Unit-1
Types of Network, Mesh, and Nodal analysis
1) Unilateral and Bilateral Elements:The unilateral elements allow the flow of current in only one direction. Ex. Diode.
The bilateral element allows the flow of current in both directions. Ex. Resistor, inductor, and capacitor.
2) Linear and Non-Linear: The elements which obey ohms law or have linear relationships are known as Linear elements. Ex. Resistor, inductor, and capacitor.
The elements which do not obey ohms law are known as Non-Linear elements.
3) Time variant and Time invariant: An element whose VI characteristics change with time are time-variant and the element whose VI characteristics do not change with time is time-invariant.
4) Lumped and Distributed: Lumped elements are Physical dimensions of the circuit are such that voltage across and current through conductors connecting elements does not vary. The current in the two-terminal lumped circuit element does not vary. In distributed elements, the current varies along conductors and elements. The voltage across points along the conductor or within element varies.
Key takeaway:
The unilateral elements allow the flow of current in only one direction. Ex. Diode.
The bilateral element allows the flow of current in both directions. Ex. Resistor, inductor, and capacitor.
Q1) For the characteristics given below comment about the type of element?
Sol: It is a Linear, Bilateral and Passive element
Q2) For the figure below comment on the type of element?
Sol: The element is Non-Linear, Passive, and Unilateral.
Voltage and current sources
1. Voltage Source: A voltage source always have resistance in series.
Ideal source
R = 0
Practical = 2 - 5Ω
I = 0
Fig 1 Voltage source
2. Current Source: Always have R parallel with the source.
Fig 2 Current Source
Dependent and Independent sources
few either
AC or DC
dc
Fig 3 Independent Source
Fig 4 Independent Source
Key takeaway
For voltage source, R is always in series
For the current source, R is always in parallel
1.3.1 Concept of voltage and current divider
VDF - Voltage Division Formula
Fig 5 VDF
CDF - Current Division Formula
Fig 6 CDF
1.3.2 Source transformation and shifting.
Fig 7 Voltage to current Transformation
2. From IV:
Fig 8 Current to voltage source Transformation
Key takeaway
The current source is replaced by a voltage source and vice versa in source transformation. In voltage, source resistance is in series.
LOOP ANALYSIS
Fig 9 Circuit for loop analysis
For Loop 1 with il
For Loop 2 with i2
For Loop 3 with i3
Example 1. For the circuits given below write the voltage equations:
Solution: Let current i1be in loop 1 current and i2 for loop 2
Que Write the loop equations for the given circuit below?
Solution:
For loop 1
For loop 2
NODE ANALYSIS
For these, we assume every node as a voltage point and write the current equation for every element. For the current source, the current entering is negative.
Fig 10 Circuit for Node analysis
For node V1
For node V2
For V
Example: Using nodal analysis find the voltage across 5resistor.
Solution:
For V1
-1
For V2
-2
Solving 1 and 2:
For 5 voltage =
= -50.9 + 57.27
= 6.37V
Key takeaway
We use Nodal analysis in those circuits where we need to find the voltage. The loop analysis is applied to the circuits when we want to calculate the branch or loop currents.
Concept of Supernodes
We are primarily interested in node voltages. We may avoid utilizing the voltage source branch that is causing difficulty. To do this we may treat N2, N3, and the voltage source together as a type of supernode and apply KCL to both nodes at the same time.
This is possible since the total current leaving node 2 is zero and the total current leaving node 3 is also zero. Thus, the current leaving both the nodes in totality must be zero.
Fig 11 Circuit with supernode
A simple way of identifying the supernode is to think of the voltage source as a short circuit physically joining the two nodes together.
Solving the above circuit for finding node voltage
KCL for V1
2+5+ [(V1-V2)/(1/3)]+[(V1-V3)(1/2)]=0
5V1 -3V2-2V3+7=0 (1)
KCL at V4
-5+6+ [(V4-V2)/(1/4)]+[(V4-V3)/1] = 0
4V2+V3+4V4+1=0 (2)
Applying KCL for both N2 and N3 simultaneously we get
[(V2-V1)/(1/3)]-2+(4V2)+[2(V3-V1)]+V3-6=0
-5V1 + 7V2 + 3V3 = 8 (3)
We get one more relation as
V3-V2 = 4 (4)
Solving equations 1,2,3 and 4 we get
V2= -3/5 V
V3 = 17/5 V
V1= -2/5 V
Concept of Super mesh
If the current source is also present in the network then we face difficulty in writing mesh equations. A better approach is we assume the current source to be open-circuited and create a new mesh called super mesh. We then apply KVL only to those meshes in the modified network.
