Kirchoff Thevenin Norton Max Power transfer
Kirchhoff’s Laws KVL ,KCL
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Kirchoffs Circuit Law
We saw in the Resistors tutorial that a single equivalent resistance, ( RT ) can be found when
two or more resistors are connected together in either series, parallel or combinations of both,
and that these circuits obey Ohm’s Law.
However, sometimes in complex circuits such as bridge or T networks, we can not simply use
Ohm’s Law alone to find the voltages or currents circulating within the circuit. For these
types of calculations we need certain rules which allow us to obtain the circuit equations and
for this we can use Kirchoffs Circuit Law.
In 1845, a German physicist, Gustav Kirchoff developed a pair or set of rules or laws which
deal with the conservation of current and energy within Electrical Circuits. These two rules
are commonly known as: Kirchoffs Circuit Laws with one of Kirchoffs laws dealing with the
current flowing around a closed circuit, Kirchoffs Current Law, (KCL) while the other law
deals with the voltage sources present in a closed circuit, Kirchoffs Voltage Law, (KVL).
Kirchoffs First Law – The Current Law, (KCL)
Kirchoffs Current Law or KCL, states that the “total current or charge entering a junction
or node is exactly equal to the charge leaving the node as it has no other place to go except
to leave, as no charge is lost within the node“. In other words the algebraic sum of ALL the
currents entering and leaving a node must be equal to zero, I(exiting) + I(entering) = 0. This idea by
Kirchoff is commonly known as the Conservation of Charge.
Kirchoffs Current Law
Here, the 3 currents entering the node, I1, I2, I3 are all positive in value and the 2 currents
leaving the node, I4 and I5 are negative in value. Then this means we can also rewrite the
equation as;
I1 + I2 + I3 – I4 – I5 = 0
The term Node in an electrical circuit generally refers to a connection or junction of two or
more current carrying paths or elements such as cables and components. Also for current to
flow either in or out of a node a closed circuit path must exist. We can use Kirchoff’s current
law when analysing parallel circuits.
Kirchoffs Second Law – The Voltage Law, (KVL)
Kirchoffs Voltage Law or KVL, states that “in any closed loop network, the total voltage
around the loop is equal to the sum of all the voltage drops within the same loop” which is
also equal to zero. In other words the algebraic sum of all voltages within the loop must be
equal to zero. This idea by Kirchoff is known as the Conservation of Energy.
Kirchoffs Voltage Law
Starting at any point in the loop continue in the same direction noting the direction of all the
voltage drops, either positive or negative, and returning back to the same starting point. It is
important to maintain the same direction either clockwise or anti-clockwise or the final
voltage sum will not be equal to zero. We can use Kirchoff’s voltage law when analysing
series circuits.
When analysing either DC circuits or AC circuits using Kirchoffs Circuit Laws a number of
definitions and terminologies are used to describe the parts of the circuit being analysed such
as: node, paths, branches, loops and meshes. These terms are used frequently in circuit
analysis so it is important to understand them.
Common DC Circuit Theory Terms:
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• Circuit – a circuit is a closed loop conducting path in which an electrical current flows.
• Path – a single line of connecting elements or sources.
• Node – a node is a junction, connection or terminal within a circuit were two or more
circuit elements are connected or joined together giving a connection point between two or
more branches. A node is indicated by a dot.
• Branch – a branch is a single or group of components such as resistors or a source which
are connected between two nodes.
• Loop – a loop is a simple closed path in a circuit in which no circuit element or node is
encountered more than once.
• Mesh – a mesh is a single open loop that does not have a closed path. There are no
components inside a mesh.
Note that:
Components are said to be connected in Series if the same current flows through
component.
Components are said to be connected in Parallel if the same voltage is applied across them.
A Typical DC Circuit
Kirchoffs Circuit Law Example No1
Find the current flowing in the 40Ω Resistor, R3
The circuit has 3 branches, 2 nodes (A and B) and 2 independent loops.
