network theory(19ec33) 2020-21
TRANSCRIPT
Network Theory(19EC33)2020-21
Class-12: Dependent sources & Mesh Analysis
MOHANKUMAR V.
ASSISTANT PROFESSORDEPARTMENT OF ELECTRONICS AND COMMUNICAT ION ENGINEER ING
DR . AMBEDKAR INST ITUTE OF TECHNOLOGY , BENGALURU -56
E-MAIL : MOHANKUMAR.V@DR -A IT .ORG
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Dependent/Controlled Sources
Dependent Sources
Current and voltage of source depends on some other current and voltage.
Example:
Applications: Analysis of amplifiers
Types:
1. Voltage Dependent Voltage source
2. Current Dependent Voltage source
3. Voltage Dependent Current Source
4. Current dependent Current source
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Examples:
1. Find the mesh currents for the electrical circuit shown in figure.
Obtain the control variables in terms of
mesh currents.
π°π¨ = π°π; π°π© = π°πKVL to Mesh-1
βπ + ππ°π + πππ°π© + ππ π°π β π°π + ππ°π¨ = πβπ + ππ°π + πππ°π + ππ π°π β π°π + ππ°π = π
πππ°π = πTherefore π°π = π. πππ¨
KVL to Mesh-2
βππ°π¨ + ππ π°π β π°π + ππ°π + ππ = πβππ°π + ππ π°π β π°π + ππ°π + ππ = π
βπππ°π + πππ°π = βππW.K.T., π°π = π. πππ¨Therefore π°π = βπ. ππππ¨
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Examples:
2. Find the mesh currents for the electrical circuit shown in figure.
Obtain the control variables in terms of
mesh currents.
π°π± = ππ; ππ² = ππ β ππKVL to mesh-1
βπ + ππ°π + π°π + π π°π β π°π = π
βπ + π°π + π°π β π°π + π°π β π°π = πππ°π β π°π = π ββ β(π)
KVL to mesh-2
π ππ β ππ β ππ² + πππ + ππ± + π ππ β ππ = π
ππ β ππ β ππ + ππ + ππ + ππ + ππ β ππ = ππππ = πππ = ππ.
Mesh-3
ππ = βππ
π°π = ππ¨ , π°π = ππ¨ ππ§π ππ = βππ
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Examples:
3. Find the mesh currents for the electrical circuit shown in figure.
Apply KVL to mesh-2
ππ½π + π π°π β π°π + ππ°π + ππ = πβπππ°π + π°π β π°π + ππ°π + ππ = πβπππ°π + ππ°π = βππ ββ β(π)
π°π = π. πππ π¨ πππ π°π = βπ. π π¨
Obtain the control variables in terms of
mesh currents.
π½π = βππ°π; π½π = ππ°π;Apply KVL to mesh-1
+π + ππ°π + ππ½π + ππ°π + π π°π β π°π β ππ½π = ππ + ππ°π + ππ°π + ππ°π + π°π β π°π + πππ°π = π
πππ°π + ππ°π = βπ ββ β(π)
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Examples:
Obtaining the relationship between control variables and mesh currents
π½π = πΉπ°π π½π = πΉ(βπ°π) π½π = πΉ(π°π β π°π) π½π = πΉ(π°π β π°π)
πͺπππ(π) πͺπππ(ππ) πͺπππ(πππ) πͺπππ(ππ)
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Examples:
4. Find the mesh currents I1, I2, I3 and I4 for the electrical circuit shown in figure.
Obtain the control variables in terms of
mesh currents.
π°π = πππ¨ , π°π = πππ¨, π°π = πππ¨ πππ π°π = πππ¨
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Examples:
5. Find the mesh currents I1 and I2 for the electrical circuit shown in figure.
Obtain the control variables in terms of
mesh currents.
π°π = ππ¨, π°π = βπ π¨ πππ π°π = ππ¨
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Disclaimer
Some Contents and Images showed in this PPT have
been taken from the various internet sources and
from books for educational purpose only.
Thank You
?
Network Theory(19EC33)2020-21
Class-13: Dependent sources & Mesh Analysis
MOHANKUMAR V.
