Chapter 5
Department of Mechanical Engineering
Chapter 5
◆
Source Transformation
By KVL:
Vs=iRs + v
Vs/Rs=i + v/Rs
is=Vs/Rs
Rs=Rp
is=i + v/Rp
By KCL:
is=i + v/Rp
Two circuits have
the same terminal
voltage and current
Department of Mechanical Engineering
Chapter 5
◆
Source Transformation
Department of Mechanical Engineering
Chapter 5
◆
Source Transformation
Example 1: Find the values of is and R in two circuits if they are
equivalent
R=10 Ω
is=12/R=1.2 A
Department of Mechanical Engineering
Chapter 5
◆
Source Transformation
Example 2: Find current i in circuit (a)
i=(5V-1.2V)/(5Ω+12Ω)=0.224A
Department of Mechanical Engineering
Chapter 5
◆
Superposition
Example1:(a) A circuit containing two independent sources. (b) The circuit after
the ideal ammeter has been replaced by the equivalent short circuit and a label
has been added to indicate the current measured by the ammeter im.
De-activate the current source
De-activate the voltage source
im=i1+ i2=1.33 Α
i1=6/(3+6)=0.67 Α
i2=[3/(3+6)]% 2=0.67 Α Department of Mechanical Engineering
Chapter 5
◆
Superposition
Example2: a) A circuit containing two independent sources. (b) The circuit after
the ideal voltmeter has been replaced by the equivalent open circuit and a label
has been added to indicate the voltage measured by the voltmeter vm.
With only the
voltage source
With only the
current source
Department of Mechanical Engineering
Chapter 5
◆
Superposition
Example3: (a) A circuit containing two independent sources. (b) The circuit after
the ideal ammeter has been replaced by the equivalent short circuit and a label has
been added to indicate the current measured by the ammeter im.
With only the
voltage source
With only the
current source
Department of Mechanical Engineering
Chapter 5
◆
Superposition
Example4: (a) A circuit containing two independent sources. (b) The circuit after
the ideal voltmeter has been replaced by the equivalent open circuit and a label
has been added to indicate the voltage measured by the voltmeter vm.
With only the
current source
With only the
voltage source
Department of Mechanical Engineering
Chapter 5
◆
Superposition
Example5:Find current i.
De-activate the current source
De-activate the voltage source
Department of Mechanical Engineering
Chapter 5
◆
Superposition
Example5: Find current i.
For circuit (a): by KVL, we have
For circuit (b): we use node voltage
analysis at node a
-i2 –7+ (va-3i2)/2=0
24-(3+2)i1-3i1=0
Then
i1=3 (A)
For 3Ω resistor, we have
-i2=va/3
Then
i2=-7/4
i=i1 +i2 =1.25 (A)
Department of Mechanical Engineering
Chapter 5
◆
Thévenin’s Theorem
(a) A circuit partitioned into two parts: circuit A and circuit B.
(b) Replacing circuit A by its Thévenin equivalent circuit.
A: Driving circuit
B: Load
Department of Mechanical Engineering
Chapter 5
◆
Thévenin’s Theorem
The Thévenin equivalent circuit involves three parameters:
(a) the open-circuit voltage, voc,
(b) the short-circuit current isc, and
(c) the Thévenin resistance, Rt.
voc=Rt isc
Department of Mechanical Engineering
Chapter 5
◆
Thévenin’s Theorem
(a) The Thévenin resistance, Rt,
(b) A method for measuring or calculating the Thévenin
resistance, Rt.
Rt=vt/it
Department of Mechanical Engineering
Chapter 5
◆
Thévenin’s Theorem
Example1: Find current i using
Thévenin’s Theorem
+
voc
Rt=4+5//20=8Ω
-
Voc=
20
20+5
=40 V
Steps for determining the Thévenin equivalent
circuit for the circuit left of the terminals
% 50
40
i= R+8
Department of Mechanical Engineering
Chapter 5
◆
Thévenin’s Theorem
Example2: Find the Thévenin equivalent circuit for:
Keep in mind:
The Thévenin equivalent circuit involves
three parameters:
(a) the open-circuit voltage, voc,
(b) the short-circuit current isc, and
(c) the Thévenin resistance, Rt.
voc=Rt isc
Department of Mechanical Engineering
Chapter 5
◆
Thévenin’s Theorem
First, find Rt:
Circuit reduction by
de-activate all ideal
sources
Then find the
equivalent resistant
Rt =10//40 + 4=12Ω
Department of Mechanical Engineering
Chapter 5
◆
Thévenin’s Theorem
Then, find voc:
+
voc
-
Using Node voltage mothod to find vc, since 1-b
is open circuit, no voltage drop for 4Ω resistor,
voc= vc
vc -10
vc
10
+
40
+ 2=0
Solve for vc
vc =-8V
Department of Mechanical Engineering
Chapter 5
◆
Thévenin’s Theorem
For circuit with dependent sources, we can not directly
obtain the Rt from simple circuit reduction.
