Relationship

Any basic electrical or electronic circuit consists of three separate but very much related quantities,Voltage, ( V ), Current, ( I ) and Resistance, ( Ω ).

Voltage.

Voltage is the potential energy of an electrical supply stored in the form of an electrical charge, and the greater the voltage the greater is its ability to produce an electrical current flowing through a given circuit. As energy has the ability to do work this potential energy can be described as the work required in joules to move the electrical current around a circuit from one point or node to another. The difference in voltage between any two nodes in a circuit is known as the Potential Difference, p.d. or sometimes called Electromotive Force, (EMF) and is measured in Volts with the circuit symbol V, or lowercase "v", although Energy, E lowercase "e" is sometimes used.

A constant voltage source is called a DC Voltage with a voltage that varies periodically with time is called an AC voltage. Voltage is measured in volts, and one volt can be defined as the electrical pressure required to force an electrical current of one ampere through a resistance of one Ohm. Voltages are generally expressed in Volts with prefixes used to denote microvolts (μV = 10-6V), millivolts (mV = 10-3V) or kilovolts (kV = 103V). Voltage can be either positive or negative.

Batteries or power supplies are mostly used to produce a steady D.C. (direct current) voltage source such as 5v, 12v, 24v etc in electronic circuits and systems. While A.C. (alternating current) voltage sources are available for domestic house and industrial power and lighting as well as power transmission. The mains voltage supply in the United Kingdom is currently 230 volts a.c. and 110 volts a.c. in the USA with general electronic circuits operating on a voltage supply of between 1.5V and 24V d.c. The circuit symbol for a constant voltage source usually given as a battery symbol with a positive, +and negative, - sign indicating the direction of the polarity. The circuit symbol for an alternating voltage source is a circle with a sine wave inside.

A simple relationship can be made between a tank of water and a voltage supply. The higher the water tank above the outlet the greater the pressure of the water as more energy is released, the higher the voltage the greater the potential energy as more electrons are released. Voltage is always measured as the difference between any two points in a circuit and the voltage between these two points is generally referred to as the "Voltage drop". Any voltage source whether DC or AC likes an open or semi-open circuit condition but hates any short circuit condition as this can destroy it.

Electric Current.

Electrical Current is the movement of electrical charge and is measured in Amperes, symbol I, forIntensity). It is the continuous and uniform flow of electrons (negative particles of an atom) around a circuit that are being "pushed" by the voltage source. In reality, electrons flow from the negative (-ve) terminal to the positive (+ve) terminal of the supply and for ease of circuit understanding conventional current flow assumes that the current flows from the positive to the negative terminal. Generally in circuit diagrams the flow of current through the circuit usually has an arrow associated with the symbol, I, or lowercase I to indicate the actual direction of the current flow. However, this arrow usually indicates the direction of conventional current flow and not necessarily the direction of the actual flow.

In electronic circuits, a current source is a circuit element that provides a specified amount of current for example, 1A, 5A 10Amps etc, with the circuit symbol for a constant current source given as a circle with an arrow inside indicating its direction. Current is measured in Amps and an amp or ampere is defined as the number of electrons or charge (Q in Coulombs) passing a certain point in the circuit in one second, (t in Seconds). Current is generally expressed in Amps with prefixes used to denote micro amps (μA = 10-6A) or milli amps (mA = 10-3A). Electrical current can be either positive or negative.

Current that flows in a single direction is called Direct Current, or D.C. and current that alternates back and forth through the circuit is known as Alternating Current, or A.C.. Whether AC or DC current only flows through a circuit when a voltage source is connected to it with its "flow" being limited to both the resistance of the circuit and the voltage source pushing it. Also, as AC currents (and voltages) are periodic and vary with time the "effective" or "RMS", (Root Mean Squared) value given as Irms produces the same average power loss equivalent to a DC current Iaverage . Current sources are the opposite to voltage sources in that they like short or closed circuit conditions but hate open circuit conditions as no current will flow.

Using the tank of water relationship, current is the equivalent of the flow of water through the pipe with the flow being the same throughout the pipe. The faster the flow of water the greater the current. Any current source whether DC or AC likes a short or semi-short circuit condition but hates any open circuit condition as this prevents it from flowing.

Resistance.

The Resistance of a circuit is its ability to resist or prevent the flow of current (electron flow) through it making it necessary to apply a bigger voltage to the circuit to cause the current to flow again. Resistance is measured in Ohms, Greek symbol (Ω, Omega) with prefixes used to denote Kilo-ohms (kΩ = 103Ω) and Mega-ohms (MΩ = 106Ω). Resistance cannot be negative only positive.

