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Unit-1(A)
Circuit Analysis Techniques
Basic Terms used in a Circuit
1. Node :- It is a point in a circuit where two or more circuit elements are connected
together.
2. Branch :- It is that part of a network which lies between two junctions.
3. Loop :- It is a close path in a circuit in which no element or node is encountered more
than once.
4. Mesh :- It is a loop that contains no other loop within it.
5. Junction:- It is a point in a circuit where three or more circuit elements are connected
together
Consider the circuit of Fig. (a).
It has six branches, five nodes, two meshes
And the circuit of Fig (b) has four branches, two nodes three meshes.
To find the no of branches or loops or nodes by using this relation= + ( − )b = no. of branches
l = no. of loops
= no. of nodes
Kirchhoff`s Laws
Kirchhoff’s laws are more comprehensive than Ohm's law and are used for solving electrical
networks which may not be readily solved by the latter.
Kirchhoff`s laws, two in number, are particularly useful in determining the equivalent resistance
of a complicated network of conductors and for calculating the currents flowing in the various
conductors.
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I. Kirchhoff`s Point Law or Current Law (KCL)
In any electrical network, the algebraic sum of the currents meeting at a point (or junction) is
equal to Zero.
(or)
The total current entering a junction is equal to the total current leaving that junction.
Consider the case of a network shown in Fig (a).A is the
I1+(-I2)+(I3)+(+I4)+(-I5) = 0
I1+I4-I2-I3-I5 = 0
Or
I1+I4 = I2+I3+I5
Or
Incoming currents = Outgoing currents
II Kirchhoff's Mesh Law or Voltage Law (KVL)
Kirchhoff's voltage Law states that the algebraic sum of voltages around any closed path is
equal to zero
- It should be noted that if we start from a particular junction and go round the mesh till
we come back to the starting point, then we must be at the same potential with which we
started
- Hence, it means that all the sources of emf met on the way must necessarily be equal to
the voltage drops in the resistances, every voltage being given its proper sign, plus or
minus.
(a)Determination of Voltage Sign
A rise in voltage should be given a + ve sign and a fall in voltage a -ve sign. That is, if we go
from the -ve terminal of a battery to its +ve terminal there is a rise in potential, hence this
voltage should be given a + ve sign.
And on the other hand, we go from +ve terminal to -ve terminal, then there is a fall in
potential, hence this voltage should be preceded by a -ve sign.
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The sign of the battery e.m.f is independent of the direction
(b) Sign of IR Drop
Now, take the case of a resistor (Fig.). If we go through a resistor in the same direction as the
current, then there is a fall in potential because current flows from a higher to a lower
potential..
Hence, this voltage fall should be taken -ve. However, if we go in a direction opposite to that of
the
current, then there is a rise in voltage. Hence, this voltage rise should be given a positive sign
Consider the closed path ABCDA in Fig
As we travel around the mesh in the clockwise direction, different voltage drops will have the
following signs :
I1R1 is - ve (fall in potential)
I2R2 is - ve (fall in potential)
I3R3 is + ve (rise in potential)
I4R4 is - ve (fall in potential)
E2 is - ve (fall in potential)
E1 is + ve (rise in potential)
Using Kirchhoff's voltage law, we get
-I1R1 – I2R2 – I3R3 – I4 R4 – E2 + E1 = 0
Or I1R1 + I2R2 – I3R3 + I4R4 = E1 –E2
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G.Purushotham,Asst.Professor, Dept. of EEE @ SVCE TPT. 3
The sign of the battery e.m.f is independent of the direction
(b) Sign of IR Drop
Now, take the case of a resistor (Fig.). If we go through a resistor in the same direction as the
current, then there is a fall in potential because current flows from a higher to a lower
potential..
