EE101: Basics
KCL, KVL, power, Thevenin’s theorem
M. B. Patil
mbpatil@ee.iitb.ac.in
www.ee.iitb.ac.in/~sequel
Department of Electrical Engineering
Indian Institute of Technology Bombay
M. B. Patil, IIT Bombay
Kirchhoff’s laws
v3
A
i2
R2
v2
i3
E
R3
i6
v6
α v4
v4
B
V0
C
i4
i5
v5
I0
R1
v1
i1
D
M. B. Patil, IIT Bombay
Kirchhoff’s laws
v3
A
i2
R2
v2
i3
E
R3
i6
v6
α v4
v4
B
V0
C
i4
i5
v5
I0
R1
v1
i1
D
* Kirchhoff’s
current law (KCL):
P
ik = 0 at each node.
M. B. Patil, IIT Bombay
Kirchhoff’s laws
v3
A
i2
R2
v2
i3
E
R3
i6
v6
α v4
v4
B
V0
C
i4
i5
v5
I0
R1
v1
i1
D
* Kirchhoff’s
current law (KCL):
P
ik = 0 at each node.
e.g., at node B, −i3 + i6 + i4 = 0.
M. B. Patil, IIT Bombay
Kirchhoff’s laws
v3
A
i2
R2
v2
i3
E
R3
i6
v6
α v4
v4
B
V0
C
i4
i5
v5
I0
R1
v1
i1
D
* Kirchhoff’s
current law (KCL):
P
ik = 0 at each node.
e.g., at node B, −i3 + i6 + i4 = 0.
(We have followed the convention that current leaving a node is positive.)
M. B. Patil, IIT Bombay
Kirchhoff’s laws
v3
A
i2
R2
v2
i3
E
R3
i6
v6
α v4
v4
B
V0
C
i4
i5
v5
I0
R1
v1
i1
D
* Kirchhoff’s
current law (KCL):
P
ik = 0 at each node.
e.g., at node B, −i3 + i6 + i4 = 0.
(We have followed the convention that current leaving a node is positive.)
* Kirchhoff’s
voltage law (KVL):
P
vk = 0 for each loop.
M. B. Patil, IIT Bombay
Kirchhoff’s laws
v3
A
i2
R2
v2
i3
E
R3
i6
v6
α v4
v4
B
V0
C
i4
i5
v5
I0
R1
v1
i1
D
* Kirchhoff’s
current law (KCL):
P
ik = 0 at each node.
e.g., at node B, −i3 + i6 + i4 = 0.
(We have followed the convention that current leaving a node is positive.)
* Kirchhoff’s
voltage law (KVL):
P
vk = 0 for each loop.
e.g., v3 + v6 − v1 − v2 = 0.
M. B. Patil, IIT Bombay
Kirchhoff’s laws
v3
A
i2
R2
v2
i3
E
R3
i6
v6
α v4
v4
B
V0
C
i4
i5
v5
I0
R1
v1
i1
D
* Kirchhoff’s
current law (KCL):
P
ik = 0 at each node.
e.g., at node B, −i3 + i6 + i4 = 0.
(We have followed the convention that current leaving a node is positive.)
* Kirchhoff’s
voltage law (KVL):
P
vk = 0 for each loop.
e.g., v3 + v6 − v1 − v2 = 0.
(We have followed the convention that voltage drop across a branch is positive.)
M. B. Patil, IIT Bombay
Circuit elements
Element
Symbol
v
Resistor
Equation
v =Ri
i
v
Inductor
i
v =L
di
dt
i =C
dv
dt
v
Capacitor
i
v
Diode
to be discussed
i
C
BJT
to be discussed
B
E
M. B. Patil, IIT Bombay
Sources
Element
Independent
Voltage source
Current source
Dependent
VCVS
VCCS
CCVS
CCCS
Symbol
v
i
v
i
v
i
v
i
v
i
v
i
Equation
v (t) = vs (t)
i(t) = is (t)
v (t) = α vc (t)
i(t) = g vc (t)
v (t) = r ic (t)
i(t) = β ic (t)
* α, β: dimensionless, r : Ω, g : Ω−1 or f (“mho”)
* The subscript ‘c’ denotes the controlling voltage or current.
M. B. Patil, IIT Bombay
Instantaneous power absorbed by an element
i2
i1
V1
VN
iN
V2
V3
i3
P(t) = V1 (t) i1 (t) + V2 (t) i2 (t) + · · · + VN (t) iN (t) ,
where V1 , V2 , etc. are “node voltages” (measured
with respect to a reference node).
M. B. Patil, IIT Bombay
Instantaneous power absorbed by an element
i2
i1
V1
V2
i3
P(t) = V1 (t) i1 (t) + V2 (t) i2 (t) + · · · + VN (t) iN (t) ,
V3
where V1 , V2 , etc. are “node voltages” (measured
with respect to a reference node).
VN
iN
* two-terminal element:
P = V1 i1 + V2 i2
v
V1
i1
i2
V2
= V1 i1 + V2 (−i1 )
= [V1 − V2 ] i1 = v i1
M. B. Patil, IIT Bombay
Instantaneous power absorbed by an element
i2
i1
V1
V2
i3
P(t) = V1 (t) i1 (t) + V2 (t) i2 (t) + · · · + VN (t) iN (t) ,
V3
where V1 , V2 , etc. are “node voltages” (measured
with respect to a reference node).
VN
iN
* two-terminal element:
P = V1 i1 + V2 i2
v
V1
i2
i1
V2
= V1 i1 + V2 (−i1 )
= [V1 − V2 ] i1 = v i1
* three-terminal element:
VC
VB
iB
VE
P = VB iB + VC iC + VE (−iE )
iC
iE
= VB iB + VC iC − VE (iB + iC )
= (VB − VE ) iB + (VC − VE ) iC
= VBE iB + VCE iE
M. B. Patil, IIT Bombay
Instantaneous power
* A resistor can only absorb power (from the circuit) since v and i have the same
sign, making P > 0. The energy “absorbed” by a resistor goes in heating the
resistor and the rest of the world.
M. B. Patil, IIT Bombay
Instantaneous power
* A resistor can only absorb power (from the circuit) since v and i have the same
sign, making P > 0. The energy “absorbed” by a resistor goes in heating the
resistor and the rest of the world.
* Often, a “heat sink” is provided to dissipate the thermal energy effectively so
that the device temperature does not become too high.
M. B. Patil, IIT Bombay
Instantaneous power
* A resistor can only absorb power (from the circuit) since v and i have the same
sign, making P > 0. The energy “absorbed” by a resistor goes in heating the
resistor and the rest of the world.
* Often, a “heat sink” is provided to dissipate the thermal energy effectively so
that the device temperature does not become too high.
