Download (b) a

Electrical Engineering and
Industrial Electronics
Review
What is superposition principle?
The general steps to apply superposition principle?
What is the source transformation?
Solution for the Wednesday assignment.
2.7 Thevenin and Norton Equivalent Circuits
Now
that we have been introduced to the
superposition principle and source transformation, it is
possible to develop two more techniques that will
greatly simplify the analysis of many linear circuits.
1.Thevenin’s Theorem
It often occurs in practice that a particular element in a
circuit is variable (usually called the load) while other
elements are fixed.
As a typical example, a household outlet terminal may be
connected to different appliances constituting a variable load.
Each time the variable element is changed, the entire circuit
has to be analyzed all over again.
To avoid this problem, Thevenin’s theorem provides a
technique by which the fixed part of the circuit is replaced by
an equivalent circuit.
According to
Thevenin’s theorem,
the linear circuit in
Fig.(a) can be replaced
by that in Fig.(b).
The load in the figure
may be a single resistor
or another circuit.
The circuit to the left of
the terminals a-b in Fig.(b)
is known as the Thevenin
equivalent circuit;
Linear
Two-terminal
Circuit
I
Figure (a)
RTh
I
Linear
+
Two-terminal
– vTh
Circuit
a
+
v
_
Load
b
a
+
v
_
Load
b
Figure (b)
It was developed in 1883 by M.Leon Thevenin (1857-1926), a
French telegraph engineer.
 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 equivalent resistance at the terminals when the
independent sources are turned off.
Our major concern right now is how to find the Thevenin
equivalent voltage vTh and resistance RTh.
a
a
Linear circuit with
Linear
+
all independent
Req
Two-terminal
vOC sources set
_
Circuit
equal to zero
b
b
vTh = vOC
RTh = Req
2.Norton’s Theorem
In 1926, about 43 years after Thevenin published his
theorem, E.L. Norton, an American engineer at Bell
Telephone Laboratories, proposed a similar theorem.
 Norton’s theorem states that a linear two-terminal
circuit can be replaced by an equivalent circuit
consisting of a current source IN in parallel with a
resistor RN, where IN is the short-circuit current
through the terminals and RN is the equivalent
resistance at the terminals when the independent
sources are turned off.
According to
Norton’s theorem, the
linear circuit in Fig.(a)
can be replaced by
that in Fig.(c).
Linear
Two-terminal
Circuit
The circuit to the left
of the terminals a-b in
Fig.(c) is known as the
Norton equivalent
circuit;
Linear
+ v
Two-terminal
– Th
Circuit
I
Figure (a)
RTh
I
Figure (b)
LinearIN
Two-terminalRN
Circuit
I
Figure (c)
a
+
v_
b
a
+
v_
b
a
+
v_
b
Load
Load
Load
For now, we are mainly concerned with how to
get the Norton equivalent current IN and resistance
RN.
Linear
Two-terminal
Circuit
IN = ISC
Please notice the
current direction!
a
ISC
Linear circuit with
all independent
sources set
equal to zero
b
a
Req
b
RN = Req
We find RN in the same
way we find RTh.
RTh
In fact, from what we
know about source
transformation, for the
external circuit load, Fig.(b)
is equivalent to Fig.(c).
+ v
– Th
For this reason, source
IN
transformation is often
called Thevenin-Norton
transformation.
According to source transformation
RN = RTh = Req
∵ vTh = vOC
IN
v
 Th
RTh
IN = ISC
∴
a
+
v
_
Figure (b) b
a
I
+
RN
v
_
b
I
Load
Load
Figure (c)
RN  RTh  Req 
vOC
I SC
We can calculate any two of the three using the method that takes
3. The general steps for finding the Thevenin (or
Norton) equivalent circuit
① Find vOC or ISC
Three
steps
By any method we have learned before
② Find Req
③ Draw the equivalent circuit, please notice the
directions about vOC or ISC.
Example ：Find the Thevenin equivalent circuit of the circuit
shown in the figure, to the left of the terminals a-b.
2Ω
1A
2Ω
+
_
a
3V
4Ω
Three steps
b
Solution: ①Find voc
By source transformation,
we can obtain
By KVL
2Ω
-2+(2+2+4)I+3 = 0
I = -0.125A
2V
By Ohm’s law
voc = 4I = -0.5V
3V
2Ω
+ _
I
+
_
4Ω
+
a
voc
-b
2 Find Req
By turning off the independent sources
Req= 4∥4 = 4/2 = 2Ω
2Ω
a
Req
4Ω
2Ω
b
③Draw the equivalent
a circuit
2Ω
-0.5V
a
Or
2Ω
+
_
0.5V
b
_
+
b
If it is desired to Find the Norton equivalent circuit
of the same circuit, to the left of the terminals a-b.
①Find ISC ISC = -0.25A
a
② Find Req Req = 2Ω
③Draw the equivalent circuit
2Ω
0.25A
b
Simulation :
video:
Assignment:
P142. practice 4.8; P147,practice 4.11