ASSISTANT PROFESSORDEPARTMENT OF ELECTRONICS AND COMMUNICAT ION ENGINEER ING
DR . AMBEDKAR INST ITUTE OF TECHNOLOGY , BENGALURU -56
E-MAIL : MOHANKUMAR.V@DR -A IT .ORG
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Examples:
Obtaining the relationship between control variables and mesh currents
π½π = πΉπ°π π½π = πΉ(βπ°π) π½π = πΉ(π°π β π°π) π½π = πΉ(π°π β π°π)
πͺπππ(π) πͺπππ(ππ) πͺπππ(πππ) πͺπππ(ππ)
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Examples:
4. Find the mesh currents I1, I2, I3 and I4 for the electrical circuit shown in figure.
Obtain the control variables in terms of
mesh currents.
π½π =π
ππ°π β π°π ββ β π
KVL at mesh-1
+π +π
πππ°π +
π
πππ β ππ +
π
πππ β ππ = π
π. πππ°π + π. ππ°π β π. ππππ°π = βπ βββ(π)5Vx current source is common to mesh-2
and 3.
π°π β π°π = ππ½π
π°π β π°π = π.π
ππ°π β π°π
π°π β ππ°π + π°π = π ββ β πKVL to supermeshπ
πππ°π +
π
πππ°π +
π
ππ°π β π°π +
π
ππ°π β π°π = π
βπ. ππ°π + π. πππ°π + π. πππ°π β π. ππ°π = π βββ β π
At mesh 4
π°π = πππ¨
π°π = πππ¨ , π°π = πππ¨, π°π = πππ¨ πππ π°π = πππ¨
Supermesh
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Examples:
5. Find the mesh currents I1 and I2 for the electrical circuit shown in figure.
Obtain the control variables in terms of
mesh currents.
π½π = π π°π β π°π ββ β π
KVL to mesh 1
βπππ + πππ°π + ππ°π + π π°π β π°π = ππππ°π β ππ°π = πππ ββ β π
0.5V1 current source is common to
mesh-2 and 3,
π. ππ½π = π°π β π°ππ. π π π°π β π°π = π°π β π°π
π°π = π°π β β πKVL to supermesh
πππ°π + ππ°π + π π°π β π°π = πβππ°π + ππ°π + πππ°π = π β β(π)
Solve equations (2, 3 and 4)
We get.
π°π = ππ¨, π°π = βπ π¨ πππ π°π = ππ¨
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis
β’ Node: A junction or a point where two or more elements
are connected.
β’ Example:
β’ Fundamental Node : A node where Current division takes
place
β’ Number of unknowns is equal to the number of nodes-1.
β’ V1, V2, V3β¦are node voltages
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis
Procedure to apply Node Analysis:
Step-1: As for as possible try to convert voltage sources into current sources, without
affecting the load elements.
Step-2: Identify the number of fundamental nodes.
Step-3: Name the nodes and assign node voltages as V1, V2, V3β¦
NOTE: Ground potential is always zero.
Step-4: Assign branch currents to each branch as I1, I2, I3 etc., and choose the directions
randomly.
Step-5: Apply KCL at each node
NOTE: Number of KCL equations is equal to the number of nodes-1/number of
unknowns.
Step-6: Replace branch currents in terms of node voltages.
Step-7: Solve KCL equations using any mathematical technique to find node voltages.
Step-8: Find the branch currents/branch voltages/power from node voltages using ohmβs
law.
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Disclaimer
Some Contents and Images showed in this PPT have
been taken from the various internet sources and
from books for educational purpose only.
Thank You
?
Network Theory(19EC33)2020-21
Class-14&15: Node Analysis and Examples
MOHANKUMAR V.
ASSISTANT PROFESSORDEPARTMENT OF ELECTRONICS AND COMMUNICAT ION ENGINEER ING
DR . AMBEDKAR INST ITUTE OF TECHNOLOGY , BENGALURU -56
E-MAIL : MOHANKUMAR.V@DR -A IT .ORG
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis
β’ Node: A junction or a point where two or more elements
are connected.