Fig 12 Circuit (a) 3A in a circuit (b) with super mesh
Let the mesh current be i1, i2, and i3. The current source is mentally open-circuited. Thus, creating a new mesh called super mesh shown by dotted lines. Applying KVL around this mesh we get
-4+ (i1-i3)1 + 2(i2-i3)+i2 = 0
i1+3i2-3i3=4 (1)
2(i3-i2)+i3-i1+3i3=0
i1+2i2-6i3=0 (2)
i1-i2=3 (3)
Solving the above equations we get
i1= 4A
i2= 1A
i3= 1A
Key takeaway
When a current source creates a problem while applying mesh we use super mesh by open circuiting the current source. When a voltage source creates a problem while applying the node we use a supernode by short-circuiting the voltage source.
Fig 13 Mutual Inductance
M12= mutual inductance on 1 due to 2
M21= mutual inductance on 2 due to 1
Fig 14 Current entering both dots
the voltage on 2 due to 1 =
the voltage on 1 due to 2 =
Fig 15 Current entering both dots
⎼⎼⎼⎼⎼⎼ 1
⎼⎼⎼⎼⎼⎼2
⎼⎼⎼⎼⎼⎼⎼1
⎼⎼⎼⎼⎼⎼⎼2
2. When current is entering from one dot and leaving the other
Fig 16 Current entering one dot and leaving other
⎼⎼⎼⎼⎼⎼⎼⎼1
⎼⎼⎼⎼⎼⎼⎼⎼2
Find =?
Fig 17 Current enters both dots
If M12= M21= M
II. When current leaves both dots
Find
Fig 18 Current Leaving both dots
III. When current enters from L1 and leaves L2
Find
Fig 19 current enters from L1 and leaves L2
IV. When current leaves L2and enters L1
Fig 20 current leaves L2and enters L1
Key takeaway
The current entering the dot is taken as positive.
Example 1. Find
= 2H + 3H +10H - 2(10H)+ 2(4H) + 2(6H)
= 15H + - 20H + 20H
= 0
Example 2. Find
5H + 6H + 9H - 2(1H) -2(5H) -2(3H)
= 2H
2. PARALLEL COMBINATION: -
Fig 21 Parallel Combination
The two networks are said to be dual if the mesh equation of one network is equal to the node equation of the other.
Let us consider the RLC circuit
Fig 22 RLC Circuit
Applying KVL we have
v(t) = ri(t) + L +
So, dual of the above network will be
Fig 23 Dual of RLC circuit
i(t) = Gv(t) + C +
Steps for construction of Dual Network:-
i) In each loop place a dot. This internal dot corresponds to an independent node in the dual network.
ii) Outside network place a dot. This is called the datum node is a dual network
iii) Connect all the adjacent loops by dashed lines crossing common branches.
iv) In a dual network, this common branch is replaced by its dual element.
Fig 24 Circuit whose dual is to make
Fig 25 Dual of the above circuit
Key takeaway
Some important duals are given below.
Loop | Node | Loop | Node |
Current | Voltage | Number of Loops | Number of nodes |
Inductance | Capacitance | Link | Tree branch |
Resistance | Conductance | Link current | Tree branch voltage |
Branch Current | Node Voltage | Tree branch current | Link voltage |
Mesh or Loop | Node or pair | Tie Set | Cut Set |
Loop Current | Node pair Voltage | Short Circuit | Open Circuit |
Mesh Current | Node Potential | Parallel path | Series path |
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Reference
[1] Network Analysis Third Edition by M. E. Van Valkenburg, Prentice Hall of India Private Limited.
[2] Network Analysis & Synthesis by G. K. Mittal, Khanna Publication.
[3] Network Analysis and Synthesis by Ravish R Singh, McGraw Hill.
[4] Introduction to Electric Circuits by Alexander & Sadiku, McGraw Hill.
[5] Introduction to Electric Circuits by S. Charkarboorty, Dhanpat Rai & Co.
[6] Fundamentals of Electrical Networks by B.R.Gupta & Vandana Singhal- S.Chand Publications
[7] Electrical Circuit Analysis 2nd Edition by P. Ramesh Babu, Scitech Publication
India Pvt Ltd.