The amount of resistance determines whether the circuit is a "good conductor" - low resistance, or a "bad conductor" - high resistance. Low resistance, for example 1Ω or less implies that the circuit is a good conductor made from materials such as copper, aluminium or carbon while a high resistance, 1MΩ or more implies the circuit is a bad conductor made from insulating materials such as glass, porcelain or plastic. A "semiconductor" on the other hand is a material whose resistance is half way between that of a good conductor and a good insulator such as silicon and germanium and is used to make Diodes and Transistors etc.

Resistance in a circuit prevents short circuits (unless its very low) by limiting and controlling the amount of current flowing in a circuit by the voltage supply connected to it and therefore the transfer of power from source to load. Resistance is not affected by frequency and the AC impedance of a pure resistance is equal to its DC resistance. Resistance also has the ability to change the characteristics of a circuit by the effect of load resistance or by temperature which changes its resistivity.

For very low values of resistance, for example milli-ohms, (mΩ´s) it is sometimes more easier to use the reciprocal of resistance (1/R) rather than resistance (R) itself. The reciprocal of resistance is calledConductance, symbol (G) and it is the ability of a conductor or device to conduct electricity with high values of conductance implying a good conductor and low values of conductance implying a bad conductor. The unit of conductance is the Siemen, symbol (S).

Again, using the water relationship, resistance is the diameter or the length of the pipe the water flows through. The smaller the diameter of the pipe the larger the resistance to the flow of water, and therefore the larger the resistance.

Relationship between Voltage and Current in a circuit of constant resistance.

Summary

This relationship between Voltage, Current and Resistance forms the basis of Ohms Law which is discussed in the next tutorial and a basic summary of the three units is given below.

Voltage or potential difference is the measure of potential energy between two points in a circuit and is commonly referred to as its "volt drop".

When a voltage source is connected to a closed loop circuit the voltage will produce a current flowing around the circuit.

In D.C. voltage sources the symbols +ve (positive) and -ve (negative) are used to denote the polarity of the voltage supply.

Voltage is measured in "Volts" and has the symbol "V" for voltage or "E" for energy. Current flow is a combination of electron flow and hole flow through a circuit. Current is the continuous and uniform flow of charge around the circuit and is measured in

"Amperes" or "Amps" and has the symbol "I". The effective (rms) value of an AC current has the same average power loss equivalent to a DC

current flowing through a resistive element. Resistance is the opposition to current flowing around a circuit. Low values of resistance implies a conductor and high values of resistance implies an insulator.

Resistance is measured in "Ohms" and has the Greek symbol "Ω" or the letter "R".

Quantity Symbol Unit of Measure Abbreviation

Voltage V or E Volt V

Current I Amp A

Resistance R Ohms Ω

Ohms Law

The relationship between Voltage, Current and Resistance in any DC electrical circuit was firstly discovered by the German physicist Georg Ohm, (1787 - 1854). Georg Ohm found that, at a constant temperature, the electrical current flowing through a fixed linear resistance is directly proportional to the voltage applied across it, and also inversely proportional to the resistance. This relationship between theVoltage, Current and Resistance forms the bases of Ohms Law and is shown below.

Ohms Law Relationship

By knowing any two values of the Voltage, Current or Resistance quantities we can use Ohms Law to find the third missing value. Ohms Law is used extensively in electronics formulas and calculations so it is "very important to understand and accurately remember these formulas".

To find Voltage (V)

[V = I x R]      V (volts)  = I (amps) x R (Ω)

To find Current (I)

[I = V ÷ R]      I (amps)  = V (volts) ÷ R (Ω)

To find Resistance (R)

[R = V ÷ I]      R  (Ω) = V (volts) ÷ I (amps)

It is sometimes easier to remember Ohms law relationship by using pictures. Here the three quantities have been superimposed into a triangle giving voltage at the top with current and resistance at the bottom. This arrangement represents the position of each quantity in the ohm's law formulas.

Ohms Law Triangle

 

Then by using Ohms Law we can see that a voltage of 1V applied to a resistor of 1Ω will cause a current of 1A to flow and the greater the resistance, the less current will flow for any applied voltage. Any Electrical device or component that obeys "Ohms Law" that is, the current flowing through it is proportional to the voltage across it (I α V), such as resistors or cables, are said to be "Ohmic" in nature, and devices that do not, such as transistors or diodes, are said to be "Non-ohmic" devices.

