Basics Of Electrical Engineering

Common Terms used in Circuit Theory

• A circuit is a closed conducting path through which an electrical current either flows or is intended to flow. A circuit consists of active and passive elements.
• Parameters are the various elements of an electrical circuit (for example, resistance, capacitance, and inductance).
• Linear circuit – a circuit in which the parameters are constant with time, do not change with voltage or current, and obey Ohm’s law. In a non-linear circuit the parameters change with voltage and current.
• A passive network is a one which contains no source of EMF.
• An active network is a one which contains one or more sources of EMF.
• A bilateral circuit is one whose properties or characteristics are same in either direction of current. Example: the usual transmission line is bilateral.
• A unilateral circuit is that circuit in which properties or characteristics change with the direction of operation. Example: a diode rectifier can rectify only in one direction.
• A Node is a point in a circuit where two or more circuit elements are connected together.
• Branch is a part of a network which lies between two nodes.
• Loop is a closed path in a circuit in which no element or node is encountered more than once.
• Mesh is a loop that contains no other loop within it.

The library uses the symbol font for some of the notation and formulae. If the symbols for the letters ‘alpha beta delta’ do not appear here [a b d] then the symbol font needs to be installed before all notation and formulae will be displayed correctly.

E    G     I    R    P

voltage source conductance current resistance power

[volts, V] [siemens, S] [amps, A] [ohms, W] [watts]

V X Y Z

voltage drop reactance admittance impedance

[volts, V] [ohms, W] [siemens, S] [ohms, W]

  

 Various Laws and Theorems:-

1	Ohm’s Law
2	Kirchhoff’s Laws
3	Thévenin’s Theorem
4	Norton’s Theorem
5	Thévenin and Norton Equivalence
6	Superposition Theorem
7	Reciprocity Theorem
8	Compensation Theorem
9	Millman’s Theorem
10	Joule’s Law
11	Maximum Power Transfer Theorem
12	Star-Delta Transformation
13	Delta-Star Transformation

1. Ohm's Law

How voltage, current, and resistance relate:-

An electric circuit is formed when a conductive path is created to allow free electrons to continuously move. This continuous movement of free electrons through the conductors of a circuit is called a current, and it is often referred to in terms of “flow,” just like the flow of a liquid through a hollow pipe. The force motivating electrons to “flow” in a circuit is called voltage. Voltage is a specific measure of potential energy that is always relative between two points. When we speak of a certain amount of voltage being present in a circuit, we are referring to the measurement of how much potential energy exists to move electrons from one particular point in that circuit to another particular point. Without reference to two particular points, the term “voltage” has no meaning.

Free electrons tend to move through conductors with some degree of friction, or opposition to motion. This opposition to motion is more properly called resistance. The amount of current in a circuit depends on the amount of voltage available to motivate the electrons, and also the amount of resistance in the circuit to oppose electron flow. Just like voltage, resistance is a quantity relative between two points.
For this reason, the quantities of voltage and resistance are often stated as being “between” or “across” two points in a circuit.

To be able to make meaningful statements about these quantities in circuits, we need to be able to describe their quantities in the same way that we might quantify mass, temperature, volume, length, or any other kind of physical quantity. For mass we might use the units of “kilogram” or “gram.” For temperature we might use degrees Fahrenheit or degrees Celsius.

Here are the standard units of measurement for electrical current, voltage, and resistance:

Quantity	Symbol	Units of Measurement	Unit Abbreviation
Current	I	Ampere ‘Amp’	A
Voltage	E or V	Volt	V
Resistance	R	Ohm	Ω

The “symbol” given for each quantity is the standard alphabetical letter used to represent that quantity in an algebraic equation. Standardized letters like these are common in the disciplines of physics and engineering, and are internationally recognized. The “unit abbreviation” for each quantity represents the alphabetical symbol used as a shorthand notation for its particular unit of measurement. And, yes, that strange-looking “horseshoe” symbol is the capital Greek letter Ω, just a character in a foreign alphabet.
Each unit of measurement is named after a famous experimenter in electricity: The amp after the Frenchman Andre M. Ampere, the volt after the Italian Alessandro Volta, and the ohm after the German Georg Simon Ohm.

The mathematical symbol for each quantity is meaningful as well. The “R” for resistance and the “V” for voltage are both self-explanatory, whereas “I” for current seems a bit weird. The “I” is thought to have been meant to represent “Intensity” (of electron flow), and the other symbol for voltage, “E,” stands for “Electromotive force.” From what research I’ve been able to do, there seems to be some dispute over the meaning of “I.” The symbols “E” and “V” are interchangeable for the most part, although some texts reserve “E” to represent voltage across a source (such as a battery or generator) and “V” to represent voltage across anything else.

All of these symbols are expressed using capital letters, except in cases where a quantity (especially voltage or current) is described in terms of a brief period of time (called an “instantaneous” value). For example, the voltage of a battery, which is stable over a long period of time, will be symbolized with a capital letter “E,” while the voltage peak of a lightning strike at the very instant it hits a power line would most likely be symbolized with a lower-case letter “e” (or lower-case “v”) to designate that value as being at a single moment in time. This same lower-case convention holds true for current as well, the lower-case letter “i” representing current at some instant in time. Most direct-current (DC) measurements, however, being stable over time, will be symbolized with capital letters.

One foundational unit of electrical measurement, often taught in the beginnings of electronics courses but used infrequently afterwards, is the unit of the coulomb, which is a measure of electric charge proportional to the number of electrons in an imbalanced state. One coulomb of charge is equal to 6,250,000,000,000,000,000 electrons. The symbol for electric charge quantity is the capital letter “Q,” with the unit of coulombs abbreviated by the capital letter “C.” It so happens that the unit for electron flow, the amp, is equal to 1 coulomb of electrons passing by a given point in a circuit in 1 second of time. Cast in these terms, current is the rate of electric charge motion through a conductor.

