This document discusses Thévenin's and Norton's theorems for circuit analysis. It includes an example problem using Thévenin's theorem to find the current through a 2 ohm resistor. The key points of Thévenin's theorem are described. Norton's theorem is also introduced. An example circuit is used to find both the Thévenin and Norton equivalents. Finally, a homework problem is assigned involving finding the Thévenin and Norton equivalents of another circuit.

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Session 12C: Focus
 Thévenin’s Theorem
◦ Example Problem 2
 Norton’s Theorem
 Home Work problem

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Example 2: Thévenin’s Theorem
 Find current through 2 Ω:
4 Ω
6 Ω
4 Ω
5 Ω
9 V
4
9
4+ 4 + 6
2.571 V
6 Ω
4 Ω 4 Ω
5 Ω
2.571 V
7.857 Ω
2 Ω
voc = 2.571 V, RTH = 7.857 Ω I2Ω = 260.8 mA

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A few Key Points: Thévenin Theorem
 The equivalent circuit that we have learnt to obtain, is
completely independent of network B
◦ To which the network A is connected to
 The only restriction that we must impose on A or B is that all
dependent sources in A have their control variables in A, and
similarly for B.
 No restrictions are imposed on the complexity of A or B;
either one may contain any combination of independent
voltage or current sources, linear dependent voltage or
current sources, resistors, or any other circuit elements which
are linear.
 The dead network A can be represented by a single
equivalent resistance RTH
◦ Which we will call the Thévenin equivalent resistance.

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Norton’s Equivalent Circuit

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Example 3: Thévenin and Nortan’s equivalent
circuits
Find the Thévenin and Nortan’s equivalent circuits:
Applying Superposition to find Voc
= 5 kΩ
Voc
o
o

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Example 3: Continued.
RTH = 5 kΩ
Voc = 8 V
Isc = 1.6 mA
Thévenin Equivalent circuit Norton Equivalent circuit

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Norton’s Theorem
1. Given any linear circuit, rearrange it in the form of
two networks, A and B, connected by two wires.
◦ Network A is the network to be simplified; B will be left
untouched. As before, if either network contains a
dependent source, its controlling variable must be in the
same network.
2. Disconnect network B, and short the terminals of A.
Define a current isc as the current now flowing
through the shorted terminals of network A.
3. Turn off or “zero out” every independent source in
network A to form an inactive network (RTh).
Leave dependent sources unchanged.

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Norton’s Theorem
4. Connect an independent current source with value isc
in parallel with the inactive network.
◦ Do not complete the circuit; leave the two terminals
disconnected.
5. Connect network B to the terminals of the new
network A.
◦ All currents and voltages in B will remain unchanged.

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Home Work Problem

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HW1: Thévenin and Norton Equivalents
Note: This is Practice 5.7 on page 147, Figure 5.30 in the Ref 1 book (by Hayt)
voc = -7.86 V, RTH = 2.429 kΩ
Isc = -3.235 mA
Voc

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Session 12C: Summary
 Thévenin’s and Norton’s Equivalent Circuit
 Thévenin’s Theorem
◦ Example Problem 2
 Norton’s Theorem
 Home Work problem

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References
Ref 1 Ref 2