Power in Electrical Circuits

Electrical Power, (P) in a circuit is the amount of energy that is absorbed or produced within the circuit. A source of energy such as a voltage will produce or deliver power while the connected load absorbs it. The quantity symbol for power is P and is the product of voltage multiplied by the current with the unit of measurement being the Watt (W) with prefixes used to denote milliwatts (mW = 10-3W) or kilowatts(kW = 103W). By using Ohm's law and substituting for V, I and R the formula for electrical power can be found as:

To find Power (P)

[P = V x I]      P (watts)  = V (volts) x I (amps)

Also,

[P = V2 ÷ R]      P  (watts) = V2 (volts) ÷ R (Ω)

Also,

[P = I2 x R]      P  (watts) = I2 (amps) x R (Ω)

The Power Triangle

 

One other point about Power, if the calculated power is positive in value for any formula the component absorbs the power, but if the calculated power is negative in value the component produces power, in other words it is a source of electrical energy. Also, we now know that the unit of power is the WATT but some electrical devices such as electric motors have a power rating in Horsepower or hp. The relationship between horsepower and watts is given as: 1hp = 746W.

Ohms Law Equations

We can now take all the equations from above for finding Voltage, Current, Resistance and Power and condense them into a simple Ohms Law pie chart for use in DC circuits.

Ohms Law Pie Chart

Example No1

For the circuit shown below find the Voltage V, the Current I, the Resistance R and the Power P.

Voltage   [ V = I x R ] = 2 x 12Ω = 24V

Current   [ I = V ÷ R ] = 24 ÷ 12Ω = 2A

Resistance   [ R = V ÷ I ] = 24 ÷ 2 = 12 Ω

Power   [ P = V x I ] = 24 x 2 = 48W

Power within an electrical circuit is only present when BOTH voltage and current are present for example, In an Open-circuit condition, Voltage is present but there is no current flow I = 0 (zero), therefore V x 0 is 0so the power dissipated within the circuit must also be 0. Likewise, if we have a Short-circuit condition, current flow is present but there is no voltage V = 0, therefore 0 x I = 0 so again the power dissipated within the circuit is 0.

As electrical power is the product of V x I, the power dissipated in a circuit is the same whether the circuit contains high voltage and low current or low voltage and high current flow. Generally, power is dissipated in the form of Heat (heaters), Mechanical Work such as motors, etc or Energy in the form of radiated (Lamps) or stored energy (Batteries).

Energy in Electrical Circuits

Electrical Energy that is either absorbed or produced is the product of the electrical power measured in Watts and the time in Seconds with the unit of energy given as Watt-seconds or Joules.

Although electrical energy is measured in Joules it can become a very large value when used to calculate the energy consumed by a component. For example, a single 100 W light bulb connected for one hour will consume a total of 100 watts x 3600 sec = 360,000 Joules. So prefixes such as kilojoules (kJ = 103J) ormegajoules (MJ = 106J) are used instead. If the electrical power is measured in "kilowatts" and the time is given in hours then the unit of energy is in kilowatt-hours or kWh which is commonly called a "Unit of Electricity" and is what consumers purchase from their electricity suppliers.

Electrical Units of Measure

The standard SI units used for the measurement of voltage, current and resistance are the Volt [ V ],Ampere [ A ] and Ohms [ Ω ] respectively. Sometimes in electrical or electronic circuits and systems it is necessary to use multiples or sub-multiples (fractions) of these standard units when the quantities being measured are very large or very small. The following table gives a list of some of the prefixes used in electrical formulas and component values.

Prefix Symbol Multiplier Power of Ten

Voltage V 1 100

Current I 1 100

Resistance Ω 1 100

Capacitance F 1 100

Inductance H 1 100

Frequency Hz 1 100

Power W 1 100

Impedance Z 1 100

Giga G 1,000,000,000 109

Mega M 1,000,000 106

kilo k 1,000 103

milli m 1/1,000 10-3

micro µ 1/1,000,000 10-6

nano n 1/1,000,000,000 10-9

pico p 1/1,000,000,000,000 10-12

So to display the units or multiples of units for either Resistance, Current or Voltage we would use as an example:

1kV = 1 kilo-volt  -  which is equal to 1000 Volts.   1mA = 1 milli-amp  -  which is equal to one thousandths (1/1000) of an Ampere.   47kΩ = 47 kilo-ohms  -  which is equal to 47 thousand Ohms.   100uF = 100 micro-farads  -  which is equal to 100 millionths (1/1,000,000) of a Farad.   1kW = 1 kilo-watt  -  which is equal to 1000 Watts.   1MHz = 1 mega-hertz  -  which is equal to one million Hertz.

