Here you will know what is Kirchoff Current Law and kirchoff Voltage Law and where you need to apply in electrical engineering.with examples.

1. EE101: Basics
KCL, KVL, power, Thevenin’s theorem
M. B. Patil
mbpatil@ee.iitb.ac.in
Department of Electrical Engineering
Indian Institute of Technology Bombay
M. B. Patil, IIT Bombay

2. Kirchhoff’s laws
4
vα
v6
v3
v2
i5
V0
I0v5
i4R3i6
i3
v4
i2
R2
R1
v1
i1
A B C
DE
M. B. Patil, IIT Bombay

3. Kirchhoff’s laws
4
vα
v6
v3
v2
i5
V0
I0v5
i4R3i6
i3
v4
i2
R2
R1
v1
i1
A B C
DE
* Kirchhoff’s current law (KCL):P
ik = 0 at each node.
M. B. Patil, IIT Bombay

4. Kirchhoff’s laws
4
vα
v6
v3
v2
i5
V0
I0v5
i4R3i6
i3
v4
i2
R2
R1
v1
i1
A B C
DE
* 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

5. Kirchhoff’s laws
4
vα
v6
v3
v2
i5
V0
I0v5
i4R3i6
i3
v4
i2
R2
R1
v1
i1
A B C
DE
* 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

6. Kirchhoff’s laws
4
vα
v6
v3
v2
i5
V0
I0v5
i4R3i6
i3
v4
i2
R2
R1
v1
i1
A B C
DE
* 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

7. Kirchhoff’s laws
4
vα
v6
v3
v2
i5
V0
I0v5
i4R3i6
i3
v4
i2
R2
R1
v1
i1
A B C
DE
* 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

8. Kirchhoff’s laws
4
vα
v6
v3
v2
i5
V0
I0v5
i4R3i6
i3
v4
i2
R2
R1
v1
i1
A B C
DE
* 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

9. Circuit elements
Element Symbol Equation
Resistor
v
i
v = R i
Inductor
v
i
v = L
di
dt
Capacitor
v
i
i = C
dv
dt
Diode
v
i
to be discussed
BJT
C
E
B to be discussed
M. B. Patil, IIT Bombay

10. Sources
Element Symbol Equation
Independent Voltage source
v
i
v(t) = vs (t)
Current source
v
i
i(t) = is (t)
Dependent VCVS
v
i
v(t) = α vc (t)
VCCS
v
i
i(t) = g vc (t)
CCVS
v
i
v(t) = r ic (t)
CCCS
v
i
i(t) = β ic (t)
* α, β: dimensionless, r: Ω, g: Ω−1
or (“mho”)
* The subscript ‘c’ denotes the controlling voltage or current.
M. B. Patil, IIT Bombay

11. Instantaneous power absorbed by an element
i2
iN
i3
i1
V1
VN
V2
V3
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

12. Instantaneous power absorbed by an element
i2
iN
i3
i1
V1
VN
V2
V3
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).
* two-terminal element:
V1 i1
V2
v
i2
P = V1 i1 + V2 i2
= V1 i1 + V2 (−i1)
= [V1 − V2] i1 = v i1
M. B. Patil, IIT Bombay

13. Instantaneous power absorbed by an element
i2
iN
i3
i1
V1
VN
V2
V3
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).
* two-terminal element:
V1 i1
V2
v
i2
P = V1 i1 + V2 i2
= V1 i1 + V2 (−i1)
= [V1 − V2] i1 = v i1
* three-terminal element:
iB
iE
iC
VC
VB
VE
P = VB iB + VC iC + VE (−iE )
= VB iB + VC iC − VE (iB + iC )
= (VB − VE ) iB + (VC − VE ) iC
= VBE iB + VCE iE
M. B. Patil, IIT Bombay

14. 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

15. 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

16. 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

17. 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

18. Resistors in series
A B A B
R2R1
i R
v3
R3
v2
i
v1 v
M. B. Patil, IIT Bombay

19. Resistors in series
A B A B
R2R1
i R
v3
R3
v2
i
v1 v
v1 = i R1, v2 = i R2, v3 = i R3, ⇒ v = v1 + v2 + v3 = i (R1 + R2 + R3)
M. B. Patil, IIT Bombay

20. Resistors in series
A B A B
R2R1
i R
v3
R3
v2
i
v1 v
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

21. Resistors in series
A B A B
R2R1
i R
v3
R3
v2
i
v1 v
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

22. Resistors in parallel
A AB B
R2
R1
i R
R3
v
i1
i2
i3
i
v
M. B. Patil, IIT Bombay

23. Resistors in parallel
A AB B
R2
R1
i R
R3
v
i1
i2
i3
i
v
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

24. Resistors in parallel
A AB B
R2
R1
i R
R3
v
i1
i2
i3
i
v
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

25. Resistors in parallel
A AB B
R2
R1
i R
R3
v
i1
i2
i3
i
v
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

26. Resistors in parallel
A AB B
R2
R1
i R
R3
v
i1
i2
i3
i
v
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
Req =
R1 R2
R1 + R2
, i1 = i ×
R2
R1 + R2
, i2 = i ×
R1
R1 + R2
.
M. B. Patil, IIT Bombay