As stated before, voltage is the measure of potential energy per unit charge available to motivate electrons from one point to another. Before we can precisely define what a “volt” is, we must understand how to measure this quantity we call “potential energy.” The general metric unit for energy of any kind is the joule, equal to the amount of work performed by a force of 1 newton exerted through a motion of 1 meter (in the same direction). In British units, this is slightly less than 3/4 pound of force exerted over a distance of 1 foot. Put in common terms, it takes about 1 joule of energy to lift a 3/4 pound weight 1 foot off the ground, or to drag something a distance of 1 foot using a parallel pulling force of 3/4 pound. Defined in these scientific terms, 1 volt is equal to 1 joule of electric potential energy per (divided by) 1 coulomb of charge. Thus, a 9 volt battery releases 9 joules of energy for every coulomb of electrons moved through a circuit.
These units and symbols for electrical quantities will become very important to know as we begin to explore the relationships between them in circuits. The first, and perhaps most important, relationship between current, voltage, and resistance is called Ohm’s Law, discovered by Georg Simon Ohm and published in his 1827 paper, The Galvanic Circuit Investigated Mathematically. Ohm’s principal discovery was that the amount of electric current through a metal conductor in a circuit is directly proportional to the voltage impressed across it, for any given temperature. Ohm expressed his discovery in the form of a simple equation, describing how voltage, current, and resistance interrelate:



In this algebraic expression, voltage (E) is equal to current (I) multiplied by resistance (R). Using algebra techniques, we can manipulate this equation into two variations, solving for I and for R, respectively:






Let’s see how these equations might work to help us analyze simple circuits:



In the above circuit, there is only one source of voltage (the battery, on the left) and only one source of resistance to current (the lamp, on the right). This makes it very easy to apply Ohm’s Law. If we know the values of any two of the three quantities (voltage, current, and resistance) in this circuit, we can use Ohm’s Law to determine the third.

In this first example, we will calculate the amount of current (I) in a circuit, given values of voltage (E) and resistance (R):



What is the amount of current (I) in this circuit?


In this second example, we will calculate the amount of resistance (R) in a circuit, given values of voltage (E) and current (I):



What is the amount of resistance (R) offered by the lamp?


In the last example, we will calculate the amount of voltage supplied by a battery, given values of current (I) and resistance (R):



What is the amount of voltage provided by the battery?



Ohm’s Law is a very simple and useful tool for analyzing electric circuits. It is used so often in the study of electricity and electronics that it needs to be committed to memory by the serious student. For those who are not yet comfortable with algebra, there’s a trick to remembering how to solve for any one quantity, given the other two.

First, arrange the letters E, I, and R in a triangle like this:


If you know E and I, and wish to determine R, just eliminate R from the picture and see what’s left:


If you know E and R, and wish to determine I, eliminate I and see what’s left:


Lastly, if you know I and R, and wish to determine E, eliminate E and see what’s left:


Eventually, you’ll have to be familiar with algebra to seriously study electricity and electronics, but this tip can make your first calculations a little easier to remember. If you are comfortable with algebra, all you need to do is commit E=IR to memory and derive the other two formulae from that when you need them!

REVIEW:

• Voltage measured in volts, symbolized by the letters “E” or “V”.

• Current measured in amps, symbolized by the letter “I”.

• Resistance measured in ohms, symbolized by the letter “R”.

• Ohm’s Law: E = IR ; I = E/R ; R = E/I

      2. Kirchhoff’s Laws

Relationships between U and I:-

The two Kirchhoff’s Laws tell us about the relationships between vofltages and currents in circuits.
i) Kirchhoff’s Current Law (KCL)
ii) Kirchhoff’s Voltage Law (KVL)

i) Kirchhoff’s Current Law (KCL) states that: 

The algebraic sum of currents at a node is zero.

Two points might need further explanation:

1. A node is the technical term for a junction in a circuit, where two or more branches are joined together. Fig. 2.1 shows a node with four branches connected;
2. the phrase algebraic sum reminds us that we have to take account of the current direction, as well as magnitude, when applying Kirchhoff’s Current Law.

This Law is used in circuit analysis to define relationships between currents flowing in branches of the circuit. For example, in Fig. 2.1 the currents flowing in the four branches connected to the node have been defined as I₁, I₂, I₃, I₄ and Kirchhoff’s Current Law allows us to write down an equation relating these currents.
Looking closely at Fig. 2.1, we see that two of the currents (I₁, I₂) are flowing towards the node, while the other two currents (I₃, I₄) are flowing outwards. The ‘algebraic sum’ needs to take account of this difference in relative direction.
To apply Kirchhoff’s Current Law rigorously, we must first make an arbitrary choice of positive current direction.
Suppose currents flowing in to the node (I₁, I₂) are treated as positive contributions to the algebraic sum (and conversely currents flowing from the node are treated as negative contributions), then the algebraic sum of currents would be written: + I₁ + I₂ – I₃ – I₄ , and according to Kirchhoff’s Current Law this algebraic sum is equal to zero:
+ I₁ + I₂ – I₃ – I₄ = 0        (2.1)
The same result could be obtained with the opposite choice of positive current direction. If currents flowing from the node (I₃, I₄) are treated as positive contributions to the algebraic sum, then the algebraic sum of currents would be written: – I1 – I2 + I3 + I4 , and equating this algebraic sum to zero:
- I₁ – I₂ + I₃ + I₄ = 0        (2.2)
which is the same relationship as Eqn. 2.1 with all terms multiplied by –1.
It must be emphasised that the choice of sign convention when using Kirchhoff’s Current Law is entirely arbitrary and, of course, makes no difference to the result obtained. However, it is good practice to be consistent in your choice, because this minimises the chance of making a mistake when writing down the algebraic sum.
Eqns. 2.1 and 2.2 can be re-arranged to show that:
I₁ + I₂ = I₃ + I₄ (2.3)
and referring back to Fig. 2.1 we see that this equation is showing that current flowing into the node is equal to the current flowing out. This formulation arises naturally from physical considerations of current as the flow of charge.
Charge does not accumulate at a node and therefore any charge flowing into the node through one or more branches must flow out from the node through other branches. Therefore, current flowing into is equal to current flowing out of the node.

Worked example 2.1

Calculate the current I flowing into the node.

Solution

Choosing currents flowing into the node as positive and applying Kirchhoff’s
Current Law: +3 –2 +I = 0, so I = -1 A
The current flowing into the node is –1A, which is the same as +1A flowing out of the node

Worked example 2.2

Calculate the current I defined in the diagram.