Hence, this voltage fall should be taken -ve. However, if we go in a direction opposite to that of
the
current, then there is a rise in voltage. Hence, this voltage rise should be given a positive sign
Consider the closed path ABCDA in Fig
As we travel around the mesh in the clockwise direction, different voltage drops will have the
following signs :
I1R1 is - ve (fall in potential)
I2R2 is - ve (fall in potential)
I3R3 is + ve (rise in potential)
I4R4 is - ve (fall in potential)
E2 is - ve (fall in potential)
E1 is + ve (rise in potential)
Using Kirchhoff's voltage law, we get
-I1R1 – I2R2 – I3R3 – I4 R4 – E2 + E1 = 0
Or I1R1 + I2R2 – I3R3 + I4R4 = E1 –E2
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G.Purushotham,Asst.Professor, Dept. of EEE @ SVCE TPT. 3
The sign of the battery e.m.f is independent of the direction
(b) Sign of IR Drop
Now, take the case of a resistor (Fig.). If we go through a resistor in the same direction as the
current, then there is a fall in potential because current flows from a higher to a lower
potential..
Hence, this voltage fall should be taken -ve. However, if we go in a direction opposite to that of
the
current, then there is a rise in voltage. Hence, this voltage rise should be given a positive sign
Consider the closed path ABCDA in Fig
As we travel around the mesh in the clockwise direction, different voltage drops will have the
following signs :
I1R1 is - ve (fall in potential)
I2R2 is - ve (fall in potential)
I3R3 is + ve (rise in potential)
I4R4 is - ve (fall in potential)
E2 is - ve (fall in potential)
E1 is + ve (rise in potential)
Using Kirchhoff's voltage law, we get
-I1R1 – I2R2 – I3R3 – I4 R4 – E2 + E1 = 0
Or I1R1 + I2R2 – I3R3 + I4R4 = E1 –E2
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Assumed Direction of Current:
In applying Kirchhoff's laws to electrical networks, the direction of current flow may be
assumed either clockwise or anticlockwise. If the assumed direction of current is not the actual
direction, then on solving the question, the current will be found to have a minus sign.
If the answer is positive, then assumed direction is the same as actual direction.
However, the important point is that once a particular direction has been assumed, the same
should be used throughout the solution of the question.
Kirchhoff's laws are applicable both to d.c. and a.c. voltages and currents. However, in
the case of alternating currents and voltages, any e.m.f. of self-inductance or that existing
across a capacitor should be also taken into account.
Voltage Division Rule:
In a series circuit, same current flows through each of the given resistors and the voltage
drop varies directly with its resistance. (V α R)
By using voltage division rule we can find the different voltages across resistors from single
voltage source.
In a series circuit Voltage across resistor is given by
Current Division Rule:
In Parallel circuit, the current divides in all branches (I α 1/R)
To find branch current in parallel circuits is given by
ℎ == += +
=
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G.Purushotham,Asst.Professor, Dept. of EEE @ SVCE TPT. 4
Assumed Direction of Current:
In applying Kirchhoff's laws to electrical networks, the direction of current flow may be
assumed either clockwise or anticlockwise. If the assumed direction of current is not the actual
direction, then on solving the question, the current will be found to have a minus sign.
If the answer is positive, then assumed direction is the same as actual direction.
However, the important point is that once a particular direction has been assumed, the same
should be used throughout the solution of the question.
Kirchhoff's laws are applicable both to d.c. and a.c. voltages and currents. However, in
the case of alternating currents and voltages, any e.m.f. of self-inductance or that existing
across a capacitor should be also taken into account.
Voltage Division Rule:
In a series circuit, same current flows through each of the given resistors and the voltage
drop varies directly with its resistance. (V α R)
By using voltage division rule we can find the different voltages across resistors from single
voltage source.
In a series circuit Voltage across resistor is given by
Current Division Rule:
In Parallel circuit, the current divides in all branches (I α 1/R)
To find branch current in parallel circuits is given by
ℎ == += +
=
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G.Purushotham,Asst.Professor, Dept. of EEE @ SVCE TPT. 4
Assumed Direction of Current:
In applying Kirchhoff's laws to electrical networks, the direction of current flow may be
assumed either clockwise or anticlockwise. If the assumed direction of current is not the actual
direction, then on solving the question, the current will be found to have a minus sign.