Using Kirchoffs Current Law, KCL the equations are given as;
At node A :
I1 + I2 = I3
At node B :
I3 = I1 + I2
Using Kirchoffs Voltage Law, KVL the equations are given as;
Loop 1 is given as :
10 = R1 x I1 + R3 x I3 = 10I1 + 40I3
Loop 2 is given as :
20 = R2 x I2 + R3 x I3 = 20I2 + 40I3
Loop 3 is given as :
10 – 20 = 10I1 – 20I2
As I3 is the sum of I1 + I2 we can rewrite the equations as;
Eq. No 1 :
10 = 10I1 + 40(I1 + I2) = 50I1 + 40I2
Eq. No 2 :
20 = 20I2 + 40(I1 + I2) = 40I1 + 60I2
We now have two “Simultaneous Equations” that can be reduced to give us the values of I1
and I2
Substitution of I1 in terms of I2 gives us the value of I1 as -0.143 Amps
Substitution of I2 in terms of I1 gives us the value of I2 as +0.429 Amps
As :
I3 = I1 + I2
The current flowing in resistor R3 is given as :
-0.143 + 0.429 = 0.286 Amps
and the voltage across the resistor R3 is given as :
0.286 x 40 = 11.44 volts
The negative sign for I1 means that the direction of current flow initially chosen was wrong,
but never the less still valid. In fact, the 20v battery is charging the 10v battery.
Application of Kirchoffs Circuit Laws
These two laws enable the Currents and Voltages in a circuit to be found, ie, the circuit is said
to be “Analysed”, and the basic procedure for using Kirchoff’s Circuit Laws is as follows:
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1. Assume all voltages and resistances are given. ( If not label them V1, V2,… R1, R2, etc. )
2. Label each branch with a branch current. ( I1, I2, I3 etc. )
3. Find Kirchoff’s first law equations for each node.
4. Find Kirchoff’s second law equations for each of the independent loops of the circuit.
5. Use Linear simultaneous equations as required to find the unknown currents.
As well as using Kirchoffs Circuit Law to calculate the various voltages and currents
circulating around a linear circuit, we can also use loop analysis to calculate the currents in
each independent loop which helps to reduce the amount of mathematics required by using
just Kirchoff's laws. In the next tutorial about DC Theory we will look at Mesh Current
Analysis to do just that.
Mesh Current Analysis
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Circuit Analysis
In the previous tutorial we saw that complex circuits such as bridge or T-networks can be
solved using Kirchoff’s Circuit Laws. While Kirchoff´s Laws give us the basic method for
analysing any complex electrical circuit, there are different ways of improving upon this
method by using Mesh Current Analysis or Nodal Voltage Analysis that results in a
lessening of the math’s involved and when large networks are involved this reduction in
maths can be a big advantage.
For example, consider the Electrical Circuit example from the previous section.
Mesh Current Analysis Circuit
One simple method of reducing the amount of math’s involved is to analyse the circuit using
Kirchoff’s Current Law equations to determine the currents, I1 and I2 flowing in the two
resistors. Then there is no need to calculate the current I3 as its just the sum of I1 and I2. So
Kirchoff’s second voltage law simply becomes:
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Equation No 1 : 10 = 50I1 + 40I2
Equation No 2 : 20 = 40I1 + 60I2
therefore, one line of math’s calculation have been saved.
Mesh Current Analysis
An easier method of solving the above circuit is by using Mesh Current Analysis or Loop
Analysis which is also sometimes called Maxwell´s Circulating Currents method. Instead
of labelling the branch currents we need to label each “closed loop” with a circulating
current.
As a general rule of thumb, only label inside loops in a clockwise direction with circulating
currents as the aim is to cover all the elements of the circuit at least once. Any required
branch current may be found from the appropriate loop or mesh currents as before using
Kirchoff´s method.
For example: :
i1 = I1 , i2 = -I2 and I3 = I1 – I2
We now write Kirchoff’s voltage law equation in the same way as before to solve them but
the advantage of this method is that it ensures that the information obtained from the circuit
equations is the minimum required to solve the circuit as the information is more general and
can easily be put into a matrix form.
For example, consider the circuit from the previous section.
These equations can be solved quite quickly by using a single mesh impedance matrix Z.
Each element ON the principal diagonal will be “positive” and is the total impedance of each
mesh. Where as, each element OFF the principal diagonal will either be “zero” or “negative”
and represents the circuit element connecting all the appropriate meshes.
First we need to understand that when dealing with matrices, for the division of two matrices
it is the same as multiplying one matrix by the inverse of the other as shown.
having found the inverse of R, as V/R is the same as V x R-1, we can now use it to find the
two circulating currents.