Solution

In this problem, there are two nodes, each with three branches connected. Begin by defining the current I’ flowing in the branch between the two nodes. The direction of I’ has been chosen randomly: it may turn out to have a positive or negative value. Choosing currents flowing out of the nodes as positive and applying Kirchhoff’s Current Law at each node:
-(-4) + 2 + I’ = 0, so I’ = -6 A
and: -I’ – 6 + I = 0, so I = I’ + 6 = 0 A
but is there any easier way? Yes! We can merge the two separate nodes into a single supernode, shown in red in the lower diagram. The supernode can’t accumulate charge, so Kirchhoff’s Current Law can be applied to currents in branches connected to it.
Making the same choice of current direction:
-(-4) + 2 + I – 6 = 0, so I = 0 A

ii)Kirchhoff’s Voltage Law (KVL) states that:-

The algebraic sum of voltages around a closed circuit loop is zero.

There’s the phrase ‘algebraic sum’ again, so we must recognise that the direction of voltages matters when using Kirchhoff’s Voltage Law.
Fig. 2.2 shows a circuit loop, which is part of a larger circuit. The loop involves four nodes, ABCD, between which are connected four components. In this case the four components are resistances, but Kirchhoff’s Voltage Law can be applied no matter what components are connected in the closed circuit loop. The voltages across the four resistances comprising the circuit loop have been defined as V₁, V₂, V₃, V₄ and Kirchhoff’s Voltage Law allows us to write down an equation relating these voltages. If we think about travelling around the closed circuit loop in any direction, we note that the four voltages will be encountered in sequence.
Two of the voltage arrows will point in the direction of travel and two will oppose the travel. The algebraic sum of voltages needs to take account of this difference in relative direction.
To apply Kirchhoff’s Voltage Law correctly, we must make arbitrary choices about the direction of travel around the closed circuit loop and the contribution which the separate voltages make to the algebraic sum around the closed circuit loop. Suppose we travel around the loop in Fig. 2.2 in the clockwise direction (ABCD) and that voltages opposite to the direction of travel make a positive contribution to the algebraic sum. In travelling from A to B the voltage V₁ is encountered and it is in a direction which is opposite to the travel. Therefore, V₁ is a positive contribution to the algebraic sum.
The same comment is true of V₂, which is met when proceeding from B to C. However, travelling from C to D and back to A, the voltages V₃ and V₄ are encountered and in both cases the voltages are in the same direction as the travel, giving a negative contribution to the algebraic sum. Expressed mathematically, the algebraic sum of voltages around the closed loop ABCD is: + V₁ + V₂ – V₃ – V₄ and Kirchhoff’s Voltage Law states that this sum is equal to zero:
+ V1 + V2 – V3 – V4 = 0 (2.4)
The same result is obtained for any choice of travel direction or voltage contribution to the algebraic sum. The other three combinations are:
Clockwise around the loop (ABCD), with the arrow positive:
- V1 – V2 + V3 + V4 = 0 (2.5)
Anticlockwise around the loop (ADCB), against the arrow positive:
- V1 – V2 + V3 + V4 = 0 (2.6)
Anticlockwise around the loop (ADCB), with the arrow positive:
+ V1 + V2 – V3 – V4 = 0 (2.7)
The four equations 2.4 – 2.7 produce exactly the same relationship between the four voltages: all four can be re-arranged to show that:
V1 + V2 = V3 + V4 (2.8)
As with the Current Law, it is a good idea to be consistent in your choice of direction and polarity when applying Kirchhoff’s Voltage Law, so there is less chance of error when writing down the algebraic sum.

Worked example 2.3

Calculate the voltage V

Solution

With an arbitrary choice of clockwise travel around the loop and counting with the voltage arrow as a positive contribution to algebraic sum, Kirchhoff’s Voltage Law:
-6 – (-10) + V +7 = 0,
so V = -11 V

Worked example 2.4

Calculate the voltage V

Solution

This example is intended to demonstrate that the ‘closed circuit loop’ does not need to be defined by a continuous connection of components: the voltage V is the voltage between two nodes, which have nothing connected between them, yet Kirchhoff’s Voltage Law is still valid.
With anticlockwise travel around the loop and counting against the voltage arrow as a positive contribution to algebraic sum:
+ V + 2 – 10 – (-8) = 0, so V = 0 V
And finally, a brief note on notation. You will find many books referring to the voltage between two points in a circuit, such as A and B, using the symbol VAB.
Naturally, you will wonder how this relates to the ‘arrow’ notation used here. As shown in Fig. 2.3, the convention is that the voltage VAB means ‘the voltage at A relative to B’, so the arrow points at A from B.


 3. Thevenin’s Theorem

Thevenin’s Theorem states that:-

"It is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load" 

                                                                              or

"Any linear circuit containing several voltages and resistances can be replaced by just a Single Voltage in series with a Single Resistor"

The qualification of “linear” is identical to that found in the Superposition Theorem, where all the underlying equations must be linear (no exponents or roots). If we’re dealing with passive components (such as resistors, and later, inductors and capacitors), this is true. However, there are some components (especially certain gas-discharge and semiconductor components) which are nonlinear: that is, their opposition to current changes with voltage and/or current. As such, we would call circuits containing these types of components, nonlinear circuits.

Thevenin’s Theorem is especially useful in analyzing power systems and other circuits where one particular resistor in the circuit (called the “load” resistor) is subject to change, and re-calculation of the circuit is necessary with each trial value of load resistance, to determine voltage across it and current through it.

Let’s take another look at our example circuit:

Let’s suppose that we decide to designate R₂ as the “load” resistor in this circuit. We already have four methods of analysis at our disposal (Branch Current, Mesh Current, Millman’s Theorem, and Superposition Theorem) to use in determining voltage across R₂ and current through R₂, but each of these methods are time-consuming. Imagine repeating any of these methods over and over again to find what would happen if the load resistance changed (changing load resistance is very common in power systems, as multiple loads get switched on and off as needed. the total resistance of their parallel connections changing depending on how many are connected at a time). This could potentially involve a lot of work!
Thevenin’s Theorem makes this easy by temporarily removing the load resistance from the original circuit and reducing what’s left to an equivalent circuit composed of a single voltage source and series resistance. The load resistance can then be re-connected to this “Thevenin equivalent circuit” and calculations carried out as if the whole network were nothing but a simple series circuit:





. . . after Thevenin conversion . . .