If the answer is positive, then assumed direction is the same as actual direction.
However, the important point is that once a particular direction has been assumed, the same
should be used throughout the solution of the question.
Kirchhoff's laws are applicable both to d.c. and a.c. voltages and currents. However, in
the case of alternating currents and voltages, any e.m.f. of self-inductance or that existing
across a capacitor should be also taken into account.
Voltage Division Rule:
In a series circuit, same current flows through each of the given resistors and the voltage
drop varies directly with its resistance. (V α R)
By using voltage division rule we can find the different voltages across resistors from single
voltage source.
In a series circuit Voltage across resistor is given by
Current Division Rule:
In Parallel circuit, the current divides in all branches (I α 1/R)
To find branch current in parallel circuits is given by
ℎ == += +
=
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BASIC NODAL AND MESH ANALYSIS:
If a circuit consists of n nodes and b branches. Then to analyze a circuit using
Ohm’s law: requires n equations.
Nodal analysis: requires n – 1 equation.
Mesh analysis: requires b – (n – 1) equations.
Depending on the circuit analysis we adapt different technique among these three. The
nodal and mesh analysis are based on Kirchhoff’s laws
Nodal Analysis:
It is based on Kirchhoff’s Current Law.
A node is a point in a network, when two or more elements meet.
A voltage is always defined as the potential difference between two points. When we talk
about the voltage at a certain point of a circuit we imply that the measurement is performed
between that point and some other point in the circuit. In most cases that other point is
referred to as ground.
The procedure for analyzing a circuit with the node method:
1. Count a number of nodes of the given network.
2. Choose one node as reference node and write ground symbol at this node
3. Determine voltage variables at each node
4. If there is some voltage source in the middle of couple nodes that aren’t the
reference node, you have to reduce these couple node as one node called supernode.
5. Use KCL at all node and supernode except the reference to write the KCL equations
- The total equations no of equations will be equal to n – 1.
6. Solve the variable system
7. Determine the required value by using the solution of variable system
There are a few general guidelines that we need
to remember as we make the selection of the
reference node.
1. A useful reference node is one which has the
largest number of elements connected to it.
2. A useful reference node is one which is
connected to the maximum number of voltage sources.
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Nodal Analysis with Voltage Sources
• Voltage source between reference node and a non reference node. Set the voltage of
the non reference node to that of the voltage source.
Explanation: For the given Figure 1 for a step by step demonstration of the node method.
1) The nodes are identified and the names are given.
2) One of the identified nodes as the reference node.
The Selection of node A as the reference node is the
best choice
3) Label the voltages at the selected nodes. Figure
shows the circuit with the labeled nodal voltages. The
reference node is assigned voltage 0 Volts indicated by
the ground symbol. The remaining node voltages are labeled V1, VB, VC and V2
4) The problem is completely defined. Once we determine the values for the node
voltages V1,VB, VC and V2. .we wil l be able to complete ly characterize this circuit. So
go on to calculate the node voltages by applying KCL at the designated nodes.
5) For node 1 since the voltage of the voltage source is known we may directly label the
voltage V1
similarly V2 as a result we have reduced the number of unknowns from 4 to 2.
Applying KCL at node B, the node voltage as VB = −=
= −+ + =
Substituting the current equations in above expression we get− + + − =Similarly applying KCL at node C we get− + + − =By solving the above two equations we get unknown node voltages.
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Super Node Analysis:
If any of the branch in the network has a voltage source such that its any terminal is notconnected to reference node or ground,then it is difficult to write KCL equations at this nodesto which this voltage source is connected. One way to overcome this difficulty is to apply thesupernode technique.
In the circuit of Figure voltage source V2 is notconnected to the reference node and thus it is afloating voltage source.