To convert from one prefix to another it is necessary to either multiply or divide by the difference between the two values. For example, convert 1MHz into kHz.

Well we know from above that 1MHz is equal to one million (1,000,000) hertz and that 1kHz is equal to one thousand (1,000) hertz, so one 1MHz is one thousand times bigger than 1kHz. Then to convert Mega-hertz into Kilo-hertz we need to multiply mega-hertz by one thousand, as 1MHz is equal to 1000 kHz. Likewise, if we needed to convert kilo-hertz into mega-hertz we would need to divide by one thousand. A much simpler and quicker method would be to move the decimal point either left or right depending upon whether you need to multiply or divide.

As well as the "Standard" electrical units of measure shown above, other units are also used in electrical engineering to denote other values and quantities such as:

•  Wh − The Watt-Hour, The amount of electrical energy consumed in the circuit by a load of one watt drawing power for one hour, eg a Light Bulb. It is commonly used in the form of kWh(Kilowatt-hour) which is 1,000 watt-hours or MWh (Megawatt-hour) which is 1,000,000 watt-hours. 

•  dB − The Decibel, The decibel is a one tenth unit of the Bel (symbol B) and is used to represent gain either in voltage, current or power. It is a logarithmic unit expressed in dB and is commonly used to represent the ratio of input to output in amplifier, audio circuits or loudspeaker systems. For example, the dB ratio of an input voltage (Vin) to an output voltage (Vout) is expressed as 20log10 (Vout/Vin). The value in dB can be either positive (20dB) representing gain or negative (-20dB) representing loss with unity, ie input = output expressed as 0dB. 

•  θ − Phase Angle, The Phase Angle is the difference in degrees between the voltage waveform and the current waveform having the same periodic time. It is a time difference or time shift and depending upon the circuit element can have a "leading" or "lagging" value. The phase angle of a waveform is measured in degrees or radians. 

•  ω − Angular Frequency, Another unit which is mainly used in a.c. circuits to represent the Phasor Relationship between two or more waveforms is called Angular Frequency, symbolω. This is a rotational unit of angular frequency 2πƒ with units in radians per second,rads/s. The complete revolution of one cycle is 360 degrees or 2π, therefore, half a revolution is given as 180 degrees or π rad. 

•  τ − Time Constant, The Time Constant of an impedance circuit or linear first-order system is the time it takes for the output to reach 63.7% of its maximum or minimum output value when subjected to a Step Response input. It is a measure of reaction time.

Kirchoffs Circuit Laws

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 a circuit. For these types of calculations we need certain rules which allow us to obtain the circuit equations and for this we can useKirchoffs Circuit Laws.

In 1845, a German physicist, Gustav Kirchoff developed a pair of rules or laws which deal with the conservation of current and energy in electrical circuits, one which deals with current flow, Kirchoffs Current Law, (KCL) and one which deals with voltage, Kirchoffs Voltage Law, (KVL).

Kirchoff's First Law - The Current Law, (KCL)

Kirchoff´s current law 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 is known as the "Conservation of Charge".

Kirchoff's 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 conductors such as cables and components. Also for current to flow either in or out of a node a closed circuit path must exist.

Kirchoff's Second Law - The Voltage Law, (KVL)

Kirchoff´s voltage or loop law 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 is known as the "Conservation of Energy".

Kirchoff's 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.

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 = 20I1 + 40(I1 + I2)  =  40I1 + 60I2

We now have two "Simultaneous Equations" that can be reduced to give us the value of both 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 Kirchoffs Circuit Laws is as follows:

1. Assume all voltage sources 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 Kirchoffs first law equations for each node. 4. Find Kirchoffs second law equations for each of the independent loops of the circuit. 5. Use Linear simultaneous equations as required to find the unknown currents.

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 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 circuit from the previous section.