The “Thevenin Equivalent Circuit” is the electrical equivalent of B₁, R₁, R₃, and B₂ as seen from the two points where our load resistor (R₂) connects.
The Thevenin equivalent circuit, if correctly derived, will behave exactly the same as the original circuit formed by B₁, R₁, R₃, and B₂. In other words, the load resistor (R₂) voltage and current should be exactly the same for the same value of load resistance in the two circuits. The load resistor R₂ cannot “tell the difference” between the original network of B₁, R₁, R₃, and B₂, and the Thevenin equivalent circuit of E_Thevenin, and R_Thevenin, provided that the values for E_Thevenin and R_Thevenin have been calculated correctly.
The advantage in performing the “Thevenin conversion” to the simpler circuit, of course, is that it makes load voltage and load current so much easier to solve than in the original network. Calculating the equivalent Thevenin source voltage and series resistance is actually quite easy.

First, the chosen load resistor is removed from the original circuit, replaced with a break (open circuit):


Next, the voltage between the two points where the load resistor used to be attached is determined. Use whatever analysis methods are at your disposal to do this. In this case, the original circuit with the load resistor removed is nothing more than a simple series circuit with opposing batteries, and so we can determine the voltage across the open load terminals by applying the rules of series circuits, Ohm’s Law, and Kirchhoff’s Voltage Law:






The voltage between the two load connection points can be figured from the one of the battery’s voltage and one of the resistor’s voltage drops, and comes out to 11.2 volts. This is our “Thevenin voltage” (E_Thevenin) in the equivalent circuit:

To find the Thevenin series resistance for our equivalent circuit, we need to take the original circuit (with the load resistor still removed), remove the power sources (in the same style as we did with the Superposition Theorem: voltage sources replaced with wires and current sources replaced with breaks), and figure the resistance from one load terminal to the other:

With the removal of the two batteries, the total resistance measured at this location is equal to R₁ and R₃ in parallel: 0.8 Ω. This is our “Thevenin resistance” (R_Thevenin) for the equivalent circuit:


With the load resistor (2 Ω) attached between the connection points, we can determine voltage across it and current through it as though the whole network were nothing more than a simple series circuit:

Notice that the voltage and current figures for R₂ (8 volts, 4 amps) are identical to those found using other methods of analysis. Also notice that the voltage and current figures for the Thevenin series resistance and the Thevenin source (total) do not apply to any component in the original, complex circuit. Thevenin’s Theorem is only useful for determining what happens to a single resistor in a network: the load.
The advantage, of course, is that you can quickly determine what would happen to that single resistor if it were of a value other than 2 Ω without having to go through a lot of analysis again. Just plug in that other value for the load resistor into the Thevenin equivalent circuit and a little bit of series circuit calculation will give you the result.

REVIEW:

• Thevenin’s Theorem is a way to reduce a network to an equivalent circuit composed of a single voltage source, series resistance, and series load.
• Steps to follow for Thevenin’s Theorem:
• (1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be.
• (2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Thevenin equivalent circuit, with the Thevenin voltage source in series with the Thevenin resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for series circuits.

 4. Norton’s Theorem (Simplify any linear circuit)

Norton’s Theorem states that:-

 "It is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load."

 Just as with Thevenin’s Theorem, the qualification of “linear” is identical to that found in the Superposition Theorem: all underlying equations must be linear (no exponents or roots).
Contrasting our original example circuit against the Norton equivalent: it looks something like this:

. . . after Norton conversion . . .

Remember that a current source is a component whose job is to provide a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current.
As with Thevenin’s Theorem, everything in the original circuit except the load resistance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin’s Theorem are the steps used in Norton’s Theorem to calculate the Norton source current (I_Norton) and Norton resistance (R_Norton).
As before, the first step is to identify the load resistance and remove it from the original circuit:

Then, to find the Norton current (for the current source in the Norton equivalent circuit), place a direct wire (short) connection between the load points and determine the resultant current. Note that this step is exactly opposite the respective step in Thevenin’s Theorem, where we replaced the load resistor with a break (open circuit):

With zero voltage dropped between the load resistor connection points, the current through R₁ is strictly a function of B₁‘s voltage and R₁‘s resistance: 7 amps (I=E/R). Likewise, the current through R₃ is now strictly a function of B₂‘s voltage and R₃‘s resistance: 7 amps (I=E/R). The total current through the short between the load connection points is the sum of these two currents: 7 amps + 7 amps = 14 amps. This figure of 14 amps becomes the Norton source current (I_Norton) in our equivalent circuit:

Remember, the arrow notation for a current source points in the direction opposite that of electron flow. Again, apologies for the confusion. For better or for worse, this is standard electronic symbol notation. Blame Mr. Franklin again!
To calculate the Norton resistance (R_Norton), we do the exact same thing as we did for calculating Thevenin resistance (R_Thevenin): take the original circuit (with the load resistor still removed), remove the power sources (in the same style as we did with the Superposition Theorem: voltage sources replaced with wires and current sources replaced with breaks), and figure total resistance from one load connection point to the other:

Now our Norton equivalent circuit looks like this:

If we re-connect our original load resistance of 2 Ω, we can analyze the Norton circuit as a simple parallel arrangement:

As with the Thevenin equivalent circuit, the only useful information from this analysis is the voltage and current values for R₂; the rest of the information is irrelevant to the original circuit. However, the same advantages seen with Thevenin’s Theorem apply to Norton’s as well: if we wish to analyze load resistor voltage and current over several different values of load resistance, we can use the Norton equivalent circuit again and again, applying nothing more complex than simple parallel circuit analysis to determine what’s happening with each trial load.

REVIEW:

• Norton’s Theorem is a way to reduce a network to an equivalent circuit composed of a single current source, parallel resistance, and parallel load.
• Steps to follow for Norton’s Theorem:
• (1) Find the Norton source current by removing the load resistor from the original circuit and calculating current through a short (wire) jumping across the open connection points where the load resistor used to be.
• (2) Find the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for parallel circuits.

5. Thévenin and Norton Equivalence  (Reducing a complex network)

Since Thevenin’s and Norton’s Theorems are two equally valid methods of reducing a complex network down to something simpler to analyze, there must be some way to convert a Thevenin equivalent circuit to a Norton equivalent circuit, and vice versa. Well, the procedure is very simple.
You may have noticed that the procedure for calculating Thevenin resistance is identical to the procedure for calculating Norton resistance: remove all power sources and determine resistance between the open load connection points. As such, Thevenin and Norton resistances for the same original network must be equal.

