Electrical Circuit Article

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Relationship Between Voltage, Current Resistance

All materials are made up from atoms, and all atoms consist of protons, neutrons and

electrons. Protons, have a positive electrical charge. Neutrons have no electrical charge

while Electrons, have a negative electrical charge. Atoms are bound together bypowerful forces of attraction existing between the atoms nucleus and the electrons in its

outer shell. When these protons, neutrons and electrons are together within the atom

they are happy and stable. However, if we separate them they exert a potential of

attraction called apotential difference. If we create a circuit or conductor for the

electrons to drift back to the protons the flow of electrons is called a current. The

electrons do not flow freely through the circuit, the restriction to this flow is called

resistance. Then all basic electrical or electronic circuit consists of three separate but

very much related quantities, Voltage, ( v ), Current, ( i ) and Resistance, ( ).

Voltage

Voltage, ( V ) is the potential energy of an electrical supply stored in the form of an

electrical charge. Voltage can be thought of as the force that pushes electrons through

a conductor and the greater the voltage the greater is its ability to "push" the electrons

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 electrons in the form of an 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.

sometimes called Voltage Drop.

The Potential difference between two points is measured in Volts with the circuit

symbol V, or lowercase "v", although Energy, E lowercase "e" is sometimes used. Then

the greater the voltage, the greater is the pressure (or pushing force) and the greater is

the capacity to do work.

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, with one volt being

defined as the electrical pressure required to force an electrical current of one amperethrough a resistance of one Ohm. Voltages are generally expressed in Volts with

prefixes used to denote sub-multiples of the voltage such as microvolts ( V = 10-6 V ),

millivolts ( mV = 10-3 V ) orkilovolts ( kV = 103 V ). Voltage can be either positive or

negative.


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current flow. However, this arrow usually indicates the direction of conventional current

flow and not necessarily the direction of the actual flow.

Conventional Current Flow

Conventionally this is the flow of positive charge around a

circuit. The diagram at the left shows the movement of the

positive charge (holes) which flows from the positive terminal

of the battery, through the circuit and returns to the negative

terminal of the battery. This was the convention chosen during

the discovery of electricity in which the direction of electric

current was thought to flow in a circuit. In circuit diagrams, the

arrows shown on symbols for components such as diodes and transistors point in the

direction of conventional current flow. Conventional Current Flow is the opposite in

direction to the flow of electrons.

Electron Flow

The flow of electrons around the circuit is opposite to the

direction of the conventional current flow. The current flowing

in a circuit is composed of electrons that flow from the

negative pole of the battery (the cathode) and return to the

positive pole (the anode). This is because the charge on an

electron is negative by definition and so is attracted to thepositive terminal. The flow of electrons is called Electron

Current Flow. Therefore, electrons flow from the negative terminal to the positive.

Both conventional current flowand electron floware used by many textbooks. In fact, it

makes no difference which way the current is flowing around the circuit as long as the

direction is used consistently. The direction of current flow does not affect what the

current does within the circuit. Generally it is much easier to understand the

conventional current flow - positive to negative.

In electronic circuits, a current source is a circuit element that provides a specified

amount of current for example, 1A, 5A 10 Amps 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


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micro amps (A = 10-6A) ormilli amps (mA = 10-3A). Electrical current can be either

positive or negative.

Current that flows in a single direction is called Direct Current, orD.C. and current that

alternates back and forth through the circuit is known as Alternating Current, orA.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, ( R ) of a circuit is its ability to resist or prevent the flow of current

(electron flow) through itself making it necessary to apply a greater voltage to the

electrical circuit to cause the current to flow again. Resistance is measured in Ohms,Greek symbol ( , Omega ) with prefixes used to denote Kilo-ohms (k = 103) and

Mega-ohms (M = 106). Resistance cannot be negative only positive.

Resistor Symbols

The amount of resistance determines whether the circuit is a "good conductor" - low


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resistance, or a "bad conductor" - high resistance. Low resistance, for example 1 or

less implies that the circuit is a good conductor made from materials such as copper,

aluminium or carbon while a high resistance, 1M 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 such as silicon or germanium, is a material whoseresistance is half way between that of a good conductor and a good insulator.