27. Resistors in parallel
A AB B
R2
R1
i R
R3
v
i1
i2
i3
i
v
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
Req =
R1 R2
R1 + R2
, i1 = i ×
R2
R1 + R2
, i2 = i ×
R1
R1 + R2
.
* If Rk = 0, all of the current will go through Rk .
M. B. Patil, IIT Bombay

28. Example
(a) 3 Ω
i1
4 Ω
3 Ω
2 Ω
2.52.55
6 V
i2

29. Example
(a) 3 Ω
i1
4 Ω
3 Ω
2 Ω
2.52.55
6 V
i2

30. Example
(a) 3 Ω
i1
4 Ω
3 Ω
2 Ω
2.52.55
6 V
i2
(b)
2 Ω4 Ω i2
i1
3 Ω
3 Ω
1 Ω
6 V

31. Example
(a) 3 Ω
i1
4 Ω
3 Ω
2 Ω
2.52.55
6 V
i2
(b)
2 Ω4 Ω i2
i1
3 Ω
3 Ω
1 Ω
6 V

32. Example
(a) 3 Ω
i1
4 Ω
3 Ω
2 Ω
2.52.55
6 V
i2
(b)
2 Ω4 Ω i2
i1
3 Ω
3 Ω
1 Ω
6 V
(c)
4 Ω
i1
6 V
i2
3
6

33. Example
(a) 3 Ω
i1
4 Ω
3 Ω
2 Ω
2.52.55
6 V
i2
(b)
2 Ω4 Ω i2
i1
3 Ω
3 Ω
1 Ω
6 V
(c)
4 Ω
i1
6 V
i2
3
6

34. Example
(a) 3 Ω
i1
4 Ω
3 Ω
2 Ω
2.52.55
6 V
i2
(b)
2 Ω4 Ω i2
i1
3 Ω
3 Ω
1 Ω
6 V
(c)
4 Ω
i1
6 V
i2
3
6
(d)
i1
6 V
2 Ω
4 Ω

35. Example
(a) 3 Ω
i1
4 Ω
3 Ω
2 Ω
2.52.55
6 V
i2
(b)
2 Ω4 Ω i2
i1
3 Ω
3 Ω
1 Ω
6 V
(c)
4 Ω
i1
6 V
i2
3
6
(d)
i1
6 V
2 Ω
4 Ω
i1 =
6 V
4 Ω + 2 Ω
= 1 A .

36. Example
(a) 3 Ω
i1
4 Ω
3 Ω
2 Ω
2.52.55
6 V
i2
(b)
2 Ω4 Ω i2
i1
3 Ω
3 Ω
1 Ω
6 V
(c)
4 Ω
i1
6 V
i2
3
6
(d)
i1
6 V
2 Ω
4 Ω
i1 =
6 V
4 Ω + 2 Ω
= 1 A .
i2 = i1 ×
6 Ω
6 Ω + 3 Ω
=
2
3
A .
M. B. Patil, IIT Bombay

37. Example
(a) 3 Ω
i1
4 Ω
3 Ω
2 Ω
2.52.55
6 V
i2
(b)
2 Ω4 Ω i2
i1
3 Ω
3 Ω
1 Ω
6 V
(c)
4 Ω
i1
6 V
i2
3
6
(d)
i1
6 V
2 Ω
4 Ω
i1 =
6 V
4 Ω + 2 Ω
= 1 A .
i2 = i1 ×
6 Ω
6 Ω + 3 Ω
=
2
3
A .
Home work:
* Verify that KCL and KVL are satisfied for each node/loop.
M. B. Patil, IIT Bombay

38. Example
(a) 3 Ω
i1
4 Ω
3 Ω
2 Ω
2.52.55
6 V
i2
(b)
2 Ω4 Ω i2
i1
3 Ω
3 Ω
1 Ω
6 V
(c)
4 Ω
i1
6 V
i2
3
6
(d)
i1
6 V
2 Ω
4 Ω
i1 =
6 V
4 Ω + 2 Ω
= 1 A .
i2 = i1 ×
6 Ω
6 Ω + 3 Ω
=
2
3
A .
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

39. Nodal analysis
I
k v3
R2
V2
v3
0
R1
3R
4R
0
V
1
V3
M. B. Patil, IIT Bombay

40. Nodal analysis
I
k v3
R2
V2
v3
0
R1
3R
4R
0
V
1
V3
* Take some node as the “reference node” and denote
the node voltages of the remaining nodes by V1, V2,
etc.
M. B. Patil, IIT Bombay

41. Nodal analysis
I
k v3
R2
V2
v3
0
R1
3R
4R
0
V
1
V3
* Take some node as the “reference node” and denote
the node voltages of the remaining nodes by V1, V2,
etc.
* Write KCL at each node in terms of the node
voltages. Follow a fixed convention, e.g., current
leaving a node is positive.
M. B. Patil, IIT Bombay

42. Nodal analysis
I
k v3
R2
V2
v3
0
R1
3R
4R
0
V
1
V3
* Take some node as the “reference node” and denote
the node voltages of the remaining nodes by V1, V2,
etc.
* Write KCL at each node in terms of the node
voltages. Follow a fixed convention, e.g., current
leaving a node is positive.
1
R1
(V1 − V2) − I0 − k (V2 − V3) = 0 ,
1
R1
(V2 − V1) +
1
R3
(V2 − V3) +
1
R2
(V2) = 0 ,
k (V2 − V3) +
1
R3
(V3 − V2) +
1
R4
(V3) = 0 .
M. B. Patil, IIT Bombay