β’ Example:
β’ Fundamental Node : A node where Current division takes
place
β’ Number of unknowns is equal to the number of nodes-1.
β’ V1, V2, V3β¦are node voltages
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis
Procedure to apply Node Analysis:
Step-1: As for as possible try to convert voltage sources into current sources, without
affecting the load elements.
Step-2: Identify the number of fundamental nodes.
Step-3: Name the nodes and assign node voltages as V1, V2, V3β¦
NOTE: Ground potential is always zero.
Step-4: Assign branch currents to each branch as I1, I2, I3 etc., and choose the
directions randomly.
Step-5: Apply KCL at each node
NOTE: Number of KCL equations is equal to the number of nodes-1/number of
unknowns.
Step-6: Replace branch currents in terms of node voltages.
Step-7: Solve KCL equations using any mathematical technique to find node voltages.
Step-8: Find the branch currents/branch voltages/power from node voltages using ohmβs
law.
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Tips
ππ =ππ β ππ
π
ππ =ππ β ππ
π
I1 =VG β (βV1) β V2
R
ππ =ππ β ππ
π(ππ’π§ππ ππ = ππ)
I1 =V2 β V1 β VG
R
ππ =ππ β ππ
π(ππ’π§ππ ππ = ππ)
I1 =VG β V1
R
ππ =βπππ
(ππ’π§ππ ππ = ππ)
I1 =V1 β VG
R
ππ =πππ
(ππ’π§ππ ππ = ππ)
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Tips
I1 =V1 β (βVs) β V2
R
ππ =ππ + ππ¬ β ππ
π
ππ =ππ β ππ¬ β ππ
π
ππ =ππ β ππ¬ β ππ
π I1 =V2 β (βVs) β V1
R
ππ =ππ + ππ¬ β ππ
π
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples1. Find the branch currents and branch voltages for the electrical circuit shown in figure.
Identify the fundamental nodes
Assign branch currents
KCL at node 1
1 = πΌ1 + πΌ2Express branch currents in terms of node
voltages
1 =π1 β ππΊ
2+π1 β π2
21 = π1 β 0.5π2 ββ β(1)
KCL at node 2
πΌ2 + 2 = πΌ3π1 β π2
2+ 2 =
π2 β ππΊ1
0.5π1 β 1.5π2 = β2 ββ β(2)
Answer:
V1=2VV2=2V
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-ExamplesBranch currents
πππ = ππ
πππ = ππ =πππ= ππ
πππ = ππ =ππ β ππ
π= ππ
πππ = ππ =πππ= ππ.
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples2. Find the Node voltages for the electrical circuit shown in figure.
Identify the fundamental nodes
Assign branch currents KCL at node A
πΌ1 + πΌ2 = πΌ3Express branch currents in terms of node voltages
10 β ππ΄2
+ βππ΄10
=ππ΄ β ππ΅
5ββ β(1)
8ππ΄ β 2ππ΅ = 50 ββ β(1)KCL at node B
πΌ3 = πΌ4 + πΌ5 +1
3ππ΄ β ππ΅
5=ππ΅15
+ππ΅ β 18
3ββ β(2)
3ππ΄ β 9ππ΅ = β30 ββ β(2)ππ§π¬π°ππ«: ππ = π. πππ ππ§π ππ = ππ. πππ
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples3. Find the Node voltages for the electrical circuit shown in figure.
Identify the fundamental nodes Assign branch currents
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples
KCL at node 1, πΌ2 = πΌ1 + πΌ3 + 9π2 β π1
2=π1 β 12
4+π1 β π3
4+ 9
βπ1 + 0.5π2 + 0.25π3 = 6 ββ β 1KCL at node 2, πΌ2 + πΌ4 + πΌ5 = 0
π2 β π12
+π2100
+π2 β π3
5= 0
β0.5π1 + 0.71π2 β 0,2π3 = 0 ββ β 2KCL at node 3, 9 + πΌ3 + πΌ5 + πΌ6 = 0
9 +π1 β π3
4+π2 β π3
5+βπ320
= 0
0.25π1 + 0.2π2 β 0.5π3 = β9 ββ β(3)
ππ§π¬π°ππ«: ππ = π. πππ , ππ = ππ. πππ ππ§π ππ = ππ. πππ
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Disclaimer
Some Contents and Images showed in this PPT have
been taken from the various internet sources and
from books for educational purpose only.