Where:
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[ V ] gives the total battery voltage for loop 1 and then loop 2
[ I ] states the names of the loop currents which we are trying to find
[ R ] is the resistance matrix
[ R-1 ] is the inverse of the [ R ] matrix
and this gives I1 as -0.143 Amps and I2 as -0.429 Amps
As :
I3 = I1 – I2
The combined current of I3 is therefore given as : -0.143 – (-0.429) = 0.286 Amps
which is the same value of 0.286 amps, we found using Kirchoff´s circuit law in the previous
tutorial.
Mesh Current Analysis Summary.
This “look-see” method of circuit analysis is probably the best of all the circuit analysis
methods
with the basic procedure for solving Mesh Current Analysis equations is as follows:
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1. Label all the internal loops with circulating currents. (I1, I2, …IL etc)
2. Write the [ L x 1 ] column matrix [ V ] giving the sum of all voltage sources in each loop.
3. Write the [ L x L ] matrix, [ R ] for all the resistances in the circuit as follows;
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o
o
o
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R11 = the total resistance in the first loop.
Rnn = the total resistance in the Nth loop.
RJK = the resistance which directly joins loop J to Loop K.
4. Write the matrix or vector equation [V] = [R] x [I] where [I] is the list of currents to be
found.
As well as using Mesh Current Analysis, we can also use node analysis to calculate the
voltages around the loops, again reducing the amount of mathematics required using just
Kirchoff’s laws. In the next tutorial about DC Theory we will look at Nodal Voltage Analysis
to do just that.
Nodal Voltage Analysis
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Nodal Voltage Analysis
As well as using Mesh Analysis to solve the currents flowing around complex circuits it is
also possible to use nodal analysis methods too. Nodal Voltage Analysis complements the
previous mesh analysis in that it is equally powerful and based on the same concepts of
matrix analysis. As its name implies, Nodal Voltage Analysis uses the “Nodal” equations of
Kirchoff’s first law to find the voltage potentials around the circuit.
So by adding together all these Nodal Voltages the net result will be equal to zero. Then, if
there are “n” nodes in the circuit there will be “n-1″ independent nodal equations and these
alone are sufficient to describe and hence solve the circuit.
At each node point write down Kirchoff’s first law equation, that is: “the currents entering a
node are exactly equal in value to the currents leaving the node” then express each current in
terms of the voltage across the branch. For “n” nodes, one node will be used as the reference
node and all the other voltages will be referenced or measured with respect to this common
node.
For example, consider the circuit from the previous section.
Nodal Voltage Analysis Circuit
In the above circuit, node D is chosen as the reference node and the other three nodes are
assumed to have voltages, Va, Vb and Vc with respect to node D. For example;
As Va = 10v and Vc = 20v , Vb can be easily found by:
again is the same value of 0.286 amps, we found using Kirchoff’s Circuit Law in the previous
tutorial.
From both Mesh and Nodal Analysis methods we have looked at so far, this is the simplest
method of
solving this particular circuit. Generally, nodal voltage analysis is more appropriate when
there are a larger number of current sources around. The network is then defined as: [ I ] = [
Y ] [ V ] where [ I ] are the driving current sources, [ V ] are the nodal voltages to be found
and [ Y ] is the admittance matrix of the network which operates on [ V ] to give [ I ].
Nodal Voltage Analysis Summary.
The basic procedure for solving Nodal Analysis equations is as follows:
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1. Write down the current vectors, assuming currents into a node are positive. ie, a (N x 1)
matrices for “N” independent nodes.
2. Write the admittance matrix [Y] of the network where:
o
Y11 = the total admittance of the first node.
o
Y22 = the total admittance of the second node.
o
RJK = the total admittance joining node J to node K.
3. For a network with “N” independent nodes, [Y] will be an (N x N) matrix and that Ynn will
be
positive and Yjk will be negative or zero value.
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4. The voltage vector will be (N x L) and will list the “N” voltages to be found.
We have now seen that a number of theorems exist that simplify the analysis of linear
circuits. In the next tutorial we will look at Thevenins Theorem which allows a network
consisting of linear resistors and sources to be represented by an equivalent circuit with a
single voltage source and a series resistance.