43. Nodal analysis
I
k v3
R2
V2
v3
0
R1
3R
4R
0
V
1
V3
* Take some node as the “reference node” and denote
the node voltages of the remaining nodes by V1, V2,
etc.
* Write KCL at each node in terms of the node
voltages. Follow a fixed convention, e.g., current
leaving a node is positive.
1
R1
(V1 − V2) − I0 − k (V2 − V3) = 0 ,
1
R1
(V2 − V1) +
1
R3
(V2 − V3) +
1
R2
(V2) = 0 ,
k (V2 − V3) +
1
R3
(V3 − V2) +
1
R4
(V3) = 0 .
* Solve for the node voltages → branch voltages and
currents.
M. B. Patil, IIT Bombay

44. Nodal analysis
I
k v3
R2
V2
v3
0
R1
3R
4R
0
V
1
V3
* Take some node as the “reference node” and denote
the node voltages of the remaining nodes by V1, V2,
etc.
* Write KCL at each node in terms of the node
voltages. Follow a fixed convention, e.g., current
leaving a node is positive.
1
R1
(V1 − V2) − I0 − k (V2 − V3) = 0 ,
1
R1
(V2 − V1) +
1
R3
(V2 − V3) +
1
R2
(V2) = 0 ,
k (V2 − V3) +
1
R3
(V3 − V2) +
1
R4
(V3) = 0 .
* 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

45. Mesh analysis
R1 R2
R3Vs
is
i1 i2
r1 is
M. B. Patil, IIT Bombay

46. Mesh analysis
R1 R2
R3Vs
is
i1 i2
r1 is
* 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

47. Mesh analysis
R1 R2
R3Vs
is
i1 i2
r1 is
* 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

48. Mesh analysis
R1 R2
R3Vs
is
i1 i2
r1 is
* 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

49. 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

50. 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 i2
c + b, the resulting system will be no longer linear.
M. B. Patil, IIT Bombay

51. 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 i2
c + 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

52. 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 i2
c + 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

53. 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 i2
c + 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

54. Superposition
* Superposition refers to superposition of response due to independent sources.
M. B. Patil, IIT Bombay

55. 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

56. 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

57. 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

58. Example
2 Ω
4 Ω
i1
18 V
3 A

59. Example
2 Ω
4 Ω
i1
18 V
3 A
2 Ω
4 Ω
i1
18 V
Case 1: Keep Vs, deactivate Is.

60. Example
2 Ω
4 Ω
i1
18 V
3 A
2 Ω
4 Ω
i1
18 V
Case 1: Keep Vs, deactivate Is.
i
(1)
1 = 3 A

61. Example
2 Ω
4 Ω
i1
18 V
3 A
2 Ω
4 Ω
i1
18 V
Case 1: Keep Vs, deactivate Is.
i
(1)
1 = 3 A
2 Ω
4 Ω
i1
3 A
Case 2: Keep Is, deactivate Vs.

62. Example
2 Ω
4 Ω
i1
18 V
3 A
2 Ω
4 Ω
i1
18 V
Case 1: Keep Vs, deactivate Is.
i
(1)
1 = 3 A
2 Ω
4 Ω
i1
3 A
Case 2: Keep Is, deactivate Vs.
i
(2)
1 = 3 A ×
2 Ω
2 Ω + 4 Ω
= 1 A

63. Example
2 Ω
4 Ω
i1
18 V
3 A
2 Ω
4 Ω
i1
18 V
Case 1: Keep Vs, deactivate Is.
i
(1)
1 = 3 A
2 Ω
4 Ω
i1
3 A
Case 2: Keep Is, deactivate Vs.
i
(2)
1 = 3 A ×
2 Ω
2 Ω + 4 Ω
= 1 A
inet
1 = i
(1)
1 + i
(2)
1 = 3 + 1 = 4 A
M. B. Patil, IIT Bombay

64. Example
v
2 i
i
1 Ω
3 Ω
6 A
12 V

65. Example
v
2 i
i
1 Ω
3 Ω
6 A
12 V
v
2 i
i
1 Ω
3 Ω12 V
Case 1: Keep Vs, deactivate Is.

66. Example
v
2 i
i
1 Ω
3 Ω
6 A
12 V
v
2 i
i
1 Ω
3 Ω12 V
Case 1: Keep Vs, deactivate Is.
⇒ i = 2 A , v(1)
= 6 V .
KVL: − 12 + 3 i + 2 i + i = 0

67. Example
v
2 i
i
1 Ω
3 Ω
6 A
12 V
v
2 i
i
1 Ω
3 Ω12 V
Case 1: Keep Vs, deactivate Is.
⇒ i = 2 A , v(1)
= 6 V .
KVL: − 12 + 3 i + 2 i + i = 0
v
2 i
i
1 Ω
3 Ω
6 A
Case 2: Keep Is, deactivate Vs.