* A source (e.g., a DC voltage source) can absorb or deliver power since the signs
of v and i are independent. For example, when a battery is charged, it absorbs
energy which gets stored within.
M. B. Patil, IIT Bombay
Instantaneous power
* A resistor can only absorb power (from the circuit) since v and i have the same
sign, making P > 0. The energy “absorbed” by a resistor goes in heating the
resistor and the rest of the world.
* Often, a “heat sink” is provided to dissipate the thermal energy effectively so
that the device temperature does not become too high.
* A source (e.g., a DC voltage source) can absorb or deliver power since the signs
of v and i are independent. For example, when a battery is charged, it absorbs
energy which gets stored within.
* A capacitor can absorb or deliver power. When it is absorbing power, its charge
builds up. Similarly, an inductor can store energy (in the form of magnetic flux).
M. B. Patil, IIT Bombay
Resistors in series
v1
A
i
R1
v2
R2
v3
R3
v
B
A
i
B
R
M. B. Patil, IIT Bombay
Resistors in series
v1
A
i
R1
v2
R2
v3
R3
v
B
A
i
B
R
v1 = i R1 , v2 = i R2 , v3 = i R3 , ⇒ v = v1 + v2 + v3 = i (R1 + R2 + R3 )
M. B. Patil, IIT Bombay
Resistors in series
v1
A
i
R1
v2
R2
v3
R3
v
B
A
i
B
R
v1 = i R1 , v2 = i R2 , v3 = i R3 , ⇒ v = v1 + v2 + v3 = i (R1 + R2 + R3 )
* The equivalent resistance is Req = R1 + R2 + R3 .
M. B. Patil, IIT Bombay
Resistors in series
v1
A
i
R1
v2
R2
v3
R3
v
B
A
i
B
R
v1 = i R1 , v2 = i R2 , v3 = i R3 , ⇒ v = v1 + v2 + v3 = i (R1 + R2 + R3 )
* The equivalent resistance is Req = R1 + R2 + R3 .
* The voltage drop across Rk is v ×
Rk
.
Req
M. B. Patil, IIT Bombay
Resistors in parallel
v
i1
A
i
R1
i2
R2
i3
R3
v
B
A
i
B
R
M. B. Patil, IIT Bombay
Resistors in parallel
v
i1
A
i
R1
i2
R2
i3
R3
v
B
A
i
B
R
i1 = G1 v , i2 = G2 v , i3 = G3 v , where G1 = 1/R1 , etc.
⇒ i = i1 + i2 + i3 = (G1 + G2 + G3 ) v .
M. B. Patil, IIT Bombay
Resistors in parallel
v
i1
A
i
R1
i2
R2
i3
R3
v
B
A
i
B
R
i1 = G1 v , i2 = G2 v , i3 = G3 v , where G1 = 1/R1 , etc.
⇒ i = i1 + i2 + i3 = (G1 + G2 + G3 ) v .
* The equivalent conductance is Geq = G1 + G2 + G3 , and the equivalent
resistance is Req = 1/Geq .
M. B. Patil, IIT Bombay
Resistors in parallel
v
i1
A
i
R1
i2
R2
i3
R3
v
B
A
i
B
R
i1 = G1 v , i2 = G2 v , i3 = G3 v , where G1 = 1/R1 , etc.
⇒ i = i1 + i2 + i3 = (G1 + G2 + G3 ) v .
* The equivalent conductance is Geq = G1 + G2 + G3 , and the equivalent
resistance is Req = 1/Geq .
* The current through Rk is i ×
Gk
.
Geq
M. B. Patil, IIT Bombay
Resistors in parallel
v
i1
A
i
R1
i2
R2
i3
R3
v
B
A
i
B
R
i1 = G1 v , i2 = G2 v , i3 = G3 v , where G1 = 1/R1 , etc.
⇒ i = i1 + i2 + i3 = (G1 + G2 + G3 ) v .
* The equivalent conductance is Geq = G1 + G2 + G3 , and the equivalent
resistance is Req = 1/Geq .
* The current through Rk is i ×
Gk
.
Geq
* If N = 2, we have
R1 R2
R2
R1
Req =
, i1 = i ×
, i2 = i ×
.
R1 + R2
R1 + R2
R1 + R2
M. B. Patil, IIT Bombay
Resistors in parallel
v
i1
A
i
R1
i2
R2
i3
R3
v
B
A
i
B
R
i1 = G1 v , i2 = G2 v , i3 = G3 v , where G1 = 1/R1 , etc.
⇒ i = i1 + i2 + i3 = (G1 + G2 + G3 ) v .
* The equivalent conductance is Geq = G1 + G2 + G3 , and the equivalent
resistance is Req = 1/Geq .
* The current through Rk is i ×
Gk
.
Geq
* If N = 2, we have
R1 R2
R2
R1
Req =
, i1 = i ×
, i2 = i ×
.
R1 + R2
R1 + R2
R1 + R2
* If Rk = 0, all of the current will go through Rk .
M. B. Patil, IIT Bombay
Example
i1
2Ω
4Ω
6V
(a)
i2
3Ω
5
3Ω
2.5
2.5
Example
i1
2Ω
4Ω
6V
(a)
i2
3Ω
5
3Ω
2.5
2.5
Example
i1
6V
(a)
i1
2Ω
4Ω
4 Ω i2
i2
3Ω
5
3Ω
2.5
6V
2.5
(b)
2Ω
3Ω
1Ω
3Ω
Example
i1
6V
(a)
i1
2Ω
4Ω
4 Ω i2
i2
3Ω
5
3Ω
2.5
6V
2.5
(b)
2Ω
3Ω
1Ω
3Ω
Example
i1
6V
4 Ω i2
i2
3Ω
5
3Ω
(a)
i1
4Ω
6V
(c)
i1
2Ω
4Ω
i2
6
3
2.5
6V
2.5
(b)
2Ω
3Ω
1Ω
3Ω
Example
i1
6V
4 Ω i2
i2
3Ω
5
3Ω
(a)
i1
4Ω
6V
(c)
i1
2Ω
4Ω
i2
6
3
2.5
6V
2.5
(b)
2Ω
3Ω
1Ω
3Ω
Example
i1
6V
4 Ω i2
i2
3Ω
5
i1
(b)
i1
4Ω
6V
6V
2.5
2.5
3Ω
(a)
(c)
i1
2Ω
4Ω
i2
4Ω
6
6V
3
(d)
2Ω
2Ω
3Ω
1Ω
3Ω
Example
i1
6V
4 Ω i2
i2
3Ω
5
i1
6V
2.5
2.5
3Ω
(a)
6V
2Ω
3Ω
1Ω
3Ω
(b)
i1
4Ω
(c)
i1
2Ω
4Ω
i2
4Ω
6
6V
3
(d)
2Ω
i1 =
6V
= 1A.