Mesh Analysis Circuit

One simple method of reducing the amount of math's involved is to do Kirchoff´s First Current Law equations to determine the currents, I1 and I2 flowing in the two resistors and there is no need to calculate the current I3 as its just the sum of I1 and I2. Then the Second Voltage Law simply becomes:

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

A more easier method of solving the above circuit is by using Mesh Current Analysis or Loop Analysiswhich 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 second voltage law equations 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. This then gives us a matrix of:

 

Where:

[ 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 called the resistance matrix.

and this gives I1 as -0.143 Amps and I2 as -0.429 Amps

As :    I3 = I1 - I2

The current 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.

Page 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 Anaysis equations is as follows:

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;o   R11 = the total resistance in the first loop.o   Rnn = the total resistance in the Nth loop.o   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.

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 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. By adding together all these nodal voltage 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 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 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 Analysis Summary.

The basic procedure for solving Nodal Analysis equations is as follows:

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.

  4. The voltage vector will be (N x L) and will list the "N" voltages to be found.

Thevenins Theorem

In the previous 3 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 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, Thevenins Theorem.

Thevenins Theorem states that "Any linear circuit containing several voltages and resistances can be replaced by just a Single Voltage in series with a Single Resistor". In other words, it is possible to simplify any "Linear" circuit, no matter how complex, to an equivalent circuit with just a single voltage source in series with a resistance connected to a load as shown below. Thevenins Theorem is especially useful in analyzing power or battery systems and other interconnected circuits where it will have an effect on the adjoining part of the circuit.

Thevenins equivalent circuit.

As far as the load resistor RL is concerned, any "One-port" network consisting of resistive circuit elements and energy sources can be replaced by one single equivalent resistance Rs and voltage Vs, where Rs is the source resistance value looking back into the circuit and Vs the open circuit voltage at the terminals.

For example, consider the circuit from the previous section.

Firstly, we have to remove the centre 40Ω resistor and short out (not physically as this would be dangerous) all the emf´s connected to the circuit, or open circuit any current sources. The value of resistorRs is found by calculating the total resistance at the terminals A and B with all the emf´s removed, and the value of the voltage required Vs is the total voltage across terminals A and B with an open circuit and no load resistor Rs connected. Then, we get the following circuit.

Find the Equivalent Resistance (Rs)

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:

so the voltage drop across the 20Ω resistor can be calculated as:

VAB  =  20  -  (20Ω x 0.33amps)  =    13.33 volts.

Then the Thevenins Equivalent circuit is shown below with the 40Ω resistor connected.

and from this the current flowing in 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 tutorial.

Thevenins theorem can be used as a circuit analysis method and is particularly useful if the load is to take a series of different values. 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, Thevenins equivalent circuits of Transistors, Voltage Sources such as batteries etc, are very useful in circuit design.

Page Summary.

The basic procedure for solving Thevenins Analysis equations is as follows:

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.

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 Analysis Summary.

The basic procedure for solving Nortons Analysis equations is as follows:

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.

The Maximum Power Transfer Theorem

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 the Maximum Power Transfer Theorem.

The Maximum Power Transfer Theorem is another useful 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.

Thevenin's 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.

Example No1.

Where:RS = 25ΩRL is variable between 0 - 100ΩVS = 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 PowerRL I P0 0 05 3.3 55

10 2.8 7815 2.5 9320 2.2 97

RL I P25 2.0 10030 1.8 9740 1.5 9460 1.2 83

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 to the source resistance, RS so then: 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 occurs when the impedance of the external device matches that of the source. Improper impedance matching can lead to excessive power use and dissipation.

Transformer Impedance Matching

One very useful application of impedance matching to provide maximum power transfer is in the output stages of amplifier circuits, where the speakers impedance is matched to the amplifier output impedance to obtain maximum sound power output. This is achieved by using a Matching Transformerto couple the load to the amplifiers output as shown below.

Transformer Coupling

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,

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 and the turns ratio is given as N.

Generally, small transformers used in low power audio amplifiers are usually regarded as ideal so any losses can be ignored.

Star and 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 Starconnected 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 to Delta Transformation or Delta to Star Transformationprocess.

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 Star-connected Resistor 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.

Pi-connected and Delta-connected Resistor Network.

Having now defined exactly what is a Star and Delta connected network it is possible to transform the Υinto an equivalent Δ network and also to convert a Δ into an equivalent Υ network using aTransformation process. This process allows us to produce a mathematical relationship between the various resistors and 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 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 re-writing the previous formulas a little we can also find the transformation formulas for converting a resistive Star network to an equivalent Delta network 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 to Delta Transformations 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:   RDELTA = 3RSTAR