The procedure to get Rt :
• Find open circuit voltage voc,
voc
• Find the short-circuit current isc,
Rt =
isc
Example 3: Find the Thévenin’s equivalent circuit for the following circuit:
Department of Mechanical Engineering
Chapter 5
◆
Thévenin’s Theorem
First, find open circuit voltage Voc
Voc
For the left loop, apply KVL:
20-6i+2i-6i=0
i=2 (A)
Voc =6i=12 (V)
Department of Mechanical Engineering
Chapter 5
◆
Thévenin’s Theorem
Make a-b a short circuit, and find isc,
By mesh current method, we have
20-6i1+2i1-6(i1-i2)=0
And
-6(i2-i1)-10i2 =0
Rt =
voc
isc
i2 = isc=120/136 (A)
Department of Mechanical Engineering
Chapter 5
◆
Thévenin’s Theorem
The Thévenin’s resistance is
voc
10
=
=13.6 Ω
Rt =
isc
120/136
The Thévenin’s equivalent circuit is
Department of Mechanical Engineering
Chapter 5
◆
Norton’s equivalent Circuit
(a) A circuit partitioned into two parts: circuit A and circuit B.
(b) Replacing circuit A by its Norton equivalent circuit.
Norton equivalent is simply the source transformation of the
Thévenin equivalent
Department of Mechanical Engineering
Chapter 5
◆
Norton’s equivalent Circuit
Example1: Find the Norton Equivalent Circuit for
Find Rn by replacing the
voltage source with a short
circuit
6x12
Rn =
6+12
=
4 kΩ
Department of Mechanical Engineering
Chapter 5
◆
Norton’s equivalent Circuit
Find the short circuit current isc,
Apply KVL for the large
loop:
15-12000isc=0
isc=1.25mA
isc=1.25mA
(Note: No current for the 6kΩ resistor
—it is shorted)
Rn =4000Ω
Department of Mechanical Engineering
Chapter 5
◆
Norton’s equivalent Circuit
Example2: Find the Norton Equivalent Circuit for
First, find the open circuit voltage:
Apply KVL for the close loop:
voc
12+6ia-2ia=0
ia=-3(A)
voc=2ia=-6(V)
Department of Mechanical Engineering
Chapter 5
◆
Norton’s equivalent Circuit
Then, find the short circuit current:
isc
Apply KVL for the left loop:
12+6ia-2ia=0
ia=-3(A)
2ia
=-2 (A)
isc=
3
Rt =
voc
-6 V =3Ω
= -2 A
isc
Department of Mechanical Engineering
Chapter 5
◆
Maximum Power Transfer
For a circuit A and load resistor RL
Circuit A contains resistors
and independent and
dependent sources.
The Thévenin equivalent is
substituted for circuit A.
Here we use vs for the
Thévenin source voltage.
Department of Mechanical Engineering
Chapter 5
◆
Maximum Power Transfer
It can be proved that when
Rt=RL
maximum power transferred from circuit A to the
vs 2
load resistor, and the power is
Pmax=
4Rt
We can also use Norton’s
equivalent circuit to substitute
circuit A. Here we use is as the
Norton source current.
Again, the maximum power occurs at Rt=RL
and the maximum power is
Pmax=
Rt is2
4
Department of Mechanical Engineering
Chapter 5
◆
Maximum Power Transfer
Example: Find the Load RL that result in maximum
power delivered to the load. Also determine Pmax
First, we use the circuit (b) to obtain the
Thévenin equivalent circuit.
Find the open circuit voltage voc. Apply
KVL to the close loop:
6-6i+2vab-4i=0
And
voc
vab=4i
i=3A
voc= vab= 12 (V)
Department of Mechanical Engineering
Chapter 5
◆
Maximum Power Transfer
Find the short circuit current for circuit (c), isc.
Since ab is short, vab=0.
Apply KVL to the close loop:
6-6isc+2vab=0
isc=1 (A)
Find the equivalent resistance:
Rt =
voc
=12 Ω
isc
RL=Rt=12 Ω
voc2
Pmax=
4Rt
122
= 4(12) =3 W
Department of Mechanical Engineering