4Ω+2Ω
Example
i1
6V
4 Ω i2
i2
3Ω
5
i1
6V
2.5
2.5
3Ω
(a)
6V
2Ω
3Ω
1Ω
3Ω
(b)
i1
4Ω
(c)
i1
2Ω
4Ω
i2
4Ω
6
6V
3
2Ω
i1 =
6V
= 1A.
4Ω+2Ω
i2 = i1 ×
2
6Ω
= A.
6Ω+3Ω
3
(d)
M. B. Patil, IIT Bombay
Example
i1
6V
4 Ω i2
i2
3Ω
5
i1
6V
2.5
2.5
3Ω
(a)
6V
2Ω
3Ω
1Ω
3Ω
(b)
i1
4Ω
(c)
i1
2Ω
4Ω
i2
4Ω
6
6V
3
2Ω
i1 =
6V
= 1A.
4Ω+2Ω
i2 = i1 ×
2
6Ω
= A.
6Ω+3Ω
3
(d)
Home work:
* Verify that KCL and KVL are satisfied for each node/loop.
M. B. Patil, IIT Bombay
Example
i1
6V
4 Ω i2
i2
3Ω
5
i1
6V
2.5
2.5
3Ω
(a)
6V
2Ω
3Ω
1Ω
3Ω
(b)
i1
4Ω
(c)
i1
2Ω
4Ω
i2
4Ω
6
6V
3
2Ω
i1 =
6V
= 1A.
4Ω+2Ω
i2 = i1 ×
2
6Ω
= A.
6Ω+3Ω
3
(d)
Home work:
* Verify that KCL and KVL are satisfied for each node/loop.
* Verify that the total power absorbed by the resistors is equal to the power
supplied by the source.
M. B. Patil, IIT Bombay
Nodal analysis
V1
R1
R2
I0
V2
v3
R3
k v3
0
R4
V3
M. B. Patil, IIT Bombay
Nodal analysis
* Take some node as the “reference node” and denote
the node voltages of the remaining nodes by V1 , V2 ,
etc.
V1
R1
R2
I0
V2
v3
R3
k v3
0
R4
V3
M. B. Patil, IIT Bombay
Nodal analysis
* Take some node as the “reference node” and denote
the node voltages of the remaining nodes by V1 , V2 ,
etc.
V1
R1
R2
I0
* Write KCL at each node in terms of the node
voltages. Follow a fixed convention, e.g., current
leaving a node is positive.
V2
v3
R3
k v3
0
R4
V3
M. B. Patil, IIT Bombay
Nodal analysis
* Take some node as the “reference node” and denote
the node voltages of the remaining nodes by V1 , V2 ,
etc.
V1
R1
R2
I0
* Write KCL at each node in terms of the node
voltages. Follow a fixed convention, e.g., current
leaving a node is positive.
V2
v3
R3
k v3
0
R4
V3
1
(V1 − V2 ) − I0 − k (V2 − V3 ) = 0 ,
R1
1
1
1
(V2 − V1 ) +
(V2 − V3 ) +
(V2 ) = 0 ,
R1
R3
R2
1
1
k (V2 − V3 ) +
(V3 − V2 ) +
(V3 ) = 0 .
R3
R4
M. B. Patil, IIT Bombay
Nodal analysis
* Take some node as the “reference node” and denote
the node voltages of the remaining nodes by V1 , V2 ,
etc.
V1
R1
R2
I0
* Write KCL at each node in terms of the node
voltages. Follow a fixed convention, e.g., current
leaving a node is positive.
V2
v3
R3
k v3
0
R4
V3
1
(V1 − V2 ) − I0 − k (V2 − V3 ) = 0 ,
R1
1
1
1
(V2 − V1 ) +
(V2 − V3 ) +
(V2 ) = 0 ,
R1
R3
R2
1
1
k (V2 − V3 ) +
(V3 − V2 ) +
(V3 ) = 0 .
R3
R4
* Solve for the node voltages → branch voltages and
currents.
M. B. Patil, IIT Bombay
Nodal analysis
* Take some node as the “reference node” and denote
the node voltages of the remaining nodes by V1 , V2 ,
etc.
V1
R1
R2
I0
* Write KCL at each node in terms of the node
voltages. Follow a fixed convention, e.g., current
leaving a node is positive.
V2
v3
R3
k v3
0
R4
V3
1
(V1 − V2 ) − I0 − k (V2 − V3 ) = 0 ,
R1
1
1
1
(V2 − V1 ) +
(V2 − V3 ) +
(V2 ) = 0 ,
R1
R3
R2
1
1
k (V2 − V3 ) +
(V3 − V2 ) +
(V3 ) = 0 .
R3
R4
* Solve for the node voltages → branch voltages and
currents.
* Remark: Nodal analysis needs to be modified if there
are voltage sources.
M. B. Patil, IIT Bombay
Mesh analysis
R2
R1
R3
Vs
i1
is
r1 i s
i2
M. B. Patil, IIT Bombay
Mesh analysis
R2
R1
R3
Vs
i1
is
r1 i s
i2
* Write KVL for each loop in terms of the “mesh currents” i1 and i2 . Use a fixed
convention, e.g., voltage drop is positive. (Note that is = i1 − i2 .)
M. B. Patil, IIT Bombay
Mesh analysis
R2
R1
R3
Vs
i1
is
r1 i s
i2
* Write KVL for each loop in terms of the “mesh currents” i1 and i2 . Use a fixed
convention, e.g., voltage drop is positive. (Note that is = i1 − i2 .)
−Vs + i1 R1 + (i1 − i2 ) R3 = 0 ,
R2 i2 + r1 (i1 − i2 ) + (i2 − i1 ) R3 = 0 .
M. B. Patil, IIT Bombay
Mesh analysis
R2
R1
R3
Vs
i1
is
r1 i s
i2
* Write KVL for each loop in terms of the “mesh currents” i1 and i2 . Use a fixed
convention, e.g., voltage drop is positive. (Note that is = i1 − i2 .)
−Vs + i1 R1 + (i1 − i2 ) R3 = 0 ,
R2 i2 + r1 (i1 − i2 ) + (i2 − i1 ) R3 = 0 .
* Solve for i1 and i2 → compute other quantities of interest (branch currents and
branch voltages).
M. B. Patil, IIT Bombay
Linearity and superposition
* A circuit containing independent sources, dependent sources, and resistors is
linear, i.e., the system of equations describing the circuit is linear.
M. B. Patil, IIT Bombay
Linearity and superposition
* A circuit containing independent sources, dependent sources, and resistors is
linear, i.e., the system of equations describing the circuit is linear.