Thank You
?
Network Theory(19EC33)2020-21
Class-16: Node Analysis and Examples
MOHANKUMAR V.
ASSISTANT PROFESSORDEPARTMENT OF ELECTRONICS AND COMMUNICAT ION ENGINEER ING
DR . AMBEDKAR INST ITUTE OF TECHNOLOGY , BENGALURU -56
E-MAIL : MOHANKUMAR.V@DR -A IT .ORG
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples4. Find the V1 and V2 using Node voltage analysis for the electrical circuit shown in figure.
Identify the fundamental nodes and assign branch currents
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples
ππ‘ ππππ β 150V source is connected between non
reference node -1 to ground node.
π1 = 50π ββ β(1)Apply KCL at node-2
πΌ2 = πΌ3 + πΌ4π1 β π250
=π2 β 10
20+π210
0.02π1 β 0.17π2 = β0.5 ββ β(2)
ππ§π¬π°ππ«: ππ = πππ and V2=8.82V
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples5. Find the V1 and V2 using Node voltage analysis for the electrical circuit shown in figure.
Identify the fundamental nodes and assign branch currents
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples
ππ = 20π ββ β 1KCL at node a.
πΌ1 = πΌ2 + 280 β ππ50
=ππ β ππ10
+ 2
β0.12ππ + 0.1ππ = 0.4 ββ β 2KCL at node b
πΌ2 = πΌ3 + πΌ4ππ β ππ10
=ππ50
+ππ β ππ20
0.1ππ β π. 17ππ + 0.05ππ = 0 βββ 3
ππ§π¬π°ππ«: ππ = π. πππ, ππ = π. πππ , ππ = πππ,
π½π = π½π β π½π; π½π = π½π β π½πππ = βπ. πππ ππ§π ππ = βππ. πππ
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples6. Find Node voltages for the electrical circuit shown in figure.
Identify the fundamental nodes and assign branch currents
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples
Answer: π½π = ππ½ ; π½π = ππ½ ; π½π = ππ½.
NOTE: if an ideal voltage source is connected between two non
reference nodes, consider that common voltage source
independently and write mathematical equation corresponding
to the common voltage sources.
π2 β π3 = 2 βββ 1For further analysis, no need to consider that voltage source.
Combine nodes 2 and 3 to form a single node called as super
node.
Apply KCL at super node.
πΌ2 + 3 + 2 = πΌ3π1 β 6 β π2
2+ 5 =
π34βββ β 2 ; 0.5V1 β 0.5V2 β 0.25V3 = β2
Apply KCL at node1
πΌ1 = πΌ2 + 2
βπ12
=π1 β 6 β π2
2+ 2 ββ β 3 ;βπ1 + 0.5π2 = β1
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Practice problems1. Carryout nodal analysis and find V2.
2. Carryout nodal analysis and find node voltages.
π½π = βπ. πππ½ πππ π½π β π. πππ½
π½π = π. ππ π½ππππ
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Practice problems3. Carryout nodal analysis and find voltage across 2 Ohms resistor (Connected Vertically).
4. Find the power dissipated in 6KOhms resistor using node voltage analysis.
π· = π. πππΎππππ
π½π π½πππππππππ πππππππππ = π. ππ π½ππππ
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Practice problems5. Find the current through 50 Ohms resistor using mesh analysis.
6. Find the power dissipated in 6KOhms resistor using Mesh analysis.
π°ππ πΆπππ = π. πππ¨ (π ππ π)
π· = π. πππΎππππ
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Disclaimer
Some Contents and Images showed in this PPT have
been taken from the various internet sources and
from books for educational purpose only.
Thank You
?
Network Theory(19EC33)2020-21
Class-17&18: Node Voltage Analysis and AC circuit Analysis
MOHANKUMAR V.