Using the example circuits from the last two sections, we can see that the two resistances are indeed equal:



Considering the fact that both Thevenin and Norton equivalent circuits are intended to behave the same as the original network in suppling voltage and current to the load resistor (as seen from the perspective of the load connection points), these two equivalent circuits, having been derived from the same original network should behave identically.
This means that both Thevenin and Norton equivalent circuits should produce the same voltage across the load terminals with no load resistor attached. With the Thevenin equivalent, the open-circuited voltage would be equal to the Thevenin source voltage (no circuit current present to drop voltage across the series resistor), which is 11.2 volts in this case. With the Norton equivalent circuit, all 14 amps from the Norton current source would have to flow through the 0.8 Ω Norton resistance, producing the exact same voltage, 11.2 volts (E=IR). Thus, we can say that the Thevenin voltage is equal to the Norton current times the Norton resistance:

So, if we wanted to convert a Norton equivalent circuit to a Thevenin equivalent circuit, we could use the same resistance and calculate the Thevenin voltage with Ohm’s Law.
Conversely, both Thevenin and Norton equivalent circuits should generate the same amount of current through a short circuit across the load terminals. With the Norton equivalent, the short-circuit current would be exactly equal to the Norton source current, which is 14 amps in this case. With the Thevenin equivalent, all 11.2 volts would be applied across the 0.8 Ω Thevenin resistance, producing the exact same current through the short, 14 amps (I=E/R). Thus, we can say that the Norton current is equal to the Thevenin voltage divided by the Thevenin resistance:

This equivalence between Thevenin and Norton circuits can be a useful tool in itself, as we shall see in the next section.

REVIEW:

• Thevenin and Norton resistances are equal.
• Thevenin voltage is equal to Norton current times Norton resistance.
• Norton current is equal to Thevenin voltage divided by Thevenin resistance.

6. Superposition Theorem (Determining voltage and current, not power)

Superposition theorem is one of those strokes of genius that takes a complex subject and simplifies it in a way that makes perfect sense. A theorem like Millman’s certainly works well, but it is not quite obvious why it works so well. Superposition, on the other hand, is obvious.
The strategy used in the Superposition Theorem is to eliminate all but one source of power within a network at a time, using series/parallel analysis to determine voltage drops (and/or currents) within the modified network for each power source separately. Then, once voltage drops and/or currents have been determined for each power source working separately, the values are all “superimposed” on top of each other (added algebraically) to find the actual voltage drops/currents with all sources active. Let’s look at our example circuit again and apply Superposition Theorem to it:

Since we have two sources of power in this circuit, we will have to calculate two sets of values for voltage drops and/or currents, one for the circuit with only the 28 volt battery in effect. . .

. . . and one for the circuit with only the 7 volt battery in effect:

When re-drawing the circuit for series/parallel analysis with one source, all other voltage sources are replaced by wires (shorts), and all current sources with open circuits (breaks). Since we only have voltage sources (batteries) in our example circuit, we will replace every inactive source during analysis with a wire.
Analyzing the circuit with only the 28 volt battery, we obtain the following values for voltage and current:


Analyzing the circuit with only the 7 volt battery, we obtain another set of values for voltage and current:


When superimposing these values of voltage and current, we have to be very careful to consider polarity (voltage drop) and direction (electron flow), as the values have to be added algebraically.

Applying these superimposed voltage figures to the circuit, the end result looks something like this:

Currents add up algebraically as well, and can either be superimposed as done with the resistor voltage drops, or simply calculated from the final voltage drops and respective resistances (I=E/R). Either way, the answers will be the same. Here I will show the superposition method applied to current:

Once again applying these superimposed figures to our circuit:

Quite simple and elegant, don’t you think? It must be noted, though, that the Superposition Theorem works only for circuits that are reducible to series/parallel combinations for each of the power sources at a time (thus, this theorem is useless for analyzing an unbalanced bridge circuit), and it only works where the underlying equations are linear (no mathematical powers or roots). The requisite of linearity means that Superposition Theorem is only applicable for determining voltage and current, not power!!! Power dissipations, being nonlinear functions, do not algebraically add to an accurate total when only one source is considered at a time. The need for linearity also means this Theorem cannot be applied in circuits where the resistance of a component changes with voltage or current. Hence, networks containing components like lamps (incandescent or gas-discharge) or varistors could not be analyzed.
Another prerequisite for Superposition Theorem is that all components must be “bilateral,” meaning that they behave the same with electrons flowing either direction through them. Resistors have no polarity-specific behavior, and so the circuits we’ve been studying so far all meet this criterion.
The Superposition Theorem finds use in the study of alternating current (AC) circuits, and semiconductor (amplifier) circuits, where sometimes AC is often mixed (superimposed) with DC. Because AC voltage and current equations (Ohm’s Law) are linear just like DC, we can use Superposition to analyze the circuit with just the DC power source, then just the AC power source, combining the results to tell what will happen with both AC and DC sources in effect. For now, though, Superposition will suffice as a break from having to do simultaneous equations to analyze a circuit.

REVIEW:

• The Superposition Theorem states that a circuit can be analyzed with only one source of power at a time, the corresponding component voltages and currents algebraically added to find out what they’ll do with all power sources in effect.
• To negate all but one power source for analysis, replace any source of voltage (batteries) with a wire; replace any current source with an open (break).

7. Reciprocity Theorem

Introduction:-

The reciprocity theorem does not appear in many recent textbooks, though it was always included in earlier texts (see References) on circuits, even at an elementary level. The text by Irwin is an exception, where a good treatment is presented, and even a proof.

The omission of the reciprocity theorem is yet more evidence of the decline of understanding and its replacement by brassy ignorance in engineering education.
The theorem is important because it does not hold for all networks, and the restrictions should be remembered. Using reciprocity where it does not apply leads to seriously erroneous results. This article will explain what the reciprocity theorem is, and give examples to make its meaning clear. It also includes quite a good review of two-port networks, which are essential to a proper explanation.

In its simplest form, the reciprocity theorem states that if an emf E in one branch of a reciprocal network produces a current I in another, then if the emf E is moved from the first to the second branch, it will cause the same current in the first branch, where the emf has been replaced by a short circuit.