* The dependent sources are assumed to be linear, e.g., if we have a CCVS with
v = a ic2 + b, the resulting system will be no longer linear.
M. B. Patil, IIT Bombay
Linearity and superposition
* A circuit containing independent sources, dependent sources, and resistors is
linear, i.e., the system of equations describing the circuit is linear.
* The dependent sources are assumed to be linear, e.g., if we have a CCVS with
v = a ic2 + b, the resulting system will be no longer linear.
* For a linear system, we can apply the principle of superposition.
M. B. Patil, IIT Bombay
Linearity and superposition
* A circuit containing independent sources, dependent sources, and resistors is
linear, i.e., the system of equations describing the circuit is linear.
* The dependent sources are assumed to be linear, e.g., if we have a CCVS with
v = a ic2 + b, the resulting system will be no longer linear.
* For a linear system, we can apply the principle of superposition.
* In the context of circuits, superposition enables us to consider the independent
sources one at a time, compute the desired quantity of interest in each case, and
get the net result by adding the individual contributions.
M. B. Patil, IIT Bombay
Linearity and superposition
* A circuit containing independent sources, dependent sources, and resistors is
linear, i.e., the system of equations describing the circuit is linear.
* The dependent sources are assumed to be linear, e.g., if we have a CCVS with
v = a ic2 + b, the resulting system will be no longer linear.
* For a linear system, we can apply the principle of superposition.
* In the context of circuits, superposition enables us to consider the independent
sources one at a time, compute the desired quantity of interest in each case, and
get the net result by adding the individual contributions.
* Caution: Superposition cannot be applied to dependent sources.
M. B. Patil, IIT Bombay
Superposition
* Superposition refers to superposition of response due to independent sources.
M. B. Patil, IIT Bombay
Superposition
* Superposition refers to superposition of response due to independent sources.
* We can consider one independent source at a time, deactivate all other
independent sources.
M. B. Patil, IIT Bombay
Superposition
* Superposition refers to superposition of response due to independent sources.
* We can consider one independent source at a time, deactivate all other
independent sources.
* Deactivating a current source ⇒ is = 0, i.e., replace the current source with an
open circuit.
M. B. Patil, IIT Bombay
Superposition
* Superposition refers to superposition of response due to independent sources.
* We can consider one independent source at a time, deactivate all other
independent sources.
* Deactivating a current source ⇒ is = 0, i.e., replace the current source with an
open circuit.
* Deactivating a voltage source ⇒ vs = 0, i.e., replace the voltage source with a
short circuit.
M. B. Patil, IIT Bombay
Example
2Ω
i1
18 V
4Ω
3A
Example
Case 1: Keep Vs , deactivate Is .
2Ω
i1
2Ω
i1
18 V
4Ω
18 V
3A
4Ω
Example
Case 1: Keep Vs , deactivate Is .
2Ω
i1
2Ω
i1
18 V
4Ω
18 V
3A
4Ω
(1)
i1 = 3 A
Example
Case 1: Keep Vs , deactivate Is .
2Ω
i1
2Ω
i1
18 V
4Ω
18 V
(1)
4Ω
i1 = 3 A
3A
Case 2: Keep Is , deactivate Vs .
2Ω
i1
4Ω
3A
Example
Case 1: Keep Vs , deactivate Is .
2Ω
i1
2Ω
i1
18 V
4Ω
18 V
(1)
4Ω
i1 = 3 A
3A
Case 2: Keep Is , deactivate Vs .
2Ω
i1
4Ω
(2)
3A
i1 = 3 A ×
2Ω
= 1A
2Ω+4Ω
Example
Case 1: Keep Vs , deactivate Is .
2Ω
i1
2Ω
i1
4Ω
18 V
18 V
(1)
4Ω
i1 = 3 A
3A
Case 2: Keep Is , deactivate Vs .
2Ω
(1)
i1
(2)
inet
1 = i1 + i1 = 3 + 1 = 4 A
4Ω
(2)
3A
i1 = 3 A ×
2Ω
= 1A
2Ω+4Ω
M. B. Patil, IIT Bombay
Example
i
12 V
v
3Ω
6A
1Ω
2i
Example
Case 1: Keep Vs , deactivate Is .
i
i
12 V
12 V
v
1Ω
v
3Ω
6A
1Ω
2i
2i
3Ω
Example
Case 1: Keep Vs , deactivate Is .
i
i
12 V
12 V
v
1Ω
v
3Ω
6A
1Ω
2i
3Ω
KVL: − 12 + 3 i + 2 i + i = 0
⇒ i = 2 A , v(1) = 6 V .
2i
Example
Case 1: Keep Vs , deactivate Is .
i
i
12 V
12 V
v
3Ω
1Ω
v
3Ω
2i
6A
1Ω
2i
Case 2: Keep Is , deactivate Vs .
i
KVL: − 12 + 3 i + 2 i + i = 0
⇒ i = 2 A , v(1) = 6 V .
v
3Ω
6A
1Ω
2i
Example
Case 1: Keep Vs , deactivate Is .
i
i
12 V
12 V
v
3Ω
1Ω
v
3Ω
2i
6A
1Ω
KVL: − 12 + 3 i + 2 i + i = 0
⇒ i = 2 A , v(1) = 6 V .
2i
Case 2: Keep Is , deactivate Vs .
i
v
3Ω
6A
1Ω
2i
KVL: i + (6 + i) 3 + 2 i = 0
⇒ i = −3 A , v(2) = (−3 + 6) × 3 = 9 V .
Example
Case 1: Keep Vs , deactivate Is .
i
i
12 V
12 V
v
3Ω
1Ω
v
3Ω
2i
6A
1Ω
KVL: − 12 + 3 i + 2 i + i = 0
⇒ i = 2 A , v(1) = 6 V .
2i
Case 2: Keep Is , deactivate Vs .
vnet = v(1) + v(2) = 6 + 9 = 15 V
i
v
3Ω
6A
1Ω
KVL: i + (6 + i) 3 + 2 i = 0
⇒ i = −3 A , v(2) = (−3 + 6) × 3 = 9 V .
2i
M. B. Patil, IIT Bombay
Superposition: Why does it work?
R1
R3
V1
A
R2
Vs
V2
B
Is
0
M. B. Patil, IIT Bombay
Superposition: Why does it work?
R1
R3
V1
A
R2
Vs
V2
B
Is
0
KCL at nodes A and B:
1
1
1
(V1 − Vs ) +
V1 +
(V1 − V2 ) = 0 ,
R1
R2
R3
1
−Is +
(V2 − V1 ) = 0 .
R3
M. B. Patil, IIT Bombay
Superposition: Why does it work?