In This method,two adjacent nodes that areconnected by a voltage source are reduced to asingle node and then equations are formed byapplying Kirchoff’s current law as usual.
Due to presence of voltage source Vx in betweennodes 2 and 3it is difficult to find out current. Thistype of networks Super Node Technique is very useful to find the current.
The super node is considered to be a single node and the KCL equations are written.
The difference between the node voltages across the voltage source becomes one
of the equations to solve the network.
We can write the combined equation for node 2 and 3 as below= + + + = 0− + + = 0 _____________(1)The other equation between node 2 and 3 is− = ______________________(2)By solving the above two equations we can find the Unknown voltages.
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Mesh current analysis (or) Loop Current Method:
A mesh is a loop that does not contain any other loops within it.
It is based on Kirchhoff’s Voltage Law.
If a network has a large number of voltage sources, It is useful to use mesh analysis.
Mesh analysis is applicable only for Planar networks. A circuit said to be planar ,if it can bedrawn on a plane surface without crossover.
The procedure for analyzing a circuit with the Mesh Analysis:
1. Use for planar circuits that can be drawn on a plane surface with no wires crossing each other.
2. Count a number of mesh in the given circuit and assign mesh current for each mesh.
3. Determine current variables at each mesh with an arbitrary direction.
4.Indicate the polarities (sign) of the elements that are not indicated with polarities.
1. Use KVL at all mesh to write the KVL equations
no of mesh equations = b – (n-1)
Example:
The above circuit consists two batteries E1,E2
connected with five resistors.
Let the mesh currents for three meshes be I1, I2,& I3
For Mesh-1 : E1 – I1R1 – R4(I1 – I2) = 0
I1[R1 + R4] – I2R4 = E1--------(1)
Mesh-2: -I2R2 – R5[I2-I3] – R4[I2 – I1] = 0
I1R4 – I2(R2+R4+R5) + I3R5 = 0------(2)
Mesh-3 : -I3R3 – E2 – R5[I3 – I2] = 0
I2R5-I3[R3 + R5] = E2 -------(3)
The above three equations can be solved to find the not only three mesh currentsbut branch currents as well.
If any mesh current results in negative magnitude then the assumed current
is opposite to that of the original.
Mesh Analysis with Current Sources
Current source exists only in one mesh.
Current source between two meshes, form ’Supermesh’ by excluding the current
source any element connected in series with it.
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The Super Mesh:
A Supermesh results when two meshes have a (dependent or independent) currentsource in common.
The loop represented with a dotted linerepresents the Supermesh.
While applying KVL to solve this circuit
with mesh analysis, the two loop
equations are expressed in a single
equation.The required second expression will be the difference in loop currents at the
branch with current source.
i.e.
KVL of loop 1 + KVL of loop 2 =0
From the circuit
Supermesh equation is− − − =I1 R1 + I2R2 = VA - VB ……………eq.1.
And the second equation is
I1 - I2 = IX …………eq.2
On solving these two equations we can determine the loop currents and anyother parameters.
Properties of a Supermesh:
The current source in the Supermesh provides the constraint equation necessary to
solve for the mesh currents.
A Supermesh has no current of itsown.
A Supermesh requires the application of both KVL and KCL.
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NETWORK THEOREMS:
Thevenin's Theorem:
“Thevenin's theorem states that any two terminal linear network having a number ofvoltage current sources and resistances can be replaced by a simple equivalent circuitconsisting of a single voltage source in series with a resistance”
where the value of the voltage source is equal to the open circuit voltage across the twoterminals of the network, and resistance is equal to the equivalent resistance measuredbetween the terminals
To find Thevenin’s Voltage:
The current through resistance R2 =Hence voltage appears across the terminals A and B i.e.= =To find Thevenin’s Resistance:= +Thus the Thevenin equivalent resistance RTh is viewed from the open terminals A
and B is given as. As per Thevenin theorem, when resistance RL is connected acrossterminals A and B, the network behaves as a source of voltage VT andinternal resistance RT and this is called Thevenin equivalent circuit. The current throughRL is given by
Thevenin’s equivalent Cicuit= +
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Norton's theorem :
“Norton's theorem, states that any two terminal linear network consisting currentsources, voltage sources and resistances can be replaced by an equivalent circuitconsisting of a current source in parallel with a resistance”.