ASSISTANT PROFESSORDEPARTMENT OF ELECTRONICS AND COMMUNICAT ION ENGINEER ING
DR . AMBEDKAR INST ITUTE OF TECHNOLOGY , BENGALURU -56
E-MAIL : MOHANKUMAR.V@DR-A IT .ORG
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples7. Find V1 and V2 using Node voltage method for the electrical circuit shown in figure.
πΌ1, πΌ2 πππ π2Express control variables in terms of node voltages.
πΌ1 =π210
; πΌ2 =π110
πππ π2
Apply KCL at node-1.
2πΌ1 = πΌ2 + πΌ3 + 2π2
2π210
=π110
+π1 β π210
+ 2π2
0.2π2 = 0.1π1 + 0.1π1 β 0.1π2 + 2π2π½π = π π½ππππ
Apply KCL at node-2
πΌ3 + 2π2 + 2πΌ2 = πΌ1π1 β π210
+ 2π2 + 2π110
=π210
0.1π1 β 0.1π2 + 2π2 + 0.2π1 = 0.1π20.3π1 + 1.8π2 = 0 β β(1)
π½π = π π½ππππ.
ππ§π¬π°ππ«: ππ = ππ ππ§π ππ = ππ
I3
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Node Analysis-Examples8. Find Node voltages for the electrical circuit shown in figure.
ππ₯ πππ ππ¦πππ ππππ‘πππ π£ππππππππ .
Express control variables in terms of node voltages.
ππ₯ = π2 β π1 βββ(1)ππ¦ = π4 β π1 ββ β(2)
Node-1
π1 = β12 ππππ‘π ββ β 3Apply KCL at node 2
πΌ1 + πΌ3 + 14 = 0π1 β π20.5
+π3 β π2
2+ 14 = 0 ββ β 4 ; 2π1 β 2π2 + 0.5π3 β 0.5π2 = β14
ππ½π β π. ππ½π + π. ππ½π = βππ0.2Vy voltage source is between V3 and V4.
0.2ππ¦ = π3 β π4 ββ β 5 ; 0.2 π4 β π1 = π3 β π4 ββ β(5)
βπ. ππ½π β π½π + ππ½π = πApply KCL at super node.
πΌ2 + πΌ4 + 0.5ππ₯ = πΌ3π1 β π42.5
+ βπ4 + 0.5 π2 β π1 =π3 β π2
2ββ β(6)
βπ. ππ½π + π½π β π. ππ½π β π. ππ½π = π
ππ§π¬π°ππ«: ππ = βππ, ππ β ππ, ππ = π π ππ§π ππ = βππ
super node
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
AC QuantitiesAC quantities are represented in two different formats.
Polar form
Format=Mβ β Where, M is the magnitude and β is the phase angle
Rectangular form
x + jyWhere, x is the real part and y is the imaginary part.
Conversion from rectangular to polar
Given: x+jy, To find: M and β
M = x2 + y2 and β = tanβ1(y
x)
Conversion from polar to rectangular
Given : M and β , To find: x and y
x = Mcosβ and y = Msinβ
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
AC QuantitiesConversions are used to perform mathematical calculations.
For addition and subtraction-Rectangular form
Consider
A = x1 + jy1 and B = x2 + jy2A + B = (x1+x2) + j y1 + y2
Similarly A β B = x1 β x2 + j y1 β y2
For multiplication and division-Polar form
Consider
A = M1β β 1 and B = M2β β 2A β B = M1 β M2β β 1 + β 2
SimilarlyA
B=
M1
M2β β 1 β β 2
Also
A β B =(x1 + jy1)(x2 + jy2)= x1x2 + jx1y2 + jx2y1 + j2y1y2
= x1x2 β y1y2 + j x1y2 + x2y1 since j2 = β1
NOTE:
A = 10; A = 10β 00 β 10 + j0
A = 10β 900 β 0 + j10
A = 10β β 900 β 0 β j10
A = 10 + j0 β 10 or 10β 00
A = j10 β 10β 900
A = βj10 β 10β β 900 or β 10β 900
A =1
jβ βj
A = j β j β j2 = β1.