We shall see that any network composed of linear, bilateral elements (such as R, L and C) is reciprocal.
The circuit in the figure is a concrete example of reciprocity. The reader should solve the circuit, and determine the values of the current I in the two cases, which will be equal (0.35294 A). If E is reversed, then the direction of I is reversed, so the direction does not matter so long as both E and I are reversed at the same time.
A non-bilateral element, such as a rectifying diode, destroys reciprocity, as the circuit at the left shows. Here it is obvious that I = 0 on the right, while I ≠ 0 on the left. Even if the polarity of E is reversed so that the diode is forward-biased on the right, the currents are not reciprocal. On the left, I will be 0.21294 A, while on the right it will be 0.32824 A, as the reader can show by solving the circuits.
A nonlinear element also destroys reciprocity. The circuit at the right includes a resistor whose voltage drop is proportional to the square of the current, V = 10i². In the circuit at the left, I = 0.3798 A, while on the right I = 0.3397 A. The inclusion of controlled sources or active elements may also destroy reciprocity.

Two-Port Networks:-

Consideration of reciprocity leads naturally to two-port networks. These are networks with four terminals considered in two pairs as ports at which connections are made. The emf E in the reciprocity theorem is considered to be connected to one port, say port 1, while the current is at port 2, assumed to be short-circuited. The ports result from breaking into two of the branches of the network. One terminal of each port is denoted by (+) to specify the polarity of the voltage applied at the port, and currents are positive when they enter the (+) terminal.

The fundamental variables are V₁, I₁, V₂ and I₂. Any two of these variables are functions of the remaining two. For certain networks, some of the four choices are not admissible. In most cases, the variables appearing in the models are variations from DC bias conditions, not the DC variables themselves.
A single resistor forms two two-ports, depending on whether it is in series or shunt. For the series resistor, it is normal to take the dependent variables as I₁ and I₂, and the independent variables V₁ and V₂. The coefficients are called the admittance parameters, since admittance is the ratio of current to voltage. If the resistor is connected in shunt, the natural independent variables are I₁ and ₂, while V₁ and V₂ are the dependent variables. The coefficients in this case are the impedance parameters, since impedance is the ratio of voltage to current. In both cases, we see that the off-diagonal or transfer coefficients are equal.


Consider the admittance model, where the coefficients are not necessarily equal to the simple ones for a single resistor. If an emf E is applied at the input, E = V₁, and the output is shorted, V₂ = 0, we see that I₂ = y₂₁E. If we apply the same voltage E to the output, and short the input, we have I₁ = y₁₂E. Therefore, the currents are equal, and the network is reciprocal. Reciprocity is the result of the equality of the transfer admittances. A similar result can be obtained for the impedance model, where reciprocity holds if z₁₂ = z₂₁, as the reader can easily show. For our simple example of a single resistor, these conditions hold.
We note that there is no impedance model for the series resistor, and no admittance model for the shunt resistor. In general, it is possible to transform from impedance to admittance and vice versa, but the determinant of the coefficients is in the denominator of the transformation formula, and for a single resistor this determinant is zero. If Δ is the determinant of the admittance coefficients, then z₁₁ = y₂₂/Δ, z₂₂ = y₁₁/Δ, z₁₂ = y₂₁/Δ and z₂₁ = y₁₂/Δ. Similar formulas hold for the inverse transformation. A full list of transformations is given in the first Reference. The conclusion is that if y₁₂ = y₂₁, then z₂₁ = z₁₂ as well. This, of course, is a general result. The particular way of modeling a network does not affect the property of reciprocity.
A popular model, especially for active elements, is the hybrid model in which the independent variables are V₂ and I₁. The equations for the series resistor are V₁ = RI₁ + V₂ and I₂ = -I₁. The hybrid parameters for this case are h₁₁ = R, h₂₂ = 0, h₁₂ = -h₂₁ = 1. Note that the off-diagonal elements are of opposite sign in this case, but of the same magnitude. This is the mark of reciprocity in the hybrid model. As an exercise, the reader may show that for the shunt resistor h₁₁ = 0, h₂₂ = 1/R, and the transfer coefficients are the same as those for the series resistor.

There is a fourth model in which the output variables I₁ and V₁ are expressed in terms of the input variables V₂ and I₂: V₁ = AV₂ – BI₂, and I₁ = CV₂ – DI₂. Note the change in sign of I₂, which is conventional so it will be in the same direction as I₁ for an identity transformation A = D = 1, B = C = 0. If networks are cascaded, these matrices simply multiply. In this model, reciprocity is expressed by AD – CB = 1. This can be proved by expressing A, B, C and D in terms of the impedance paramters. A = z₁₁/z₂₁, B = z₁₁z₂₂/z₂₁ – z₁₂, C = 1/z₂₁ and D = z₂₂/z₂₁. The result is AD – CB = z₁₂/z₂₁ = 1 if the network is reciprocal. It is also clear that any cascade of reciprocal networks is also reciprocal, since the determinant of the product of matrices is the product of the determinants.



Any model can be expressed in terms of controlled sources and impedances. This is shown at the right for they hybrid model. The h₁₁ is represented by an impedance, h₂₂ by an admittance. h₁₂ and h₂₁ are the reverse and forward transfer ratios, which are pure numbers. A current source appears in the output, while a voltage source appears in the input. This is the model often used to represent a transistor, in which h₂₁ is usually denoted h_fe or β, the base to collector current gain. The input is between base and emitter, the output between collector and emitter. Since h₁₂ is much smaller than h₂₁, the transistor is not reciprocal. A good approximate model results if h₁₁ = h₂₁(25Ω/I_c), h₁₂ = h₂₂ = 0. I_c is the DC collector current. A typical value for h₂₁ is 100. For accurate work, the hybrid coefficients can be expressed as functions of the bias conditions.
A vacuum tube can be represented by an admittance model, in which the plate current I_p is a joint function of the grid voltage V_g and the plate voltage V_p. The input port is connected to grid and cathode, the output port to plate and cathode. As long as the grid is biased negatively with respect to the cathode, I_g = 0, and two of the admittance parameters are zero. The remaining equation can be written I_p = g_mV_g + V_p/r_p, where y₁₂ = g_m is the transconductance and 1/y₂₂ = r_p is the plate resistance. When I_p is held constant, -g_mr_pV_g = V_p, which defines the amplification factor μ = g_mr_p. The same model can be used for an FET. We note that none of these models are reciprocal, which is worth remembering. These models are extremely useful in circuit design.