68. Example
v
2 i
i
1 Ω
3 Ω
6 A
12 V
v
2 i
i
1 Ω
3 Ω12 V
Case 1: Keep Vs, deactivate Is.
⇒ i = 2 A , v(1)
= 6 V .
KVL: − 12 + 3 i + 2 i + i = 0
v
2 i
i
1 Ω
3 Ω
6 A
Case 2: Keep Is, deactivate Vs.
KVL: i + (6 + i) 3 + 2 i = 0
⇒ i = −3 A , v(2)
= (−3 + 6) × 3 = 9 V .

69. Example
v
2 i
i
1 Ω
3 Ω
6 A
12 V
v
2 i
i
1 Ω
3 Ω12 V
Case 1: Keep Vs, deactivate Is.
⇒ i = 2 A , v(1)
= 6 V .
KVL: − 12 + 3 i + 2 i + i = 0
v
2 i
i
1 Ω
3 Ω
6 A
Case 2: Keep Is, deactivate Vs.
KVL: i + (6 + i) 3 + 2 i = 0
⇒ i = −3 A , v(2)
= (−3 + 6) × 3 = 9 V .
vnet = v(1) + v(2) = 6 + 9 = 15 V
M. B. Patil, IIT Bombay

70. Superposition: Why does it work?
R1 R3
Vs Is
V1 V2
R2
A B
0
M. B. Patil, IIT Bombay

71. Superposition: Why does it work?
R1 R3
Vs Is
V1 V2
R2
A B
0
KCL at nodes A and B:
1
R1
(V1 − Vs ) +
1
R2
V1 +
1
R3
(V1 − V2) = 0 ,
−Is +
1
R3
(V2 − V1) = 0 .
M. B. Patil, IIT Bombay

72. Superposition: Why does it work?
R1 R3
Vs Is
V1 V2
R2
A B
0
KCL at nodes A and B:
1
R1
(V1 − Vs ) +
1
R2
V1 +
1
R3
(V1 − V2) = 0 ,
−Is +
1
R3
(V2 − V1) = 0 .
Writing in a matrix form, we get (using G1 = 1/R1, etc.),
»
G1 + G2 + G3 −G3
−G3 G3
– »
V1
V2
–
=
»
G1Vs
Is
–
M. B. Patil, IIT Bombay

73. Superposition: Why does it work?
R1 R3
Vs Is
V1 V2
R2
A B
0
KCL at nodes A and B:
1
R1
(V1 − Vs ) +
1
R2
V1 +
1
R3
(V1 − V2) = 0 ,
−Is +
1
R3
(V2 − V1) = 0 .
Writing in a matrix form, we get (using G1 = 1/R1, etc.),
»
G1 + G2 + G3 −G3
−G3 G3
– »
V1
V2
–
=
»
G1Vs
Is
–
i.e., A
»
V1
V2
–
=
»
G1Vs
Is
–
→
»
V1
V2
–
= A
−1
»
G1Vs
Is
–
.
M. B. Patil, IIT Bombay

74. Superposition: Why does it work?
R1 R3
Vs Is
V1 V2
R2
A B
0
»
V1
V2
–
= A
−1
»
G1Vs
Is
–
≡
»
m11 m12
m21 m22
– »
G1Vs
Is
–
.
M. B. Patil, IIT Bombay

75. Superposition: Why does it work?
R1 R3
Vs Is
V1 V2
R2
A B
0
»
V1
V2
–
= A
−1
»
G1Vs
Is
–
≡
»
m11 m12
m21 m22
– »
G1Vs
Is
–
.
We are now in a position to see why superposition works.
»
V1
V2
–
=
»
m11G1 m12
m21G1 m22
– »
Vs
0
–
+
»
m11G1 m12
m21G1 m22
– »
0
Is
–
≡
"
V
(1)
1
V
(1)
2
#
+
"
V
(2)
1
V
(2)
2
#
.
M. B. Patil, IIT Bombay

76. Superposition: Why does it work?
R1 R3
Vs Is
V1 V2
R2
A B
0
»
V1
V2
–
= A
−1
»
G1Vs
Is
–
≡
»
m11 m12
m21 m22
– »
G1Vs
Is
–
.
We are now in a position to see why superposition works.
»
V1
V2
–
=
»
m11G1 m12
m21G1 m22
– »
Vs
0
–
+
»
m11G1 m12
m21G1 m22
– »
0
Is
–
≡
"
V
(1)
1
V
(1)
2
#
+
"
V
(2)
1
V
(2)
2
#
.
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

77. Superposition: Why does it work?
R1 R3
Vs Is
V1 V2
R2
A B
0
»
V1
V2
–
= A
−1
»
G1Vs
Is
–
≡
»
m11 m12
m21 m22
– »
G1Vs
Is
–
.
We are now in a position to see why superposition works.
»
V1
V2
–
=
»
m11G1 m12
m21G1 m22
– »
Vs
0
–
+
»
m11G1 m12
m21G1 m22
– »
0
Is
–
≡
"
V
(1)
1
V
(1)
2
#
+
"
V
(2)
1
V
(2)
2
#
.
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

78. Thevenin’s theorem
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B

79. Thevenin’s theorem
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
VTh
M. B. Patil, IIT Bombay

80. Thevenin’s theorem
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
VTh
* VTh is simply VAB when nothing is connected on the other side, i.e., VTh = Voc .
M. B. Patil, IIT Bombay

81. Thevenin’s theorem
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
VTh
* 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

82. Thevenin’s theorem: RTh
Method 1:
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
VTh
* Deactivate all independent sources.