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
AC Quantities
Resistors:
R Ohms (Same for both DC and AC)
Voltage in phase with the current
Capacitors:
C Farads (DC analysis)
-jXπͺ Ohms (Capacitive reactance AC analysis) (Negative sign-
Voltage lags current by 90o)
Where, ππΆ =1
2πππΆβ 1/π€πΆ,π€βπππ, π ππ π‘βπ πππππ’ππππ¦.
Inductors:
L Henry (DC analysis)
+jXπ³ Ohms (Inductive reactance AC analysis) (Positive sign-
voltage leads current by 90o)
Where, ππΏ = 2πππΏ β π€πΏ,π€βπππ, π ππ π‘βπ πππππ’ππππ¦.
Resistors
Capacitors
Inductors
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Mesh Analysis-Examples
KVL at mesh-1
5I1 + j5(I1 β I2) β 30β 00 = 0π + π£π ππ β π£πππ = ππβ ππ ββ β π
KVL at mesh-2
2I2 + j3I2 + 6 I2 β I3 + j5 I2 β I1 = 0βπ£π ππ + π + π£π ππ β πππ = π ββ β π
KVL at mesh-3
4I3 + 35.36β 450 + 6 I3 β I2 = 0βπ ππ + ππ ππ = βππ. ππβ πππ ββ β π
Solve equations (1), (2) and (3), we get
I1, I2 and I3
1. Find the mesh currents using mesh analysis
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Mesh Analysis-Examples
β1= 1320 + π2400 β π750 + 750βπ= ππππ + πππππ or ππππ. ππβ ππ. πππ
Similarly evaluate β2 πππ β3
Therefore π°π =βπ
β= π. ππβ ππ. ππ ππ¨
Similarly evaluate πΌ2 =β2
βand πΌ3 =
β3
β
Cramerβs rule:
β=(5 + π5) βπ5 0βπ5 8 + π8 β60 β6 10
, π =30 + π0
0β25 β π25
β= 5 + π5 [ 8 + π8 10) β 6 6 β (βπ5)[βπ5 10 β 0]β= (5 + π5)[80 + π80 β 36] + π5[βπ50]
β= 5 + π5 44 + π80 + 250β= 220 + π400 + π220 β 400 + 250β= ππ + ππππ or πππ. πβ ππ. πππ
β1=
30 + π0 βπ5 00 8 + π8 β6
β25 β π25 β6 10
β1= 30 8 + π8 10 β 36 + π5 β6 25 + π25
β1= 30 80 + π80 β 36 + π5 β150 β π150β1= 30 44 + π80 β π750 + 750
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Mesh Analysis-Examples
2. Find the current I0 using mesh analysis
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Mesh Analysis-ExamplesKCL at node-1
πΌ1 = πΌ2 + πΌ330β 00 β π1
5=
π1π5
+π1 β π22 + π3
6 β 0.2π1 = βπ0.2π1 +π1 β 2
3.6β 56.30
6 β 0.2π1 = βπ0.2π1 + 0.277β β 56.30(π1 β π2)6 β 0.2π1 = βπ0.2π1 + 0.153π1 β π0.23π1 β 0.153π2 + π0.23π26 = 0.2π1 β π0.2π1 + 0.153π1 β π0.23π1 β 0.153π2 + π0.23π2(π. πππ β ππ. ππ)π½π + (βπ. πππ + ππ. ππ)π½π = π ββ β(π)
KCL at node-2
πΌ3 + πΌ4 = πΌ5π1 β π22 + π3
+ βπ26
= (π2β35.36β 450)/4
3. Find the Node voltages using node analysis
0.153π1 β π0.23π1 β 0.153π2 + π0.23π2 β 0.166V2 = 0.25V2 β 8.84β 450.153π1 β π0.23π1 β 0.153π2 + π0.23π2 β 0.166V2 β 0.25V2 = 8.84β 45
π. πππ β ππ. ππ π½π + βπ. πππ + ππ. ππ π½π = π. ππβ ππ ββ β(π)
πΊππππ π πππ π ππ ππππ π½π πππ π½π
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Problems4. Find IX using node analysis.