R1
R3
V1
V2
B
A
R2
Vs
Is
0
KCL at nodes A and B:
1
1
1
(V1 − Vs ) +
V1 +
(V1 − V2 ) = 0 ,
R1
R2
R3
1
−Is +
(V2 − V1 ) = 0 .
R3
Writing in a matrix form, we get (using G1 = 1/R1 , etc.),
»
G1 + G2 + G3
−G3
−G3
G3
–»
V1
V2
–
=
»
G1 Vs
Is
–
M. B. Patil, IIT Bombay
Superposition: Why does it work?
R1
R3
V1
V2
B
A
R2
Vs
Is
0
KCL at nodes A and B:
1
1
1
(V1 − Vs ) +
V1 +
(V1 − V2 ) = 0 ,
R1
R2
R3
1
−Is +
(V2 − V1 ) = 0 .
R3
Writing in a matrix form, we get (using G1 = 1/R1 , etc.),
»
i.e.,
G1 + G2 + G3
−G3
»
A
V1
V2
–
=
»
−G3
G3
G1 Vs
Is
–»
–
»
→
V1
V2
V1
V2
–
=
–
»
=A
G1 Vs
Is
−1
»
–
G1 Vs
Is
–
.
M. B. Patil, IIT Bombay
Superposition: Why does it work?
R1
R3
V1
V2
B
A
R2
Vs
Is
0
»
V1
V2
–
=A
−1
»
G1 Vs
Is
–
»
≡
m11
m21
m12
m22
–»
G1 Vs
Is
–
.
M. B. Patil, IIT Bombay
Superposition: Why does it work?
R1
R3
V1
V2
B
A
R2
Vs
Is
0
»
V1
V2
–
=A
−1
»
G1 Vs
Is
–
»
≡
m11
m21
m12
m22
–»
–»
0
Is
G1 Vs
Is
–
.
We are now in a position to see why superposition works.
»
V1
V2
–
=
»
m11 G1
m21 G1
m12
m22
–»
Vs
0
– »
m11 G1
+
m21 G1
m12
m22
"
–
≡
(1)
V1
(1)
V2
# "
+
(2)
V1
(2)
#
.
V2
M. B. Patil, IIT Bombay
Superposition: Why does it work?
R1
R3
V1
V2
B
A
R2
Vs
Is
0
»
V1
V2
–
=A
−1
»
G1 Vs
Is
–
»
≡
m11
m21
m12
m22
–»
–»
0
Is
G1 Vs
Is
–
.
We are now in a position to see why superposition works.
»
V1
V2
–
=
»
m11 G1
m21 G1
m12
m22
–»
Vs
0
– »
m11 G1
+
m21 G1
m12
m22
"
–
≡
(1)
V1
(1)
V2
# "
+
(2)
V1
(2)
#
.
V2
The first vector is the response due to Vs alone (and Is deactivated).
The second vector is the response due to Is alone (and Vs deactivated).
M. B. Patil, IIT Bombay
Superposition: Why does it work?
R1
R3
V1
V2
B
A
R2
Vs
Is
0
»
V1
V2
–
=A
−1
»
G1 Vs
Is
–
»
≡
m11
m21
m12
m22
–»
–»
0
Is
G1 Vs
Is
–
.
We are now in a position to see why superposition works.
»
V1
V2
–
=
»
m11 G1
m21 G1
m12
m22
–»
Vs
0
– »
m11 G1
+
m21 G1
m12
m22
"
–
≡
(1)
V1
(1)
V2
# "
+
(2)
V1
(2)
#
.
V2
The first vector is the response due to Vs alone (and Is deactivated).
The second vector is the response due to Is alone (and Vs deactivated).
All other currents and voltages are linearly related to V1 and V2
⇒ Any voltage (node voltage or branch voltage) or current can also be computed using
superposition.
M. B. Patil, IIT Bombay
Thevenin’s theorem
Circuit
A
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
B
Thevenin’s theorem
RTh
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
A
VTh
B
B
M. B. Patil, IIT Bombay
Thevenin’s theorem
RTh
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
A
VTh
B
B
* VTh is simply VAB when nothing is connected on the other side, i.e., VTh = Voc .
M. B. Patil, IIT Bombay
Thevenin’s theorem
RTh
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
A
VTh
B
B
* VTh is simply VAB when nothing is connected on the other side, i.e., VTh = Voc .
* RTh can be found by different methods.
M. B. Patil, IIT Bombay
Thevenin’s theorem: RTh
Method 1:
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
RTh
A
A
VTh
B
* Deactivate all independent sources.
B
Thevenin’s theorem: RTh
Method 1:
Circuit
RTh
A
A
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
B
Circuit
A
A
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
B
B
VTh
B
RTh
* Deactivate all independent sources.
Thevenin’s theorem: RTh
Method 1:
Circuit
RTh
A
A
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
B
Circuit
A
A
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
B
B
VTh
B
RTh
* Deactivate all independent sources.
* RTh can often be found by inspection.
Thevenin’s theorem: RTh
Method 1:
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
RTh
A
A
A Is
VTh
B
Vs
B
B
RTh
Circuit
A
A
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
B
B
* Deactivate all independent sources.
* RTh can often be found by inspection.
* RTh may be found by connecting a test source.
Thevenin’s theorem: RTh
Method 1:
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
RTh
A
A
A Is
VTh
B
Vs
B
B
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
RTh
A
A
A
Vs
B
B
Is
B
* Deactivate all independent sources.
* RTh can often be found by inspection.
* RTh may be found by connecting a test source.
M. B. Patil, IIT Bombay
Thevenin’s theorem: RTh
Method 2:
A
Voc
B
* Find Voc .
Thevenin’s theorem: RTh
Method 2:
A
A
Voc
B
Isc
B
* Find Voc .
* Find Isc .
M. B. Patil, IIT Bombay
Thevenin’s theorem: RTh
Method 2:
A
A
Voc
B
Isc
B
* Find Voc .
* Find Isc .
Voc
.
* RTh =
Isc
M. B. Patil, IIT Bombay
Thevenin’s theorem: RTh
Method 2:
A
A
Voc
B
Isc
B
* Find Voc .
* Find Isc .
Voc
.
* RTh =
Isc
* Note: Sources are not deactivated.