83. Thevenin’s theorem: RTh
Method 1:
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
VTh
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
* Deactivate all independent sources.

84. Thevenin’s theorem: RTh
Method 1:
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
VTh
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
* Deactivate all independent sources.
* RTh can often be found by inspection.

85. Thevenin’s theorem: RTh
Method 1:
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
VTh
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
A
B
Is
Vs
* Deactivate all independent sources.
* RTh can often be found by inspection.
* RTh may be found by connecting a test source.

86. Thevenin’s theorem: RTh
Method 1:
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
VTh
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
A
B
Is
Vs
A
B
Is
Vs
* Deactivate all independent sources.
* RTh can often be found by inspection.
* RTh may be found by connecting a test source.

87. Thevenin’s theorem: RTh
Method 1:
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
VTh
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
A
B
RTh
A
B
Is
Vs
A
B
Is
Vs
A
B
Is
Vs
* 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

88. Thevenin’s theorem: RTh
Method 2:
A
B
Voc
* Find Voc .

89. Thevenin’s theorem: RTh
Method 2:
A
B
Voc
A
B
Isc
* Find Voc .
* Find Isc .
M. B. Patil, IIT Bombay

90. Thevenin’s theorem: RTh
Method 2:
A
B
Voc
A
B
Isc
* Find Voc .
* Find Isc .
* RTh =
Voc
Isc
.
M. B. Patil, IIT Bombay

91. Thevenin’s theorem: RTh
Method 2:
A
B
Voc
A
B
Isc
* Find Voc .
* Find Isc .
* RTh =
Voc
Isc
.
* Note: Sources are not deactivated.
M. B. Patil, IIT Bombay

92. Thevenin’s theorem: example
A
B
R3
RL
R1
R2
3 Ω
2 Ω
9 V
6 Ω

93. Thevenin’s theorem: example
A
B
R3
RL
R1
R2
3 Ω
2 Ω
9 V
6 Ω A
B
RL≡
RTh
VTh

94. Thevenin’s theorem: example
A
B
R3
RL
R1
R2
3 Ω
2 Ω
9 V
6 Ω A
B
RL≡
RTh
VTh
A
B
3 Ω
2 Ω
9 V
6 Ω
Voc
VTh :

95. Thevenin’s theorem: example
A
B
R3
RL
R1
R2
3 Ω
2 Ω
9 V
6 Ω A
B
RL≡
RTh
VTh
A
B
3 Ω
2 Ω
9 V
6 Ω
Voc
VTh :
Voc = 9 V ×
3 Ω
6 Ω + 3 Ω
= 9 V ×
1
3
= 3 V

96. Thevenin’s theorem: example
A
B
R3
RL
R1
R2
3 Ω
2 Ω
9 V
6 Ω A
B
RL≡
RTh
VTh
A
B
3 Ω
2 Ω
9 V
6 Ω
Voc
VTh :
Voc = 9 V ×
3 Ω
6 Ω + 3 Ω
= 9 V ×
1
3
= 3 V
A
B
3 Ω
2 Ω6 ΩRTh :

97. Thevenin’s theorem: example
A
B
R3
RL
R1
R2
3 Ω
2 Ω
9 V
6 Ω A
B
RL≡
RTh
VTh
A
B
3 Ω
2 Ω
9 V
6 Ω
Voc
VTh :
Voc = 9 V ×
3 Ω
6 Ω + 3 Ω
= 9 V ×
1
3
= 3 V
A
B
3 Ω
2 Ω6 ΩRTh :
RTh = (R1 R2) + R3 = (3 6) + 2
= 3 ×
1 × 2
1 + 2
+ 2 = 4 Ω

98. Thevenin’s theorem: example
A
B
R3
RL
R1
R2
3 Ω
2 Ω
9 V
6 Ω A
B
RL≡
RTh
VTh
A
B
3 Ω
2 Ω
9 V
6 Ω
Voc
VTh :
Voc = 9 V ×
3 Ω
6 Ω + 3 Ω
= 9 V ×
1
3
= 3 V
A
B
3 Ω
2 Ω6 ΩRTh :
RTh = (R1 R2) + R3 = (3 6) + 2
= 3 ×
1 × 2
1 + 2
+ 2 = 4 Ω
A
B
RL 3 V≡ RL
4 Ω
M. B. Patil, IIT Bombay

99. Maximum power transfer
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
RL
iL

100. Maximum power transfer
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
RL
iL * Power “transferred” to load is,
PL = i2
L RL .

101. Maximum power transfer
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
RL
iL * Power “transferred” to load is,
PL = i2
L RL .
* For a given black box, what is the
value of RL for which PL is
maximum?

102. Maximum power transfer
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
RL
iL * Power “transferred” to load is,
PL = i2
L 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.

103. Maximum power transfer
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
RL
iL
A
B
iL
RL
RTh
VTh
* Power “transferred” to load is,
PL = i2
L 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.