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Problems
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Practice problems1. Find node voltages
2. Find node voltages and mesh currents for the electrical circuit shown in figure.
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Practice problems3. Find node voltages
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Disclaimer
Some Contents and Images showed in this PPT have
been taken from the various internet sources and
from books for educational purpose only.
Thank You
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Network Theory(19EC33)2020-21
Class-19&20: UNIT-II; Ch-1: Network Theorems
MOHANKUMAR V.
ASSISTANT PROFESSORDEPARTMENT OF ELECTRONICS AND COMMUNICAT ION ENGINEER ING
DR . AMBEDKAR INST ITUTE OF TECHNOLOGY , BENGALURU -56
E-MAIL : MOHANKUMAR.V@DR-A IT .ORG
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Network TheoremsTheorems:
Theorems are statements that can be demonstrated to be true by some accepted mathematical
arguments and functions. Generally theorems are general principles. The process of showing a
theorem to be correct is called a proof.
β’ Proved theorems can be used to analyze the given system, and theorems helps to analyze the
complex systems easily.
β’ In electrical system most popular theorems are
1. Theveninβs Theorem
2. Nortonβs Theorem
3. Superposition Theorem
4. Reciprocity theorem
5. Millmanβs Theorem and
6. Maximum Power Transfer Theorem
`
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Theveninβs Theorem1. Theveninβs Theorem
Statement: Any active linear bilateral complex electrical network between open circuited load terminals can
be replaced by a single practical voltage source between the same open circuited load terminals.
A practical voltage source is a series combination of ideal voltages source and a resistor (DC
circuit)/Impedance(AC Circuits).
The voltage source being equal to the voltage measured between the open circuited load terminals, denoted
as VTH or VOC and Resistor/Impedance being equal to the equivalent Resistance / Impedance measured
between open circuited load terminals by replacing all independent sources by their internal impedances,
denoted as RTH or ZTH.
Internal Impedance of an ideal voltage sources is zero, Hence replace it by short circuit.
Internal Impedance of an ideal current sources is infinity , hence replace it by open circuit.
`
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Theveninβs TheoremProcedure to obtain Theveninβs equivalent circuit.
Step-1: Identify the load element, remove the load element and name the load terminals.
Step-2: Find the open circuit voltage using any network analysis technique.
Step-3: Find the equivalent resistance/Impedance between the open circuited terminals by
replacing all independent sources by their internal impedance.
Step-4: Replace the given circuit between the open circuited load terminals by the Theveninβs
equivalent circuit.
Step-5: Connect the load element between the load terminals and find the required load quantity
using current division or voltage division formula.
NOTE: For the circuits with dependent sources, find RTH using the ratio=VOC/ISC
`
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Theveninβs Theorem-ExamplesExample:
1. Obtain the Theveninβs equivalent circuit between the terminals A and B.
+_20 V
5
20
10
1.5 A
A
B
5
20
10 A
B
Step-2. To find VTH
ππ΄π΅ = πππΆ = πππ» = π10 + π20πππ» = 10 πΌ2 + 20 πΌ1 ββ β(1)
Apply KVL at mesh-1
5πΌ1 + 20πΌ1 = 2025πΌ1 = 20πΌ1 = 0.8π΄.
At mesh-2
πΌ2 = 1.5π΄
πππ» = 10 1.5 + 20 0.8 β ππ π½ππππ
Step-3: To find RTH
π ππ» = π π΄π΅ = π ππ = 20||5 + 10
π ππ» = ππ πΆπππ
πΌ1
πΌ2
`
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
2. Obtain the Theveninβs equivalent circuit to find the current through R of 10 ohms
Remove the load element and name the load terminals.
Theveninβs Theorem-Examples
πππ» = π5.2 + π10.9 β 5.2 πΌ1 + 10.9 βπΌ2πππ» = π7.1 + π19.6 β 7.1 βπΌ1 + 19.6(πΌ2)
πππ» = π5.2 β 100 + π19.6 β 5.2 πΌ1 β 100 + 19.6(πΌ2)πππ» = π7.1 + 100 + π10.9 β 7.1 βπΌ1 + 100 + 10.9 βπΌ2
Apply KVL at mesh-1; 12.3πΌ1 = 100; πΌ1 = 8.13π΄Apply KVL at mesh-2; 30.5πΌ2 = 100; πΌ2 = 3.28 π΄.