Thevenin’s Theorem
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Thevenin’s Theorem
In the previous three tutorials we have looked at solving complex electrical circuits using
Kirchoff’s Circuit Laws, Mesh Analysis and finally Nodal Analysis but there are many more
“Circuit Analysis Theorems” available to choose from which can calculate the currents and
voltages at any point in a circuit. In this tutorial we will look at one of the more common
circuit analysis theorems (next to Kirchoff´s) that has been developed, Thevenin’s Theorem.
Thevenin’s Theorem states that “Any linear circuit containing several voltages and
resistances can be replaced by just one single voltage in series with a single resistance
connected across the load“. In other words, it is possible to simplify any electrical circuit, no
matter how complex, to an equivalent two-terminal circuit with just a single constant voltage
source in series with a resistance (or impedance) connected to a load as shown below.
Thevenin’s Theorem is especially useful in the Circuit Analysis of power or battery systems
and other interconnected resistive circuits where it will have an effect on the adjoining part of
the circuit.
Thevenin’s equivalent circuit.
As far as the load resistor RL is concerned, any complex “one-port” network consisting of
multiple resistive circuit elements and energy sources can be replaced by one single
equivalent resistance Rs and one single equivalent voltage Vs. Rs is the source resistance
value looking back into the circuit and Vs is the open circuit voltage at the terminals.
For example, consider the circuit from the previous section.
Firstly, to analyse the circuit we have to remove the centre 40Ω load resistor connected
across the terminals A-B, and remove any internal resistance associated with the voltage
source(s). This is done by shorting out all the voltage sources connected to the circuit, that is
v = 0, or open circuit any connected current sources making i = 0. The reason for this is that
we want to have an ideal voltage source or an ideal current source for the circuit analysis.
The value of the equivalent resistance, Rs is found by calculating the total resistance looking
back from the terminals A and B with all the voltage sources shorted. We then get the
following circuit.
Find the Equivalent Resistance (Rs)
The voltage Vs is defined as the total voltage across the terminals A and B when there is an
open circuit between them. That is without the load resistor RL connected.
Find the Equivalent Voltage (Vs)
We now need to reconnect the two voltages back into the circuit, and as VS = VAB the
current flowing around the loop is calculated as:
This current of 0.33 amperes (330mA) is common to both resistors so the voltage drop across
the 20Ω resistor or the 10Ω resistor can be calculated as:
VAB = 20 – (20Ω x 0.33amps) = 13.33 volts.
or
VAB = 10 + (10Ω x 0.33amps) = 13.33 volts, the same.
Then the Thevenin’s Equivalent circuit would consist or a series resistance of 6.67Ω’s and a
voltage source of 13.33v. With the 40Ω resistor connected back into the circuit we get:
and from this the current flowing around the circuit is given as:
which again, is the same value of 0.286 amps, we found using Kirchoff´s circuit law in the
previous circuit analysis tutorial.
Thevenin’s theorem can be used as another type of Circuit Analysis method and is
particularly useful in the analysis of complicated circuits consisting of one or more voltage or
current source and resistors that are arranged in the usual parallel and series connections.
While Thevenin’s circuit theorem can be described mathematically in terms of current and
voltage, it is not as powerful as Mesh or Nodal analysis in larger networks because the use of
Mesh or Nodal analysis is usually necessary in any Thevenin exercise, so it might as well be
used from the start. However, Thevenin’s equivalent circuits of Transistors, Voltage
Sources such as batteries etc, are very useful in circuit design.
Thevenin’s Theorem Summary
We have seen here that Thevenin’s theorem is another type of circuit analysis tool that can be
used to reduce any complicated electrical network into a simple circuit consisting of a single
voltage source, Vs in series with a single resistor, Rs. When looking back from terminals A
and B, this single circuit behaves in exactly the same way electrically as the complex circuit
it replaces. That is the i-v relationships at terminals A-B are identical.
The basic procedure for solving a circuit using Thevenin’s Theorem is as follows:
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1. Remove the load resistor RL or component concerned.
2. Find RS by shorting all voltage sources or by open circuiting all the current sources.
3. Find VS by the usual circuit analysis methods.
4. Find the current flowing through the load resistor RL.
In the next tutorial we will look at Nortons Theorem which allows a network consisting of
linear resistors and sources to be represented by an equivalent circuit with a single current
source in parallel with a single source resistance.
Nortons Theorem
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Nortons Theorem
In some ways Norton’s Theorem can be thought of as the opposite to “Thevenins Theorem”,
in that Thevenin reduces his circuit down to a single resistance in series with a single voltage.