104. Maximum power transfer
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
RL
iL
A
B
iL
RL
RTh
VTh
* Power “transferred” to load is,
PL = i2
L 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 =
VTh
RTh + RL
,
PL = V 2
Th ×
RL
(RTh + RL)2
.

105. Maximum power transfer
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
RL
iL
A
B
iL
RL
RTh
VTh
* Power “transferred” to load is,
PL = i2
L 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 =
VTh
RTh + RL
,
PL = V 2
Th ×
RL
(RTh + RL)2
.
* For
dPL
dRL
= 0 , we need
(RTh + RL)2 − RL × 2 (RTh + RL)
(RTh + RL)4
= 0 ,
i.e., RTh + RL = 2 RL ⇒ RL = RTh .

106. Maximum power transfer
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
A
B
RL
iL
A
B
iL
RL
RTh
VTh
PL
Pmax
L
RL
RL = RTh
* Power “transferred” to load is,
PL = i2
L 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 =
VTh
RTh + RL
,
PL = V 2
Th ×
RL
(RTh + RL)2
.
* For
dPL
dRL
= 0 , we need
(RTh + RL)2 − RL × 2 (RTh + RL)
(RTh + RL)4
= 0 ,
i.e., RTh + RL = 2 RL ⇒ RL = RTh .
M. B. Patil, IIT Bombay

107. Maximum power transfer: example
A
B
Find RL for which PL is maximum.
R3
RL
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω

108. Maximum power transfer: example
A
B
Find RL for which PL is maximum.
R3
RL
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
RTh:
R3
R2
R1
2 Ω3 Ω
6 Ω

109. Maximum power transfer: example
A
B
Find RL for which PL is maximum.
R3
RL
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
RTh:
R3
R2
R1
2 Ω3 Ω
6 Ω
RTh = (R1 R2) + R3 = (3 6) + 2
= 3 ×
1 × 2
1 + 2
+ 2 = 4 Ω

110. Maximum power transfer: example
A
B
Find RL for which PL is maximum.
R3
RL
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
RTh:
R3
R2
R1
2 Ω3 Ω
6 Ω
RTh = (R1 R2) + R3 = (3 6) + 2
= 3 ×
1 × 2
1 + 2
+ 2 = 4 Ω
A
B
Voc:
R3
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω

111. Maximum power transfer: example
A
B
Find RL for which PL is maximum.
R3
RL
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
RTh:
R3
R2
R1
2 Ω3 Ω
6 Ω
RTh = (R1 R2) + R3 = (3 6) + 2
= 3 ×
1 × 2
1 + 2
+ 2 = 4 Ω
A
B
Voc:
R3
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
A
B
R3R3
R2 R2
R1 R1
2 Ω 2 Ω3 Ω
12 V
6 Ω
3 Ω
2 A
6 Ω
Use superposition to find Voc:

112. Maximum power transfer: example
A
B
Find RL for which PL is maximum.
R3
RL
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
RTh:
R3
R2
R1
2 Ω3 Ω
6 Ω
RTh = (R1 R2) + R3 = (3 6) + 2
= 3 ×
1 × 2
1 + 2
+ 2 = 4 Ω
A
B
Voc:
R3
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
A
B
R3R3
R2 R2
R1 R1
2 Ω 2 Ω3 Ω
12 V
6 Ω
3 Ω
2 A
6 Ω
Use superposition to find Voc:
V
(1)
oc = 12 ×
6
9
= 8 V

113. Maximum power transfer: example
A
B
Find RL for which PL is maximum.
R3
RL
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
RTh:
R3
R2
R1
2 Ω3 Ω
6 Ω
RTh = (R1 R2) + R3 = (3 6) + 2
= 3 ×
1 × 2
1 + 2
+ 2 = 4 Ω
A
B
Voc:
R3
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
A
B
R3R3
R2 R2
R1 R1
2 Ω 2 Ω3 Ω
12 V
6 Ω
3 Ω
2 A
6 Ω
Use superposition to find Voc:
V
(1)
oc = 12 ×
6
9
= 8 V V
(2)
oc = 4 Ω × 2 A = 8 V

114. Maximum power transfer: example
A
B
Find RL for which PL is maximum.
R3
RL
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
RTh:
R3
R2
R1
2 Ω3 Ω
6 Ω
RTh = (R1 R2) + R3 = (3 6) + 2
= 3 ×
1 × 2
1 + 2
+ 2 = 4 Ω
A
B
Voc:
R3
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
A
B
R3R3
R2 R2
R1 R1
2 Ω 2 Ω3 Ω
12 V
6 Ω
3 Ω
2 A
6 Ω
Use superposition to find Voc:
V
(1)
oc = 12 ×
6
9
= 8 V V
(2)
oc = 4 Ω × 2 A = 8 V
Voc = V
(1)
oc + V
(2)
oc = 8 + 8 = 16 V