The value of the current source is the short circuit current between the twoterminals of the network and the resistance is the equivalent resistance measured theterminals of the network
In the example. we have to find out the current through RL by applying Norton theorem.
Step 1: First, we have to remove the resistor RL fromterminals A and B and make the terminals A and Bshort circuited by zero resistance.as shown in fig.b
Step 2: Second, we have to calculate the shortcircuit current or Norton equivalent current IN
through the points A and B.
the equivalent resistance of the network
. = + ( // )= + ( )+
Current supplied from the Source =Norton equivalent current = ℎ ℎ =Step 3:To determine Norton equivalent resistance RN of the network the equivalent resistance asviewed from open terminals A and B is RN
= + ( )= + ( + )
As per Norton theorem, when resistance RL is reconnected across terminals A and B, thenetwork behaves as a source of constant current IN with shunt connected internal resistanceRN and this is Norton equivalent circuit.
Norton Equivalent Circuit
The Current through RL is given as =
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Superposition theorem:
“The superposition theorem states that in any linear network containing two or moresources, the response in any element is equal to the algebraic sum of the responses caused byindividual sources acting alone, while the other sources are non-operative”
while considering the effect of individual sources, other ideal voltage sources and idealcurrent sources in the network are replaced by short circuit and open circuit across theirterminals.
This theorem is valid only for linear systems.
Suppose there are two voltage sources V1 and V2 acting simultaneously on the circuit. becauseof these two voltage sources, say current I flows throughthe resistance R.
i) When source V1 acting alone:
Now replace V2 by short circuit, keeping V1 at its positionand measure current through the resistance, R1. Say it is I1.
As shown in fig.1 = += and is given by = + ( )ii) When source V2 acting alone:
Now replace V1 by short circuit, keeping V2 at its position and measure current through theresistance, R1. Say it is I2. = += and is given by = + ( )Now if we add these two currents, I1 and I2 we will get the current which is equal to the currentwas actually flowing through R, when both voltage sources V1 and V2 were acting on the circuitsimultaneously. That is = +Limitations:
1. Super positions theorem is not valid for Power Calculations2. Super position theorem is not valid for Non-linear circuits.
Figure 1
Figure 2
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G.Purushotham,Asst.Professor, Dept. of EEE @ SVCE TPT. 13
Maximum Power Transfer Theorem:
“The maximum Power Transfer Theorem states that maximum power is delivered from asource to a load when the load resistance is equal to the source resistance.”
In circuits basically consist of sources, supplying voltage,current, or power to the load; for example, a radio speakersystem, or a microphone supplying the input signals tovoltage pre-amplifiers. Sometimes it is necessary totransfer maximum voltage, current or power from thesource to the load.
It is a fact that more voltage is delivered to the load whenthe load resistance is high as compared to the resistance ofthe source.
On the other hand, maximum current is transferred to the load when the load resistanceis small compared to the source resistance.
From the above fig
= +power delivered to load resistance =
= + .To find the maximum power, differentiate the above expression with respect to resistance RL
and equate it to zero. Thus,
= ( + )= [( + ) − ( )( + )]( + )( + ) − ( )( + ) =+ + − − == = ∴ =
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Reciprocity Theorem :
Reciprocity Theorem states that in any linear, bilateral, passive network containing singlesource (voltage or current), the ratio of excitation (V or I) to the response remains unchangedeven after interchanging the positions of excitation and response. It is also referred as “Principleof Reciprocity”
Consider example, but with only one source. Determine the current in the 160 Ω branch.Now replace the 160 Ω resistor with the source in series with it and after short- circuitthe source at the original location, find the current flowing at the original sourcelocation. Show that it verifies the Reciprocality theorem.