Norton on the other hand reduces his circuit down to a single resistance in parallel with a
constant current source.
Nortons Theorem states that “Any linear circuit containing several energy sources and
resistances can be replaced by a single Constant Current generator in parallel with a Single
Resistor“.
As far as the load resistance, RL is concerned this single resistance, RS is the value of the
resistance looking back into the network with all the current sources open circuited and IS is
the short circuit current at the output terminals as shown below.
Nortons equivalent circuit.
The value of this “constant current” is one which would flow if the two output terminals
where shorted together while the source resistance would be measured looking back into the
terminals, (the same as Thevenin).
For example, consider our now familiar circuit from the previous section.
To find the Nortons Equivalent of the above circuit we firstly have to remove the centre 40Ω
load resistor and short out the terminals A and B to give us the following circuit.
When the terminals A and B are shorted together the two resistors are connected in parallel
across their two respective voltage sources and the currents flowing through each resistor as
well as the total short circuit current can now be calculated as:
with A-B Shorted Out
If we short-out the two voltage sources and open circuit terminals A and B, the two resistors
are now effectively connected together in parallel. The value of the internal resistor Rs is
found by calculating the total resistance at the terminals A and B giving us the following
circuit.
Find the Equivalent Resistance (Rs)
Having found both the short circuit current, Is and equivalent internal resistance, Rs this then
gives us the following Nortons equivalent circuit.
Nortons equivalent circuit.
Ok, so far so good, but we now have to solve with the original 40Ω load resistor connected
across terminals A and B as shown below.
Again, the two resistors are connected in parallel across the terminals A and B which gives us
a total resistance of:
The voltage across the terminals A and B with the load resistor connected is given as:
Then the current flowing in the 40Ω load resistor can be found as:
which again, is the same value of 0.286 amps, we found using Kirchoff´s circuit law in the
previous tutorials.
Nortons Theorem Summary
The basic procedure for solving a circuit using Nortons Theorem is as follows:
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1. Remove the load resistor RL or component concerned.
2. Find RS by shorting all voltage sources or by open circuiting all the current sources.
3. Find IS by placing a shorting link on the output terminals A and B.
4. Find the current flowing through the load resistor RL.
In a circuit, power supplied to the load is at its maximum when the load resistance is equal to
the source resistance. In the next tutorial we will look at Maximum Power Transfer. The
application of the maximum power transfer theorem can be applied to either simple and
complicated linear circuits having a variable load and is used to find the load resistance that
leads to transfer of maximum power to the load.
Max Power Transfer
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Maximum Power Transfer
We have seen in the previous tutorials that any complex circuit or network can be replaced by
a single energy source in series with a single internal source resistance, RS. Generally, this
source resistance or even impedance if inductors or capacitors are involved is of a fixed value
in Ohm´s.
However, when we connect a load resistance, RL across the output terminals of the power
source, the impedance of the load will vary from an open-circuit state to a short-circuit state
resulting in the power being absorbed by the load becoming dependent on the impedance of
the actual power source. Then for the load resistance to absorb the maximum power possible
it has to be “Matched” to the impedance of the power source and this forms the basis of
Maximum Power Transfer.
The Maximum Power Transfer Theorem is another useful Circuit Analysis method to
ensure that the maximum amount of power will be dissipated in the load resistance when the
value of the load resistance is exactly equal to the resistance of the power source. The
relationship between the load impedance and the internal impedance of the energy source will
give the power in the load. Consider the circuit below.
Thevenins Equivalent Circuit.
In our Thevenin equivalent circuit above, the maximum power transfer theorem states that
“the maximum amount of power will be dissipated in the load resistance if it is equal in value
to the Thevenin or Norton source resistance of the network supplying the power“.
In other words, the load resistance resulting in greatest power dissipation must be equal in
value to the equivalent Thevenin source resistance, then RL = RS but if the load resistance is
lower or higher in value than the Thevenin source resistance of the network, its dissipated
power will be less than maximum.
For example, find the value of the load resistance, RL that will give the maximum power
transfer in the following circuit.
Maximum Power Transfer Example No1.
Where:
RS = 25Ω
RL is variable between 0 – 100Ω
VS = 100v
Then by using the following Ohm’s Law equations:
We can now complete the following table to determine the current and power in the circuit
for different values of load resistance.