M. B. Patil, IIT Bombay
Thevenin’s theorem: example
6Ω
2Ω
R1
R3
9V
3Ω
R2
A
RL
B
Thevenin’s theorem: example
6Ω
2Ω
R1
R3
9V
3Ω
R2
RTh
A
RL
B
≡
A
VTh
RL
B
Thevenin’s theorem: example
6Ω
2Ω
R1
R3
9V
3Ω
R2
RTh
A
RL
≡
B
VTh :
6Ω
9V
3Ω
2Ω
A
VTh
RL
B
A
Voc
B
Thevenin’s theorem: example
6Ω
2Ω
R1
R3
9V
R2
3Ω
RTh
A
RL
≡
B
VTh :
6Ω
9V
2Ω
A
B
Voc
3Ω
= 9V×
6Ω + 3Ω
= 9V ×
1
= 3V
3
VTh
RL
B
Voc
3Ω
A
Thevenin’s theorem: example
6Ω
2Ω
R1
R3
9V
R2
3Ω
RTh
A
RL
≡
B
VTh :
6Ω
9V
2Ω
A
B
Voc
3Ω
= 9V×
6Ω + 3Ω
= 9V ×
1
= 3V
3
VTh
RL
B
Voc
3Ω
A
RTh :
6Ω
2Ω
A
3Ω
B
Thevenin’s theorem: example
6Ω
2Ω
R1
R3
9V
R2
3Ω
RTh
A
RL
≡
B
VTh :
6Ω
9V
2Ω
A
VTh
RL
B
A
RTh :
6Ω
Voc
3Ω
2Ω
3Ω
B
Voc
3Ω
= 9V×
6Ω + 3Ω
= 9V ×
1
= 3V
3
A
B
RTh = (R1 k R2 ) + R3 = (3 k 6) + 2
=3×
1×2
+2 = 4Ω
1+2
Thevenin’s theorem: example
6Ω
2Ω
R1
R3
9V
R2
3Ω
RTh
A
RL
≡
B
VTh :
6Ω
9V
2Ω
4Ω
A
VTh
RL
3V
A
RTh :
B
6Ω
2Ω
Voc
= 9V ×
1
= 3V
3
A
3Ω
B
3Ω
= 9V×
6Ω + 3Ω
RL
B
Voc
3Ω
≡
A
B
RTh = (R1 k R2 ) + R3 = (3 k 6) + 2
=3×
1×2
+2 = 4Ω
1+2
M. B. Patil, IIT Bombay
Maximum power transfer
Circuit
A
iL
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
RL
B
Maximum power transfer
Circuit
A
iL
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
RL
B
* Power “transferred” to load is,
PL = iL2 RL .
Maximum power transfer
Circuit
A
iL
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
RL
B
* Power “transferred” to load is,
PL = iL2 RL .
* For a given black box, what is the
value of RL for which PL is
maximum?
Maximum power transfer
Circuit
A
iL
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
RL
B
* Power “transferred” to load is,
PL = iL2 RL .
* For a given black box, what is the
value of RL for which PL is
maximum?
* Replace the black box with its
Thevenin equivalent.
Maximum power transfer
A
Circuit
iL
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
RL
B
RTh
A
iL
VTh
RL
B
* Power “transferred” to load is,
PL = iL2 RL .
* For a given black box, what is the
value of RL for which PL is
maximum?
* Replace the black box with its
Thevenin equivalent.
Maximum power transfer
A
Circuit
iL
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
RL
B
RTh
A
iL
VTh
RL
* Power “transferred” to load is,
PL = iL2 RL .
* For a given black box, what is the
value of RL for which PL is
maximum?
* Replace the black box with its
Thevenin equivalent.
* iL =
B
VTh
,
RTh + RL
2 ×
PL = VTh
RL
.
(RTh + RL )2
Maximum power transfer
A
Circuit
iL
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
RL
B
RTh
A
iL
VTh
RL
* Power “transferred” to load is,
PL = iL2 RL .
* For a given black box, what is the
value of RL for which PL is
maximum?
* Replace the black box with its
Thevenin equivalent.
* iL =
B
VTh
,
RTh + RL
2 ×
PL = VTh
* For
RL
.
(RTh + RL )2
dPL
= 0 , we need
dRL
(RTh + RL )2 − RL × 2 (RTh + RL )
= 0,
(RTh + RL )4
i.e., RTh + RL = 2 RL ⇒ RL = RTh .
Maximum power transfer
A
Circuit
iL
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
* Power “transferred” to load is,
PL = iL2 RL .
RL
* For a given black box, what is the
value of RL for which PL is
maximum?
B
RTh
A
* Replace the black box with its
Thevenin equivalent.
iL
VTh
RL
* iL =
B
VTh
,
RTh + RL
2 ×
PL = VTh
PL
Pmax
L
* For
RL
.
(RTh + RL )2
dPL
= 0 , we need
dRL
(RTh + RL )2 − RL × 2 (RTh + RL )
= 0,
(RTh + RL )4
i.e., RTh + RL = 2 RL ⇒ RL = RTh .
RL
RL = RTh
M. B. Patil, IIT Bombay
Maximum power transfer: example
Find RL for which PL is maximum.
3Ω
2Ω
R1
R3
6Ω
12 V
R2
A
RL
2A
B
Maximum power transfer: example
Find RL for which PL is maximum.
3Ω
2Ω
R1
R3
6Ω
12 V
A
RL
R2
2A
B
3Ω
RTh :
2Ω
R1
R3
A
6Ω
R2
B
Maximum power transfer: example
Find RL for which PL is maximum.
3Ω
2Ω
R1
R3
6Ω
12 V
A
RL
R2
2A
B
3Ω
RTh :
2Ω
R1
R3
A
6Ω
R2
B
RTh = (R1 k R2 ) + R3 = (3 k 6) + 2
=3×
1×2
+2 = 4Ω
1+2
Maximum power transfer: example
Find RL for which PL is maximum.
3Ω
2Ω
R1
R3
6Ω
12 V
A
B
3Ω
2Ω
R3
A
6Ω
R2
B
RTh = (R1 k R2 ) + R3 = (3 k 6) + 2
=3×
R1
R3
1×2
+2 = 4Ω
1+2
A
6Ω
12 V
2A
R1
2Ω
RL
R2
RTh :
3Ω
Voc :
R2
2A
B
Maximum power transfer: example
Find RL for which PL is maximum.
3Ω
2Ω
R1
R3
6Ω
12 V
A
3Ω
Voc :
2Ω
R1
R3
6Ω
RL
R2
12 V
2A
R2
2A
B
B
3Ω
RTh :
2Ω
R1
R3
A
6Ω
R2
RTh
B
= (R1 k R2 ) + R3 = (3 k 6) + 2
=3×
1×2
+2 = 4Ω
1+2
A
Use superposition to find Voc :
3Ω
2Ω
R1
R3
3Ω
A
2Ω
R1
R3
6Ω
12 V
A
6Ω
R2
R2
B
2A
B
Maximum power transfer: example
Find RL for which PL is maximum.