Proof of the Reciprocity Theorem:-

We wish to show that in a network of linear, bilinear elements, that is, in one constructed of of ordinary impedances, that if when a voltage V is inserted in one loop the current I in another loop due to the insertion of this voltage is the same as the current at the first position due to the insertion of a voltage V in the second loop, or that the network is reciprocal. We shall do this by explicitly calculating the currents in the two cases, and observing that they are equal.
Consider the network as made up of N independent loops, of which loop 1 contains the input port, and loop 2 the output port. Let there be no emf’s in the network. If there are, they can be taken care of by superposition, and we can set them all to zero for our purposes. The loop currents are I_k, k = 1 to N. If emf V is at port 1 and port 2 is shorted, then Kirchhoff’s laws give Σz_1kI_k = V, and Σz_jkI_k = 0, j = 2 to N. The current I₂ can then be expressed in terms of determinants as I₂ = -V det A/Δ, where the determinant in the numerator has been expanded by minors of the second column, which is all zeros except for V in the first position. A is the matrix of the z’s with the first row and second column taken away. Δ is the determinant of the matrix of the z_jk. Write out this solution in detail if it is not clear from this description.
Now connect V in port 2, and short port 1. The solution by determinants for I₁ is -V det B/Δ, where B is the matrix of the z’s with the second row and first column taken away. If we compare matrices A and B, we see that they are transposes of each other, provided that z_jk = z_kj. However, the off-diagonal z’s are just the mutual impedances of the current loops (impedances common to a pair of loops), and do not depend on the order of the subscripts. (They are not the coefficients of an impedance model here, simply actual impedances.) Since the determinant of the transpose of a matrix is equal to the determinant of the matrix, the two solutions are the same, and I₁ = I₂. This proves the reciprocity theorem for networks made from linear, bilateral elements, Q.E.D.
We have proved the reciprocity theorem for a relatively limited class of networks, but it is possible to extend the theorem more widely.
The figure at the left shows that reciprocity holds for ideal transformers at least.
Since real transformers may be modeled by networks containing only impedances and ideal transformers, real transformers must also be reciprocal, and this extends the utility of the theorem further. Reciprocity crops up in many unexpected places, such as in electromagnetic fields and microwaves.

8. Compensation Theorem (Behaviour Of Circuit)

Compensation theorem is a very useful theorem. It reflects the changes that can occur due to incremental changes in the value of a component. This theorem is illustrated by the circuit in Fig. 45.
.


Let the resistance change by a small amount.  Its effect can be computed as shown.  The effect can be computed by the response due to a voltage source V_C , the value of which is obtained as shown below.
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From equation (73), cancel the common term on both sides of equation.  Then we obtain equation (74).  The value of compensating voltage V_C is obtained as shown by equation (75).  The change in current is also obtained, as shown.

Compensation theorem is illustrated by an example.  For the circuit in Fig. 46, obtain current I₃.  Let the value of resistor R₃ change by a small value.  Compute the change in current I₃.

The values of components are specified.  The value of current I₃ is obtained as shown.  The value of compensating source V_C is then obtained and the change in current I₃, denoted by DI₃, is obtained and its value has been shown above.

9. Millman’s Theorem  (Set of parallel-connected branches)

In Millman’s Theorem, the circuit is re-drawn as a parallel network of branches, each branch containing a resistor or series battery/resistor combination. Millman’s Theorem is applicable only to those circuits which can be re-drawn accordingly. Here again is our example circuit used for the last two analysis methods:

And here is that same circuit, re-drawn for the sake of applying Millman’s Theorem:

By considering the supply voltage within each branch and the resistance within each branch, Millman’s Theorem will tell us the voltage across all branches. Please note that I’ve labeled the battery in the rightmost branch as “B₃” to clearly denote it as being in the third branch, even though there is no “B₂” in the circuit!
Millman’s Theorem is nothing more than a long equation, applied to any circuit drawn as a set of parallel-connected branches, each branch with its own voltage source and series resistance:

Substituting actual voltage and resistance figures from our example circuit for the variable terms of this equation, we get the following expression:

The final answer of 8 volts is the voltage seen across all parallel branches, like this:

The polarity of all voltages in Millman’s Theorem are referenced to the same point. In the example circuit above, I used the bottom wire of the parallel circuit as my reference point, and so the voltages within each branch (28 for the R₁ branch, 0 for the R₂ branch, and 7 for the R₃ branch) were inserted into the equation as positive numbers. Likewise, when the answer came out to 8 volts (positive), this meant that the top wire of the circuit was positive with respect to the bottom wire (the original point of reference). If both batteries had been connected backwards (negative ends up and positive ends down), the voltage for branch 1 would have been entered into the equation as a -28 volts, the voltage for branch 3 as -7 volts, and the resulting answer of -8 volts would have told us that the top wire was negative with respect to the bottom wire (our initial point of reference).
To solve for resistor voltage drops, the Millman voltage (across the parallel network) must be compared against the voltage source within each branch, using the principle of voltages adding in series to determine the magnitude and polarity of voltage across each resistor:

To solve for branch currents, each resistor voltage drop can be divided by its respective resistance (I=E/R):

The direction of current through each resistor is determined by the polarity across each resistor, not by the polarity across each battery, as current can be forced backwards through a battery, as is the case with B₃ in the example circuit. This is important to keep in mind, since Millman’s Theorem doesn’t provide as direct an indication of “wrong” current direction as does the Branch Current or Mesh Current methods. You must pay close attention to the polarities of resistor voltage drops as given by Kirchhoff’s Voltage Law, determining direction of currents from that.

Millman’s Theorem is very convenient for determining the voltage across a set of parallel branches, where there are enough voltage sources present to preclude solution via regular series-parallel reduction method. It also is easy in the sense that it doesn’t require the use of simultaneous equations. However, it is limited in that it only applied to circuits which can be re-drawn to fit this form. It cannot be used, for example, to solve an unbalanced bridge circuit. And, even in cases where Millman’s Theorem can be applied, the solution of individual resistor voltage drops can be a bit daunting to some, the Millman’s Theorem equation only providing a single figure for branch voltage.
As you will see, each network analysis method has its own advantages and disadvantages. Each method is a tool, and there is no tool that is perfect for all jobs. The skilled technician, however, carries these methods in his or her mind like a mechanic carries a set of tools in his or her tool box. The more tools you have equipped yourself with, the better prepared you will be for any eventuality.