π½π»π― = π. πππ½
`
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
To find RTH Re arranging the resistors
π ππ» = (5.2| 7.1 + (10.9||19.6)π ππ» = 10 πβππ
Theveninβs Equivalent circuit
Given R=10 Ohms
π°ππ =π½π»π―
πΉππ + πΉβ π. πππ¨.
Theveninβs Theorem-Examples
`
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
3. Obtain the Theveninβs equivalent circuit and find the current through RL of 20 Ohms
Theveninβs Theorem-Examples
`
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Nortonβs Theorem1. Nortonβs Theorem
Statement: Any active linear complex bilateral electrical network between open circuited load
terminals can be replaced by a single practical current source between the same open circuited
load terminals.
A practical current source is a parallel combination of ideal current source and a resistor (DC
circuit)/Impedance(AC Circuits).
The Current source being equal to the current measured through the short circuited load
terminals, denoted as IN or ISC and Resistor/Impedance being equal to the equivalent Resistance
/ Impedance measured between open circuited load terminals, denoted as RN or ZN
`
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Nortonβs Theorem
Procedure to obtain Nortonβs equivalent circuit.
Step-1: Identify the load element, remove the load element, name the load terminals and short the
load terminals.
Step-2: Find the short circuit current using any network analysis technique.
Step-3: Find the equivalent resistance/Impedance between the open circuited terminals by replacing
all independent sources by their internal impedances. (NOTE: RTH=RN)
Step-4: Replace the given circuit between the open circuited load terminals by the Nortonβs
equivalent circuit.
Step-5: Connect the load element between the load terminals and find the required load quantity
using current division or voltage division formula.
NOTE:
β’ For the circuits with dependent sources, find RN using the ratio=VOC/ISC; RTH=RN
β’ Theveninβs Theorem is the dual of Nortonβs Theorem
β’ Theveninβs equivalent circuit can be converted into Nortonβs Equivalent circuit and vice-
versa using source transformation
`
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Nortonβs TheoremExample:
2. Obtain the Nortonβs equivalent circuit between the terminals A and B.
+_20 V
5
20
10
1.5 A
A
B
5
20
10 A
B
To find IN
π°π΅ = π°π¨π© = π°πΊπͺ = π°π
25I1 β 20I3 = 20 ββ β(1)β20πΌ1 β 10πΌ2 + 30πΌ3 = 0 ββ β 2
πΌ2 = 1.5π΄ ββ β(3)π°π = π. π. π¨
π°π΅ = π°πΊπͺ = π°π = π. ππ¨
To find RN
RAB = RTH = RN = 5||20 + 10ππ = πππ
πΌ1
πΌ2
πΌ3
`
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Norton's Theorem3. Obtain the Theveninβs and Norton's equivalent circuit
between the terminals A and B.
To find IN
4πΌ1 β 6πΌπ₯ + 6(πΌ1 β πΌ2) = 20 ββ β(1)4πΌ1 = 20; πΌ1 = 5π΄.
β6πΌ1 + 6πΌ2 = 0 ββ β 2πΌ1 = πΌ2
πΌπ₯ = πΌ1 β πΌ2From (1), π°π = π°π = ππ¨
π°ππ = π°π΅ = π°π = ππ¨.To find RN
RN =πππΆπΌππΆ
To find VOC
4πΌ β 6πΌπ₯ + 6πΌ β 20 = 0πΌπ₯ = πΌπ° = ππ¨.
π½πΆπͺ = π β π = πππ½.
πΉπ΅ =ππ
π= π πΆπππ
πΌ1
πΌ2
πΌ3
πΌ1
πΌ2
Dept. of ECE, Dr. Ambedkar Institute of Technology, Bengaluru
Network Theory (19EC33)
Disclaimer
Some Contents and Images showed in this PPT have
been taken from the various internet sources and
from books for educational purpose only.
Thank You
?