3Ω
2Ω
R1
R3
6Ω
12 V
A
3Ω
Voc :
2Ω
R1
R3
6Ω
RL
R2
12 V
2A
R2
2A
B
B
3Ω
RTh :
2Ω
R1
R3
A
6Ω
R2
RTh
B
= (R1 k R2 ) + R3 = (3 k 6) + 2
=3×
1×2
+2 = 4Ω
1+2
A
Use superposition to find Voc :
3Ω
2Ω
R1
R3
3Ω
A
2Ω
R1
R3
6Ω
12 V
6Ω
R2
V(1)
oc = 12 ×
R2
6
=8V
9
A
B
2A
B
Maximum power transfer: example
Find RL for which PL is maximum.
3Ω
2Ω
R1
R3
6Ω
12 V
A
3Ω
Voc :
2Ω
R1
R3
6Ω
RL
R2
12 V
2A
R2
2A
B
B
3Ω
RTh :
2Ω
R1
R3
A
6Ω
R2
RTh
B
= (R1 k R2 ) + R3 = (3 k 6) + 2
=3×
1×2
+2 = 4Ω
1+2
A
Use superposition to find Voc :
3Ω
2Ω
R1
R3
3Ω
A
2Ω
R1
R3
6Ω
12 V
6Ω
R2
V(1)
oc = 12 ×
R2
6
=8V
9
A
B
V(oc2) = 4 Ω × 2 A = 8 V
2A
B
Maximum power transfer: example
Find RL for which PL is maximum.
3Ω
2Ω
R1
R3
6Ω
12 V
A
3Ω
Voc :
2Ω
R1
R3
6Ω
RL
R2
12 V
2A
R2
2A
B
B
3Ω
RTh :
2Ω
R1
R3
A
6Ω
R2
RTh
B
= (R1 k R2 ) + R3 = (3 k 6) + 2
=3×
1×2
+2 = 4Ω
1+2
A
Use superposition to find Voc :
3Ω
2Ω
R1
R3
3Ω
A
2Ω
R1
R3
6Ω
12 V
6Ω
R2
R2
B
V(1)
oc = 12 ×
A
6
=8V
V(oc2) = 4 Ω × 2 A = 8 V
9
(2)
(1)
Voc = Voc + Voc = 8 + 8 = 16 V
2A
B
Maximum power transfer: example
Find RL for which PL is maximum.
3Ω
2Ω
R1
R3
6Ω
12 V
A
3Ω
Voc :
2Ω
R1
R3
6Ω
RL
R2
12 V
2A
R2
2A
B
B
3Ω
RTh :
2Ω
R1
A
R3
B
= (R1 k R2 ) + R3 = (3 k 6) + 2
=3×
1×2
+2 = 4Ω
1+2
RTh
R3
3Ω
A
2Ω
R1
R3
6Ω
R2
RTh
Use superposition to find Voc :
3Ω
2Ω
R1
6Ω
A
A
6Ω
R2
R2
B
V(1)
oc = 12 ×
2A
6
=8V
V(oc2) = 4 Ω × 2 A = 8 V
9
(2)
(1)
Voc = Voc + Voc = 8 + 8 = 16 V
B
PL is maximum when RL = RTh = 4 Ω
iL
VTh
12 V
A
RL
⇒ iL = VTh /(2 RTh ) = 2 A
Pmax
= 22 × 4 = 16 W .
L
B
M. B. Patil, IIT Bombay
Thevenin’s theorem: example
4Ω
A B
4Ω
6A
2Ω
48 V
12 Ω
12 Ω
Thevenin’s theorem: example
4Ω
A B
4Ω
6A
48 V
12 Ω
12 Ω
2Ω
RTh :
4Ω
2Ω
A B
12 Ω
C
4Ω
12 Ω
Thevenin’s theorem: example
4Ω
A B
4Ω
6A
48 V
12 Ω
12 Ω
2Ω
RTh :
4Ω
A B
12 Ω
2Ω
C
A B
≡
3Ω
4Ω
C
4Ω
12 Ω
Thevenin’s theorem: example
4Ω
A B
4Ω
6A
48 V
12 Ω
12 Ω
2Ω
RTh :
4Ω
A B
12 Ω
2Ω
C
4Ω
12 Ω
A B
≡
3Ω
4Ω
C
⇒
RTh = 7 Ω
Thevenin’s theorem: example
4Ω
A B
4Ω
6A
Voc :
4Ω
48 V
A B
12 Ω
2Ω
RTh :
4Ω
A B
12 Ω
2Ω
C
4Ω
12 Ω
A B
≡
3Ω
4Ω
C
Voc
48 V
12 Ω
12 Ω
2Ω
4Ω
⇒
RTh = 7 Ω
6A
C i
12 Ω
Thevenin’s theorem: example
4Ω
A B
4Ω
6A
Voc :
4Ω
48 V
Voc
12 Ω
6A
2Ω
RTh :
4Ω
48 V
12 Ω
12 Ω
2Ω
4Ω
A B
A B
4Ω
C i
12 Ω
Note: i = 0 (since there is no return path).
VAB = VA − VB
= (VA − VC ) + (VC − VB )
12 Ω
2Ω
C
12 Ω
A B
≡
3Ω
4Ω
C
⇒
RTh = 7 Ω
= VAC + VCB
= 24 V + 36 V = 60 V
Thevenin’s theorem: example
4Ω
A B
4Ω
6A
Voc :
4Ω
48 V
Voc
12 Ω
6A
2Ω
RTh :
4Ω
48 V
12 Ω
12 Ω
2Ω
4Ω
A B
A B
4Ω
C i
12 Ω
Note: i = 0 (since there is no return path).
VAB = VA − VB
= (VA − VC ) + (VC − VB )
12 Ω
2Ω
C
12 Ω
= VAC + VCB
= 24 V + 36 V = 60 V
A B
≡
3Ω
4Ω
C
⇒
VTh = 60 V
RTh = 7 Ω
RTh = 7 Ω
Thevenin’s theorem: example
4Ω
A B
4Ω
6A
Voc :
4Ω
48 V
Voc
12 Ω
6A
2Ω
RTh :
4Ω
48 V
12 Ω
12 Ω
2Ω
4Ω
A B
A B
4Ω
C i
12 Ω
Note: i = 0 (since there is no return path).
VAB = VA − VB
= (VA − VC ) + (VC − VB )
12 Ω
2Ω
C
12 Ω
= VAC + VCB
= 24 V + 36 V = 60 V
A B
A B
≡
3Ω
4Ω
⇒
VTh = 60 V
RTh = 7 Ω
RTh = 7 Ω
7Ω
60 V
C
M. B. Patil, IIT Bombay
Graphical method for finding VTh and RTh
SEQUEL file: ee101 thevenin 1.sqproj
4Ω
A B
4Ω
6A
2Ω
48 V
12 Ω
12 Ω
Graphical method for finding VTh and RTh
SEQUEL file: ee101 thevenin 1.sqproj
4Ω
A B
4Ω
6A
48 V
12 Ω
12 Ω
2Ω
Connect a voltage source between A and B.