Solution
It is seen that the identical current has appeared verifying the reciprocity theorem.Theadvantage of the theorem is when a circuit has already been analysed for one solution, it maybe possible to find a corresponding solution without further work
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Tellegen theorem :
This theorem has been introduced in the year of 1952 by Dutch Electrical EngineerBernard D.H. Tellegen. This is a very useful theorem in network analysis.
Tellegen's theorem is valid for any lumped network which may be linear or non linear,passive or active, time-varying or time-invarient. All branch currents and voltages in thatnetwork must satisfy Kirchhoff's laws.
“This theorem states that in an arbitrary lumped network, the algebraic sum of thepowers in all branches at any instant is zero”.
(or)
“the algebraic sum of the powers delivered by all sources is equal to the algebraic sum ofthe powers absorbed by all elements”.
Example: Verifying Tellegens theorem for the circuit.
-100+50+50+0-0=0 hence verified.
Millman’s Theorem:
Millman’s theorem states that in any network, if the voltage sources V1, V2…. Vn in series withinternal resistances R1,R2…..Rn,respectively,are in parallel, then these sourcesmay be replaced by a single voltage source V’ in series with Req.
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G.Purushotham,Asst.Professor, Dept. of EEE @ SVCE TPT. 15
Tellegen theorem :
This theorem has been introduced in the year of 1952 by Dutch Electrical EngineerBernard D.H. Tellegen. This is a very useful theorem in network analysis.
Tellegen's theorem is valid for any lumped network which may be linear or non linear,passive or active, time-varying or time-invarient. All branch currents and voltages in thatnetwork must satisfy Kirchhoff's laws.
“This theorem states that in an arbitrary lumped network, the algebraic sum of thepowers in all branches at any instant is zero”.
(or)
“the algebraic sum of the powers delivered by all sources is equal to the algebraic sum ofthe powers absorbed by all elements”.
Example: Verifying Tellegens theorem for the circuit.
-100+50+50+0-0=0 hence verified.
Millman’s Theorem:
Millman’s theorem states that in any network, if the voltage sources V1, V2…. Vn in series withinternal resistances R1,R2…..Rn,respectively,are in parallel, then these sourcesmay be replaced by a single voltage source V’ in series with Req.
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G.Purushotham,Asst.Professor, Dept. of EEE @ SVCE TPT. 15
Tellegen theorem :
This theorem has been introduced in the year of 1952 by Dutch Electrical EngineerBernard D.H. Tellegen. This is a very useful theorem in network analysis.
Tellegen's theorem is valid for any lumped network which may be linear or non linear,passive or active, time-varying or time-invarient. All branch currents and voltages in thatnetwork must satisfy Kirchhoff's laws.
“This theorem states that in an arbitrary lumped network, the algebraic sum of thepowers in all branches at any instant is zero”.
(or)
“the algebraic sum of the powers delivered by all sources is equal to the algebraic sum ofthe powers absorbed by all elements”.
Example: Verifying Tellegens theorem for the circuit.
-100+50+50+0-0=0 hence verified.
Millman’s Theorem:
Millman’s theorem states that in any network, if the voltage sources V1, V2…. Vn in series withinternal resistances R1,R2…..Rn,respectively,are in parallel, then these sourcesmay be replaced by a single voltage source V’ in series with Req.
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= + +. … … .+ +⋯+Here Gn is the conductance of the nth branch,
= 1+ +⋯+Miller Theorem:
The Miller theorem states that in a linear circuit,if there exists a branch with impedanceZ,connecting two nodes with nodal voltages V1 and V2, we can replace this branch by two
branches connecting the corresponding nodes to ground by impedances respectively Z1 and Z2.
Where = ( )= ( − 1) ( . − )= ( − 1)
=
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