Table of Current against Power
RL (Ω)
I (amps)
P (watts)
RL (Ω)
I (amps)
P (watts)
0
4.0
0
25
2.0
100
5
3.3
55
30
1.8
97
10
2.8
78
40
1.5
94
15
2.5
93
60
1.2
83
20
2.2
97
100
0.8
64
Using the data from the table above, we can plot a graph of load resistance, RL against power,
P for different values of load resistance. Also notice that power is zero for an open-circuit
(zero current condition) and also for a short-circuit (zero voltage condition).
Graph of Power against Load Resistance
From the above table and graph we can see that the Maximum Power Transfer occurs in the
load when the load resistance, RL is equal in value to the source resistance, RS that is:
RS = RL = 25Ω. This is called a “matched condition” and as a general rule, maximum power
is transferred from an active device such as a power supply or battery to an external device
when the impedance of the external device exactly matches the impedance of the source.
One good example of impedance matching is between an audio amplifier and a loudspeaker.
The output impedance, ZOUT of the amplifier may be given as between 4Ω and 8Ω, while the
nominal input impedance, ZIN of the loudspeaker may be given as 8Ω only.
Then if the 8Ω speaker is attached to the amplifiers output, the amplifier will see the speaker
as an 8Ω load. Connecting two 8Ω speakers in parallel is equivalent to the amplifier driving
one 4Ω speaker and both configurations are within the output specifications of the amplifier.
Improper impedance matching can lead to excessive power loss and heat dissipation. But how
could you impedance match an amplifier and loudspeaker which have very different
impedances. Well, there are loudspeaker impedance matching transformers available that can
change impedances from 4Ω to 8Ω, or to 16Ω’s to allow impedance matching of many
loudspeakers connected together in various combinations such as in PA (public address)
systems.
Transformer Impedance Matching
One very useful application of impedance matching in order to provide maximum power
transfer between the source and the load is in the output stages of amplifier circuits. Signal
transformers are used to match the loudspeakers higher or lower impedance value to the
amplifiers output impedance to obtain maximum sound power output. These audio signal
transformers are called “matching transformers” and couple the load to the amplifiers output
as shown below.
Transformer Impedance Matching
The maximum power transfer can be obtained even if the output impedance is not the same as
the load impedance. This can be done using a suitable “turns ratio” on the transformer with
the corresponding ratio of load impedance, ZLOAD to output impedance, ZOUT matches that of
the ratio of the transformers primary turns to secondary turns as a resistance on one side of
the transformer becomes a different value on the other.
If the load impedance, ZLOAD is purely resistive and the source impedance is purely resistive,
ZOUT then the equation for finding the maximum power transfer is given as:
Where: NP is the number of primary turns and NS the number of secondary turns on the
transformer. Then by varying the value of the transformers turns ratio the output impedance
can be “matched” to the source impedance to achieve maximum power transfer. For example,
Maximum Power Transfer Example No2.
If an 8Ω loudspeaker is to be connected to an amplifier with an output impedance of 1000Ω,
calculate the turns ratio of the matching transformer required to provide maximum power
transfer of the audio signal. Assume the amplifier source impedance is Z1, the load
impedance is Z2 with the turns ratio given as N.
Generally, small transformers used in low power audio amplifiers are usually regarded as
ideal so any losses can be ignored.
In the next tutorial about DC Theory we will look at Star Delta Transformation which allows
us to convert balanced star connected circuits into equivalent delta and vice versa.
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Star Delta Transformation
Star Delta Transformations allow us to convert impedances connected together from one
type of connection to another. We can now solve simple series, parallel or bridge type
resistive networks using Kirchoff´s Circuit Laws, mesh current analysis or nodal voltage
analysis techniques but in a balanced 3-phase circuit we can use different mathematical
techniques to simplify the analysis of the circuit and thereby reduce the amount of math’s
involved which in itself is a good thing.
Standard 3-phase circuits or networks take on two major forms with names that represent the
way in which the resistances are connected, a Star connected network which has the symbol
of the letter, Υ (wye) and a Delta connected network which has the symbol of a triangle, Δ
(delta).
If a 3-phase, 3-wire supply or even a 3-phase load is connected in one type of configuration,
it can be easily transformed or changed it into an equivalent configuration of the other type
by using either the Star Delta Transformation or Delta Star Transformation process.