115. Maximum power transfer: example
A
B
Find RL for which PL is maximum.
R3
RL
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
RTh:
R3
R2
R1
2 Ω3 Ω
6 Ω
RTh = (R1 R2) + R3 = (3 6) + 2
= 3 ×
1 × 2
1 + 2
+ 2 = 4 Ω
A
B
Voc:
R3
R2
R1
2 Ω3 Ω
12 V 2 A
6 Ω
A
B
A
B
R3R3
R2 R2
R1 R1
2 Ω 2 Ω3 Ω
12 V
6 Ω
3 Ω
2 A
6 Ω
Use superposition to find Voc:
V
(1)
oc = 12 ×
6
9
= 8 V V
(2)
oc = 4 Ω × 2 A = 8 V
Voc = V
(1)
oc + V
(2)
oc = 8 + 8 = 16 V
A
B
iL
RL
Pmax
L = 22 × 4 = 16 W .
PL is maximum when RL = RTh = 4 Ω
⇒ iL = VTh/(2 RTh) = 2 A
RTh
VTh
M. B. Patil, IIT Bombay

116. Thevenin’s theorem: example
A B
2 Ω
6 A
12 Ω
48 V
4 Ω
12 Ω
4 Ω

117. Thevenin’s theorem: example
A B
2 Ω
6 A
12 Ω
48 V
4 Ω
12 Ω
4 Ω
C
A B
RTh:
2 Ω 12 Ω
4 Ω 4 Ω
12 Ω

118. Thevenin’s theorem: example
A B
2 Ω
6 A
12 Ω
48 V
4 Ω
12 Ω
4 Ω
C
A B
RTh:
2 Ω 12 Ω
4 Ω 4 Ω
12 Ω
C
A B
≡
4 Ω
3 Ω

119. Thevenin’s theorem: example
A B
2 Ω
6 A
12 Ω
48 V
4 Ω
12 Ω
4 Ω
C
A B
RTh:
2 Ω 12 Ω
4 Ω 4 Ω
12 Ω
C
A B
≡
4 Ω
3 Ω
RTh = 7 Ω⇒

120. Thevenin’s theorem: example
A B
2 Ω
6 A
12 Ω
48 V
4 Ω
12 Ω
4 Ω
C
A B
RTh:
2 Ω 12 Ω
4 Ω 4 Ω
12 Ω
C
A B
≡
4 Ω
3 Ω
RTh = 7 Ω⇒
A B
C
Voc
Voc:
2 Ω 12 Ω
48 V
4 Ω 4 Ω
12 Ω
6 A
i

121. Thevenin’s theorem: example
A B
2 Ω
6 A
12 Ω
48 V
4 Ω
12 Ω
4 Ω
C
A B
RTh:
2 Ω 12 Ω
4 Ω 4 Ω
12 Ω
C
A B
≡
4 Ω
3 Ω
RTh = 7 Ω⇒
A B
C
Voc
Voc:
2 Ω 12 Ω
48 V
4 Ω 4 Ω
12 Ω
6 A
i
VAB = VA − VB
= 24 V + 36 V = 60 V
= VAC + VCB
= (VA − VC) + (VC − VB)
Note: i = 0 (since there is no return path).

122. Thevenin’s theorem: example
A B
2 Ω
6 A
12 Ω
48 V
4 Ω
12 Ω
4 Ω
C
A B
RTh:
2 Ω 12 Ω
4 Ω 4 Ω
12 Ω
C
A B
≡
4 Ω
3 Ω
RTh = 7 Ω⇒
A B
C
Voc
Voc:
2 Ω 12 Ω
48 V
4 Ω 4 Ω
12 Ω
6 A
i
VAB = VA − VB
= 24 V + 36 V = 60 V
= VAC + VCB
= (VA − VC) + (VC − VB)
Note: i = 0 (since there is no return path).
VTh = 60 V
RTh = 7 Ω

123. Thevenin’s theorem: example
A B
2 Ω
6 A
12 Ω
48 V
4 Ω
12 Ω
4 Ω
C
A B
RTh:
2 Ω 12 Ω
4 Ω 4 Ω
12 Ω
C
A B
≡
4 Ω
3 Ω
RTh = 7 Ω⇒
A B
C
Voc
Voc:
2 Ω 12 Ω
48 V
4 Ω 4 Ω
12 Ω
6 A
i
VAB = VA − VB
= 24 V + 36 V = 60 V
= VAC + VCB
= (VA − VC) + (VC − VB)
Note: i = 0 (since there is no return path).
VTh = 60 V
RTh = 7 Ω
A B
7 Ω
60 V
M. B. Patil, IIT Bombay

124. Graphical method for finding VTh and RTh
SEQUEL file: ee101 thevenin 1.sqproj
A B
2 Ω
6 A
12 Ω
48 V
4 Ω 4 Ω
12 Ω

125. Graphical method for finding VTh and RTh
SEQUEL file: ee101 thevenin 1.sqproj
A B
2 Ω
6 A
12 Ω
48 V
4 Ω 4 Ω
12 Ω
i
v
A B
2 Ω
4 Ω
48 V6 A
12 Ω
4 Ω
12 Ω
Connect a voltage source between A and B.
Plot i versus v.
Voc = intercept on the v-axis.
Isc = intercept on the i-axis.