Plot i versus v.
A
4Ω
6A
2Ω
B
4Ω
i
48 V
v
12 Ω
Voc = intercept on the v-axis.
Isc = intercept on the i-axis.
12 Ω
Graphical method for finding VTh and RTh
SEQUEL file: ee101 thevenin 1.sqproj
4Ω
A B
4Ω
6A
8
i (Amp)
12 Ω
12 Ω
2Ω
10
48 V
6
4
2
0
Connect a voltage source between A and B.
0
Plot i versus v.
A
4Ω
6A
2Ω
B
4Ω
i
48 V
v
12 Ω
Voc = intercept on the v-axis.
Isc = intercept on the i-axis.
12 Ω
20
40
v (Volt)
60
Graphical method for finding VTh and RTh
SEQUEL file: ee101 thevenin 1.sqproj
4Ω
A B
4Ω
6A
8
i (Amp)
12 Ω
12 Ω
2Ω
10
48 V
6
4
2
0
Connect a voltage source between A and B.
0
Plot i versus v.
A
4Ω
6A
2Ω
B
48 V
v
12 Ω
Voc = intercept on the v-axis.
Isc = intercept on the i-axis.
12 Ω
60
Voc = 60 V, Isc = 8.57 A
4Ω
i
20
40
v (Volt)
RTh = Vsc /Isc = 7 Ω
Graphical method for finding VTh and RTh
SEQUEL file: ee101 thevenin 1.sqproj
4Ω
A B
4Ω
6A
8
i (Amp)
12 Ω
12 Ω
2Ω
10
48 V
6
4
2
0
Connect a voltage source between A and B.
0
Plot i versus v.
A
4Ω
6A
2Ω
B
48 V
v
12 Ω
RTh = Vsc /Isc = 7 Ω
A B
12 Ω
VTh = 60 V
Voc = intercept on the v-axis.
60
Voc = 60 V, Isc = 8.57 A
4Ω
i
20
40
v (Volt)
RTh = 7 Ω
7Ω
60 V
Isc = intercept on the i-axis.
M. B. Patil, IIT Bombay
Norton equivalent circuit
RTh
A
VTh
B
Norton equivalent circuit
RTh
A
A
IN
VTh
B
RN
B
Norton equivalent circuit
RTh
A
A
IN
VTh
B
* Consider the open circuit case.
RN
B
Norton equivalent circuit
RTh
A
A
IN
VTh
B
RN
B
* Consider the open circuit case.
Thevenin circuit: VAB = VTh .
Norton equivalent circuit
RTh
A
A
IN
VTh
B
RN
B
* Consider the open circuit case.
Thevenin circuit: VAB = VTh .
Norton circuit: VAB = IN RN .
Norton equivalent circuit
RTh
A
A
IN
VTh
B
RN
B
* Consider the open circuit case.
Thevenin circuit: VAB = VTh .
Norton circuit: VAB = IN RN .
⇒ VTh = IN RN .
Norton equivalent circuit
RTh
RTh
A
A
IN
VTh
B
RN
VTh
B
* Consider the open circuit case.
Thevenin circuit: VAB = VTh .
Norton circuit: VAB = IN RN .
⇒ VTh = IN RN .
* Consider the short circuit case.
A
Isc
B
A
IN
RN
Isc
B
Norton equivalent circuit
RTh
RTh
A
A
IN
VTh
B
RN
A
VTh
B
* Consider the open circuit case.
Thevenin circuit: VAB = VTh .
Norton circuit: VAB = IN RN .
⇒ VTh = IN RN .
* Consider the short circuit case.
Thevenin circuit: Isc = VTh /RTh .
Isc
B
A
IN
RN
Isc
B
Norton equivalent circuit
RTh
RTh
A
A
IN
VTh
B
RN
A
VTh
B
* Consider the open circuit case.
Thevenin circuit: VAB = VTh .
Norton circuit: VAB = IN RN .
⇒ VTh = IN RN .
* Consider the short circuit case.
Thevenin circuit: Isc = VTh /RTh .
Norton circuit: Isc = IN .
Isc
B
A
IN
RN
Isc
B
Norton equivalent circuit
RTh
RTh
A
A
IN
VTh
B
RN
A
VTh
B
Isc
B
A
IN
RN
Isc
B
* Consider the open circuit case.
Thevenin circuit: VAB = VTh .
Norton circuit: VAB = IN RN .
⇒ VTh = IN RN .
* Consider the short circuit case.
Thevenin circuit: Isc = VTh /RTh .
Norton circuit: Isc = IN .
⇒ RTh = RN .
M. B. Patil, IIT Bombay
Example
5Ω
20 V
i
10 Ω
1A
Example
A
5Ω
i
10 Ω
20 V
B
1A
Example
A
5Ω
i
10 Ω
20 V
B
1A
RN = 5 Ω
IN =
20 V
= 4A
5Ω
Example
A
5Ω
i
1A
IN =
10 Ω
20 V
B
A
i
4A
5Ω
10 Ω
B
RN = 5 Ω
1A
20 V
= 4A
5Ω
Example
A
5Ω
i
1A
IN =
10 Ω
20 V
B
A
i
4A
5Ω
10 Ω
B
RN = 5 Ω
1A
20 V
= 4A
5Ω
i
3A
5Ω
10 Ω
Example
A
5Ω
i
1A
IN =
10 Ω
20 V
RN = 5 Ω
i
3A
20 V
= 4A
5Ω
5Ω
10 Ω
B
A
i
4A
5Ω
1A
i = 3A ×
5
5 + 10
= 1A
10 Ω
B
M. B. Patil, IIT Bombay
Example
A
5Ω
i
1A
IN =
10 Ω
20 V
RN = 5 Ω
i
3A
20 V
= 4A
5Ω
5Ω
10 Ω
B
A
i
4A
5Ω
1A
i = 3A ×
5
5 + 10
= 1A
10 Ω
B
Home work:
* Find i by superposition and compare.
M. B. Patil, IIT Bombay
Example
A
5Ω
i
1A
IN =
10 Ω
20 V
RN = 5 Ω
i
3A
20 V
= 4A
5Ω
5Ω
10 Ω
B
A
i
4A
5Ω
1A
i = 3A ×
5
5 + 10
= 1A
10 Ω
B
Home work:
* Find i by superposition and compare.
* Compute the power absorbed by each element, and verify that
P
Pi = 0 .
M. B. Patil, IIT Bombay