Linearity
An element is said to be linear if it satisfies homogeneity (scaling)
property and additive (superposition) property.
1. homogeneity property
Let x be the input and y be the output of an element.
x →y
If kx is applied to the element, the output must be ky .
kx → ky
2. additivity property
x1 → y1 ,
x2 → y2
If (x1 + x2 ) is applied to the element, the output must be
y1 + y2 .
(x1 + x2 ) → y1 + y2
Linear Elements
I
Resistor
It is a linear element because its voltage-current relation
satisfies both homogeneity and additivity property.
v = iR
I
Inductor
It is a linear element.. Its voltage-current relationship1 is
v =L
I
Capacitor
It is also a linear element
2
. It i − v relationship is
i =C
1
2
di
dt
dv
dt
(differential operator is linear)
Inductor and capacitor should not have any initial energy
I
Dependent Sources
Dependent voltage and current sources are linear as long as
their input-output relationship is linear.
vo = kvi ,
vo = rii ,
io = kii ,
io = gii
A linear circuit consists of only linear elements and linear
dependent sources.
In linear circuit, the output and input are related by a linear
relationship.
−
+
I ∝V
V
Linear
Network
R
I
What about Independent Sources?
They are nonlinear as their i − v relationship does not satisfy the
linearity.
i
i
I
−
+
V
V
v
I
v
Figure: Independent voltage source
Figure: Independent current source
v =V
i =I
Test yourself
Find whether the element is linear or not.
i (A)
i
+
2
v
v (V)
-2
−
i =v −2
v1 → i1 = v1 − 2,
v2 → i2 = v2 − 2
(v1 + v2 ) → i3 = v1 + v2 − 2
i3 6= i1 + i2
It is nonlinear.
For an element to be linear, its i − v characteristics must not only
be a straight line but also pass through origin.
Superposition
It states that in a linear network with a number of independent
sources, the response can be found by summing the responses to
each independent source acting alone, with all other independent
sources set to zero.
1. Consider one independent source at a time while all other
independent sources are turned off. It means that voltage
source is replaced by short circuit and current source by open
circuit.
2. Leave dependent sources in the circuit as they are controlled
by other variables.
R1
−
+
V
e
R2
I
Let us find e using nodal analysis.
By KCL,
e −V
e
+
=I
R1
R2
e=
R2
R1 R2
V+
I
R1 + R2
R1 + R2
e = αV + βI
where α and β are constants.
Notice that e has two components as there are two independent
sources.
Let us analyze the same circuit using superposition.
1. Voltage source acting alone:
R1
eV
−
+
V
eV =
R2
V
R1 + R2
eI =
R1 R2
I
R1 + R2
R2
2. Current source acting alone:
R1
eI
R2
I
By superposition,
e = eV + eI =
R2
R1 R2
V+
I
R1 + R2
R1 + R2
Test yourself
Use superposition to find the vlaue of current ix .
−
+
8V
2A
3Ω
3ix
−
+
6Ω i
x
Voltage source acting alone:
6Ω i
3Ω
x
−8 + 6ix + 3ix + 3ix = 0
3ix
−
+
−
+
8V
2
ix = A
3
Current source acting alone:
By KCL,
3Ω
2A
3ix
3ix − e
−e
+2+
=0
6
3
−
+
6Ω i
x e
ix = −
e
6
By solving this, ix = −
By superposition,
ix = ixV + ixI
ix =
2 1
1
− = A
3 2
6
1
A.
2
Thevenin’s Theorem
Any 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 sources are turned off.
R
RTh
i
+
v
−
+
−
+
Vm In
i
VTh
v
−
−
+
Proof
Let us apply a current source and find the voltage.
i
R
+
v
Vm In
i
−
+
−
Since the network is linear, its terminal voltage can be found by
superposition.
The voltage v will have as many components as the number of
independent sources.
X
X
v=
αm Vm +
βn In + Ri
m
n
v = VTh + RTh i
This equation can be represented in a network form as shown
below.
RTh
i
+
−
+
VTh
v
i
−
This makes us replace any linear network with its Thevenin
equivalent.
−
+
Vm
In
i
RTh
+
v
−
−
+
R
VTh
i
+
v
−
The Thevenin equivalent circuit for any linear network at a given
pair of terminals consists of a voltage source VTh in series with a
resistor RTh . The voltage VTh and RTh can be obtained as follows.
1. VTh can be found by calculating the open circuit voltage at
the designated terminal pair on the original network.
2. RTh can be found by calculating the resistance of the
open-circuit network seen from the designated terminal pair
with all independent sources internal to the network set to
zero.
If the network has dependent sources, RTh can be found as follows.
1. Turn off all independent sources. Do not turn off dependent
sources.
2. Apply a test voltage vo at the terminals and calculate the
current io .
io
Network with
−
+
all independent
sources set to zero
RTh =
vo
io
vo
Example
R1
−
+
V
e
R2
I
Let us find e using Thevenin equivalent.
To find VTh :
R1
e
+
−
+
V
Voc
−
I
VTh = V + IR1
To find RTh :
R1
RTh = R1
RTh
Thevenin equivalent is
RTh
−
+
VTh
e
R2
By voltage division,
e = VTh
R2
R2
R2
R1 R2
= (V + IR1 )
=V
+I
R2 + RTh
R2 + R1
R1 + R2
R1 + R2
Norton’s Theorem
It 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 input or equivalent resistance at the terminals
when the independent sources are turned off.
R
i
i
+
Vm In
v
−
RN = RTh
+
IN
RN
v
−
−
+
The Norton equivalent circuit for any linear network at a given pair
of terminals consists of a current source IN in parallel with a resistor
RN . The current IN and resistance RN can be obtained as follows:
1. IN can be found by applying a short at the designated
terminal pair on the original network and calculating or
measuring the current through the short circuit.
R
Vm In
Isc
−
+
IN = Isc
2. RN can be found in the same manner as RTh .
Proof
Let us apply a voltage source and find the current.
R
i
v
−
+
Vm In
−
+
Since the network is linear, the current can be found by
superposition.
The current i will have as many components as the number of
independent sources.
!
X
X
v
i =−
αm Vm +
β n In +
R
m
n
v
RN
This equation can be represented in a network form as shown
below.
i
i = −IN +
RN
−
+
IN
v
This makes us replace any linear network with its Norton
equivalent.
R
i
i
+
Vm In
v
−
+
IN
RN
v
−
−
+
Source Transformation
a
a
R
I
−
+
V
R
b
b
(a)
(b)
These two circuits are equal as long as they have same i − v
characteristics at their terminals.
V = IR
I =
V
R
Thevenin and Norton Transformation
a
a
RTh
IN
−
+
VTh
RTh
b
b
(c)
(d)
VTh = IN RTh
IN =
VTh
RTh
Maximum Power Transfer
RTh
−
+
VTh
RL
The power delivered to the load is
p = i 2 RL =
2
VTh
RL
(RTh + RL )2
p
pmax
RTh
RL
dp
=0
dRL
2
VTh
(RTh + RL )2 − 2RL (RTh + RL )
(RTh + RL )4
=0
RL = RTh
Maximum power is transferred to the load when the load resistance
equals the Thevenin resistance as seen from the load (RL = RTh ).
The maximum power transferred to load is
pmax =
2
VTh
4RTh