN = 0A.
 It is also the resistance of the linear circuit when
the load is an open circuit (RL = ∞ ).
R in
RN
V OC I N

13. Steps to Determine IN and RN
Identify the load, which may be a resistor or a part of
the circuit.
Replace the load with a short circuit .
Calculate ISC. This is IN.
Turn off all independent voltage and currents sources
in the linear 2-terminal circuit.
Calculate the equivalent resistance of the circuit.
This is RN.
1.
2.
3.
4.
5.

The current through and voltage across the load in
parallel with IN and RN is the load’s actual current and
voltage in the original circuit.

14. Source Conversion
 A Thévenin equivalent circuit can easily be
transformed to a Norton equivalent circuit (or visa
versa).
 If RTh = RN, then V Th = RNIN and IN = V Th/RTh

15. Voltage Polarity and Current Flow

16. Value of Theorems
 Simplification of complex circuits.
 Used to predict the current through and voltage across
any load attached to the two terminals.
 Provides information to users of the circuit.

17. Example #1

18. Example #1 (con’t)
Find IN and RN

19. Example #1 (con’t)
 Calculation for IN
 Look at current divider equation:
I load
R eq
R load
R load R N
IN
R load
RN
2 mA
2k
RN
1
R N R load
IN
If RTh = RN= 1k , then IN = 6mA
IN

20. Why chose RTh = RN?
 Suppose V Th = 0V and IN = 0mA
 Replace the voltage source with a short circuit.
 Replace the current source with an open circuit.
 Looking towards the source, both circuits have the identical
resistance (1k ).

22. Source Transformation
Equations for Thévenin/Norton Transformations
V Th = IN RTh
IN = V Th/RTh
RTh= RN

23. Example #1: Norton’s Theorem
IN is the current that flows when a short circuit is used as
the load with a voltage source
IN = VTh/RTh = 6mA

24. Example #1: Norton’s Theorem
RN is the resistance of the linear circuit when the
power sources in the original circuit are turned off
(V Th is replaced with a short circuit).

25. Example #1: Norton’s Theorem
 The Norton equivalent circuit is:

26. Check: Thévenin Theorem
V Th is the voltage across the load when an open short
circuit is used as the load with a current source
VTh = IN RTh = 6V

27. Check: Iload and Vload
I load
1k
1k
6 mA
2k
I load
2 mA
V load
2 mA ( 2 k
V load
4V
)

28. Example #2
Simplification through Transformation

29. Example #2 (con’t)

30. Example #2 (con’t)
Find Req to obtain a Norton equivalent circuit

31. Example #2 (con’t)
Current Source to Voltage Source
RTh = 3
VTh = 0.1A (3 ) = 0.3V
0.3V

32. Example #2 (con’t)
0.3V

33. Example #2 (con’t)
Voltage Source to Current Source
RTh = 2
IN = 3V/2 = 1.5A

34. Example #2 - Solution 1
 Simplify to Minimum Number of Current Sources
0.3V

35. Example #2 (con’t)
Voltage Source to Current Source
RTh = 6
IN = 0.3V/6 = 50.0mA
0.3V

36. Example #2 (con’t)

37. Example #2 (con’t)
Current Sources in Parallel Add

38. Example #2 - Solution 2
 Simplify to Minimum Number of Voltage Sources
0.3V

39. Example #2 (con’t)
Transform solution for Norton circuit to Thévenin
circuit to obtain single voltage source/single equivalent
resistor in series with load.

40. PSpice

41. Example #2 - Solution 1

42. Example #2 – Solution 2

43. Summary
 Thévenin and Norton transfomrations are performed
to simplify a circuit for analysis and design.
 Two techniques were described.

Examples using the source transformation technique were
given.