A resistive network consisting of three impedances can be connected together to form a T or
“Tee” configuration but the network can also be redrawn to form a Star or Υ type network as
shown below.
T-connected and Equivalent Star Network
As we have already seen, we can redraw the T resistor network above to produce an
electrically equivalent Star or Υ type network. But we can also convert a Pi or π type resistor
network into an electrically equivalent Delta or Δ type network as shown below.
Pi-connected and Equivalent Delta Network.
Having now defined exactly what is a Star and Delta connected network it is possible to
transform the Υ into an equivalent Δ circuit and also to convert a Δ into an equivalent Υ
circuit using a the transformation process. This process allows us to produce a mathematical
relationship between the various resistors giving us a Star Delta Transformation as well as
a Delta Star Transformation.
These Circuit Transformations allow us to change the three connected resistances (or
impedances) by their equivalents measured between the terminals 1-2, 1-3 or 2-3 for either a
star or delta connected circuit. However, the resulting networks are only equivalent for
voltages and currents external to the star or delta networks, as internally the voltages and
currents are different but each network will consume the same amount of power and have the
same power factor to each other.
Delta Star Transformation
To convert a delta network to an equivalent star network we need to derive a transformation
formula for equating the various resistors to each other between the various terminals.
Consider the circuit below.
Delta to Star Network.
Compare the resistances between terminals 1 and 2.
Resistance between the terminals 2 and 3.
Resistance between the terminals 1 and 3.
This now gives us three equations and taking equation 3 from equation 2 gives:
Then, re-writing Equation 1 will give us:
Adding together equation 1 and the result above of equation 3 minus equation 2 gives:
From which gives us the final equation for resistor P as:
Then to summarize a little about the above maths, we can now say that resistor P in a Star
network can be found as Equation 1 plus (Equation 3 minus Equation 2) or Eq1 + (Eq3 –
Eq2).
Similarly, to find resistor Q in a star network, is equation 2 plus the result of equation 1
minus equation 3 or Eq2 + (Eq1 – Eq3) and this gives us the transformation of Q as:
and again, to find resistor R in a Star network, is equation 3 plus the result of equation 2
minus equation 1 or Eq3 + (Eq2 – Eq1) and this gives us the transformation of R as:
When converting a delta network into a star network the denominators of all of the
transformation formulas are the same: A + B + C, and which is the sum of ALL the delta
resistances. Then to convert any delta connected network to an equivalent star network we
can summarized the above transformation equations as:
Delta to Star Transformations Equations
If the three resistors in the delta network are all equal in value then the resultant resistors in
the equivalent star network will be equal to one third the value of the delta resistors, giving
each branch in the star network as: RSTAR = 1/3RDELTA
Delta – Star Example No1
Convert the following Delta Resistive Network into an equivalent Star Network.
Star Delta Transformation
Star Delta transformation is simply the reverse of above. We have seen that when converting
from a delta network to an equivalent star network that the resistor connected to one terminal
is the product of the two delta resistances connected to the same terminal, for example
resistor P is the product of resistors A and B connected to terminal 1.
By rewriting the previous formulas a little we can also find the transformation formulas for
converting a resistive star network to an equivalent delta network giving us a way of
producing a star delta transformation as shown below.
Star to Delta Transformation
The value of the resistor on any one side of the delta, Δ network is the sum of all the twoproduct combinations of resistors in the star network divide by the star resistor located
“directly opposite” the delta resistor being found. For example, resistor A is given as:
with respect to terminal 3 and resistor B is given as:
with respect to terminal 2 with resistor C given as:
with respect to terminal 1.
By dividing out each equation by the value of the denominator we end up with three separate
transformation formulas that can be used to convert any Delta resistive network into an
equivalent star network as given below.
Star Delta Transformation Equations
One final point about converting a star resistive network to an equivalent delta network. If all
the resistors in the star network are all equal in value then the resultant resistors in the
equivalent delta network will be three times the value of the star resistors and equal,
giving: RDELTA = 3RSTAR
Star – Delta Example No2
Convert the following Star Resistive Network into an equivalent Delta Network.
Both Star Delta Transformation and Delta Star Transformation allows us to convert one
type of circuit connection into another type in order for us to easily analyse the circuit. These
transformation techniques can be used to good effect for either star or delta circuits
containing resistances or impedances.