126. Graphical method for finding VTh and RTh
SEQUEL file: ee101 thevenin 1.sqproj
A B
2 Ω
6 A
12 Ω
48 V
4 Ω 4 Ω
12 Ω
i
v
A B
2 Ω
4 Ω
48 V6 A
12 Ω
4 Ω
12 Ω
Connect a voltage source between A and B.
Plot i versus v.
Voc = intercept on the v-axis.
Isc = intercept on the i-axis.
0
2
4
6
8
10
v (Volt)
0 20 40 60
i(Amp)

127. Graphical method for finding VTh and RTh
SEQUEL file: ee101 thevenin 1.sqproj
A B
2 Ω
6 A
12 Ω
48 V
4 Ω 4 Ω
12 Ω
i
v
A B
2 Ω
4 Ω
48 V6 A
12 Ω
4 Ω
12 Ω
Connect a voltage source between A and B.
Plot i versus v.
Voc = intercept on the v-axis.
Isc = intercept on the i-axis.
0
2
4
6
8
10
v (Volt)
0 20 40 60
i(Amp)
Voc = 60 V, Isc = 8.5714 A
RTh = Vsc/Isc = 7 Ω

128. Graphical method for finding VTh and RTh
SEQUEL file: ee101 thevenin 1.sqproj
A B
2 Ω
6 A
12 Ω
48 V
4 Ω 4 Ω
12 Ω
i
v
A B
2 Ω
4 Ω
48 V6 A
12 Ω
4 Ω
12 Ω
Connect a voltage source between A and B.
Plot i versus v.
Voc = intercept on the v-axis.
Isc = intercept on the i-axis.
0
2
4
6
8
10
v (Volt)
0 20 40 60
i(Amp)
Voc = 60 V, Isc = 8.5714 A
RTh = Vsc/Isc = 7 Ω
A B
7 Ω
VTh = 60 V
RTh = 7 Ω
60 V
M. B. Patil, IIT Bombay

129. Norton equivalent circuit
A
B
RTh
VTh

130. Norton equivalent circuit
A
B
RTh
VTh
A
B
IN
RN

131. Norton equivalent circuit
A
B
RTh
VTh
A
B
IN
RN
* Consider the open circuit case.

132. Norton equivalent circuit
A
B
RTh
VTh
A
B
IN
RN
* Consider the open circuit case.
Thevenin circuit: VAB = VTh .

133. Norton equivalent circuit
A
B
RTh
VTh
A
B
IN
RN
* Consider the open circuit case.
Thevenin circuit: VAB = VTh .
Norton circuit: VAB = IN RN .

134. Norton equivalent circuit
A
B
RTh
VTh
A
B
IN
RN
* Consider the open circuit case.
Thevenin circuit: VAB = VTh .
Norton circuit: VAB = IN RN .
⇒ VTh = IN RN .

135. Norton equivalent circuit
A
B
RTh
VTh
A
B
IN
RN
A
B
A
B
Isc IscVTh
IN
RN
RTh
* Consider the open circuit case.
Thevenin circuit: VAB = VTh .
Norton circuit: VAB = IN RN .
⇒ VTh = IN RN .
* Consider the short circuit case.

136. Norton equivalent circuit
A
B
RTh
VTh
A
B
IN
RN
A
B
A
B
Isc IscVTh
IN
RN
RTh
* 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 .

137. Norton equivalent circuit
A
B
RTh
VTh
A
B
IN
RN
A
B
A
B
Isc IscVTh
IN
RN
RTh
* 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 .

138. Norton equivalent circuit
A
B
RTh
VTh
A
B
IN
RN
A
B
A
B
Isc IscVTh
IN
RN
RTh
* 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

139. Example
20 V
1 Ai5 Ω
10 Ω

140. Example
20 V
1 Ai5 Ω
10 Ω
A
B

141. Example
20 V
1 Ai5 Ω
10 Ω
A
B
RN = 5 Ω
IN =
20 V
5 Ω
= 4 A

142. Example
20 V
1 Ai5 Ω
10 Ω
A
B
RN = 5 Ω
IN =
20 V
5 Ω
= 4 A
A
B
1 Ai
10 Ω5 Ω
4 A

143. Example
20 V
1 Ai5 Ω
10 Ω
A
B
RN = 5 Ω
IN =
20 V
5 Ω
= 4 A
A
B
1 Ai
10 Ω5 Ω
4 A
i
10 Ω5 Ω
3 A

144. Example
20 V
1 Ai5 Ω
10 Ω
A
B
RN = 5 Ω
IN =
20 V
5 Ω
= 4 A
A
B
1 Ai
10 Ω5 Ω
4 A
i
10 Ω5 Ω
3 A
= 1 A
i = 3 A ×
5
5 + 10
M. B. Patil, IIT Bombay

145. Example
20 V
1 Ai5 Ω
10 Ω
A
B
RN = 5 Ω
IN =
20 V
5 Ω
= 4 A
A
B
1 Ai
10 Ω5 Ω
4 A
i
10 Ω5 Ω
3 A
= 1 A
i = 3 A ×
5
5 + 10
Home work:
* Find i by superposition and compare.
M. B. Patil, IIT Bombay

146. Example
20 V
1 Ai5 Ω
10 Ω
A
B
RN = 5 Ω
IN =
20 V
5 Ω
= 4 A
A
B
1 Ai
10 Ω5 Ω
4 A
i
10 Ω5 Ω
3 A
= 1 A
i = 3 A ×
5
5 + 10
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