REVIEW:

• Millman’s Theorem treats circuits as a parallel set of series-component branches.
• All voltages entered and solved for in Millman’s Theorem are polarity-referenced at the same point in the circuit (typically the bottom wire of the parallel network).

10.Joule’s Law

(Amount of heat (energy) delivered to something)

Joule’s Law states that H (Heat) = I (Current) x V (Voltage) x T (Time the current is allowed to flow).

                                                                                Or

written differently,

H (Heat) = I2 (Current squared) x R (Resistance) x T (Time the current is allowed to flow).

Note: V (Voltage) = I (Current) x R (Resistance), so the two equations are the same, just stated differently. The second version of this law is probably more common in the field.
Joule’s Law is an equation that gives the amount of heat (energy) delivered to something. It would seem sensible to assume that it’s the amount of heat delivered to the weld. However, it is important to consider all the factors in the equation: Current, Voltage, and Time. Joule’s Law assumes that each of these factors remains constant in the secondary of the welding transformer. A weld controller or weld timer may indeed provide a constant amount of current at the electrodes, but recall Ohm’s Law: Voltage equals Current times Resistance, or written differently, Current equals Voltage divided by Resistance.

Factors like pitting or mushrooming of the electrodes, dirty workpieces, changes in force, etc. all have an effect on the surface area (the area of contact) between the electrode and the workpiece. Since changes in the surface area affect the contact resistance (resistance of the surface area), it is reasonable to say that the resistance at the workpiece is not constant, but rather a factor that can change depending on a number of other conditions.

If Resistance is not constant, then according to Ohm’s Law, Current is not constant either. This means that the I-squared version Joule’s Law will not reveal the amount of heat generated at the workpiece unless the resistance at the tips is known.

Simply put, to determine how much heat is being generated at the workpiece using Joule’s Law, current, voltage or resistance must be measured at the workpiece. Although a weld controller may be programmed to deliver 20 KA at 10 Volts, if there is significant resistance in the secondary weld loop, the heat will go there and not to the workpiece. Likewise, if the electrodes are worn or the workpiece is dirty, resistance and current density will be affected. In such a situation, a controller might indicate 10 Volts at the secondary, however there might actually be only 5 Volts at the weld tips.
Such a disparity could easily cause bad welds.

11. Maximum Power Transfer Theorem (Power dissipated by a load resistance)

The Maximum Power Transfer Theorem is not so much a means of analysis as it is an aid to system design. Simply stated, the maximum amount of power will be dissipated by a load resistance when that load resistance is equal to the Thevenin/Norton resistance of the network supplying the power. If the load resistance is lower or higher than the Thevenin/Norton resistance of the source network, its dissipated power will be less than maximum.

This is essentially what is aimed for in radio transmitter design , where the antenna or transmission line “impedance” is matched to final power amplifier “impedance” for maximum radio frequency power output. Impedance, the overall opposition to AC and DC current, is very similar to resistance, and must be equal between source and load for the greatest amount of power to be transferred to the load. A load impedance that is too high will result in low power output. A load impedance that is too low will not only result in low power output, but possibly overheating of the amplifier due to the power dissipated in its internal (Thevenin or Norton) impedance.

Taking our Thevenin equivalent example circuit, the Maximum Power Transfer Theorem tells us that the load resistance resulting in greatest power dissipation is equal in value to the Thevenin resistance (in this case, 0.8 Ω):

With this value of load resistance, the dissipated power will be 39.2 watts:

If we were to try a lower value for the load resistance (0.5 Ω instead of 0.8 Ω, for example), our power dissipated by the load resistance would decrease:

Power dissipation increased for both the Thevenin resistance and the total circuit, but it decreased for the load resistor. Likewise, if we increase the load resistance (1.1 Ω instead of 0.8 Ω, for example), power dissipation will also be less than it was at 0.8 Ω exactly:

If you were designing a circuit for maximum power dissipation at the load resistance, this theorem would be very useful. Having reduced a network down to a Thevenin voltage and resistance (or Norton current and resistance), you simply set the load resistance equal to that Thevenin or Norton equivalent (or vice versa) to ensure maximum power dissipation at the load. Practical applications of this might include radio transmitter final amplifier stage design (seeking to maximize power delivered to the antenna or transmission line), a grid tied inverter loading a solar array, or electric vehicle design (seeking to maximize power delivered to drive motor).

The Maximum Power Transfer Theorem is not: 

Maximum power transfer does not coincide with maximum efficiency. Application of The Maximum Power Transfer theorem to AC power distribution will not result in maximum or even high efficiency. The goal of high efficiency is more important for AC power distribution, which dictates a relatively low generator impedance compared to load impedance.

Similar to AC power distribution, high fidelity audio amplifiers are designed for a relatively low output impedance and a relatively high speaker load impedance. As a ratio, “output impdance” : “load impedance” is known as damping factor, typically in the range of 100 to 1000.
Maximum power transfer does not coincide with the goal of lowest noise. For example, the low-level radio frequency amplifier between the antenna and a radio receiver is often designed for lowest possible noise. This often requires a mismatch of the amplifier input impedance to the antenna as compared with that dictated by the maximum power transfer theorem.

REVIEW:

• The Maximum Power Transfer Theorem states that the maximum amount of power will be dissipated by a load resistance if it is equal to the Thevenin or Norton resistance of the network supplying power.
• The Maximum Power Transfer Theorem does not satisfy the goal of maximum efficiency.

11. Star to Delta Transforms

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 to produce an equivalent Star or Υ type network. But we can also convert a Pi or π type resistor network into an equivalent Delta or Δ type network as shown below.

12. 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 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:-

Example No1:-

Convert the following Delta Resistive Network into an equivalent Star Network.

Star Delta Transformation

We have seen above 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 Network.

The value of the resistor on any one side of the delta, Δ network is the sum of all the two-product 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 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:   R_DELTA = 3R_STAR.