Semiconductors are used to make Diodes and Transistors etc.

Resistance can be linear in nature or non-linear in nature. Linear resistance obeys

Ohm's Law and controls or limits the amount of current flowing within a circuit in

proportion to the voltage supply connected to it and therefore the transfer of power to

the load. Non-linear resistance, does not obey Ohm's Law but has a voltage drop

across it that is proportional to some power of the current. Resistance is pure and is not

affected by frequency with the AC impedance of a resistance being equal to its DC

resistance and as a result can not be negative. resistance is always positive. Also,

resistance is an attenuator which 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, (ms) it is sometimes more

easier to use the reciprocal of resistance (1/R) rather than resistance (R) itself. The

reciprocal of resistance is called Conductance, 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.
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Summary

Hopefully by know you have an idea of how voltage, current and resistance are related.

The relationship between Voltage, Current and Resistance forms the basis of Ohm's

law which in a linear circuit states that if we increase the voltage, the current goes up

and if we increase the resistance, the current goes down. 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".


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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 "" or the letter

"R".

Quantity SymbolUnit of

MeasureAbbreviation

Voltage V orE Volt V

Current I Amp A

Resistance R Ohms

In the next tutorial about DC Theory we will look at Ohms Law which is a mathematical

equation explaining the relationship between Voltage, Current, and Resistance within

electrical circuits and is the foundation of electronics and electrical engineering. Ohm's

Law is defined as: E = I x R.

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 Ohmfound 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 the Voltage, Current

and Resistance forms the bases ofOhms Law and is shown below.

Ohms Law Relationship

By knowing any two values of the Voltage, Current orResistance 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".
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To find Voltage (V)

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

To find Current (I)

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

To find Resistance (R)

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

It is sometimes easier to remember Ohms law relationship by using pictures. Here the

three quantities ofV, I and R have been superimposed into a triangle (affectionately

called the Ohms Law Triangle) giving voltage at the top with current and resistance atthe bottom. This arrangement represents the actual position of each quantity in the

Ohms law formulas.

Ohms Law Triangle

Then by using Ohms Law we can see that a voltage of 1V applied to a resistor of 1 will

cause a current of 1A to flow and the greater the resistance, the less current will flow for


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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 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) orkilowatts (kW = 103W).

By using Ohm's law and substituting forV, 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 ()

Also,

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

Again, the three quantities have been superimposed into a triangle this time called the

Power Triangle with power at the top and current and voltage at the bottom. Again, this

arrangement represents the actual position of each quantity in the Ohms law power

formulas.

The Power Triangle


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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 WATTbut some electrical devices such as

electric motors have a power rating in Horsepowerorhp. The relationship between

horsepower and watts is given as: 1hp = 746W.

Ohms Law Pie Chart

We can now take all the equations from above for finding Voltage, Current, Resistance

and Powerand condense them into a simple Ohms Law pie chart for use in DC

circuits and calculations.

Ohms Law Pie Chart


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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 12 = 24V

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

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

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


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current flow I = 0 (zero), therefore V x 0 is 0 so 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 ofV 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 ofHeat (heaters), Mechanical Work

such as motors, etc Energy in the form of radiated (Lamps) or as 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 orJoules.

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 givenin hours then the unit of energy is in kilowatt-hours orkWh which is commonly called a

"Unit of Electricity" and is what consumers purchase from their electricity suppliers.

Now that we know what is the relationship between voltage, current and resistance in a

circuit, in the next tutorial about DC Theory we will look at the Standard Electrical

Units used in electrical and electronic engineering to enable us to calculate these

values and see that each value can be represented by either multiples or sub-multiples

of the unit.
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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 [ ] 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 standard units used in

electrical formulas and component values.

Standard Electrical Units

Parameter SymbolMeasuring

UnitDescription

Voltage Volt V or E

Unit of Electrical Potential

V = I R

Current Ampere I or iUnit of Electrical Current

I = V R

Resistance Ohm R or Unit of DC Resistance

R = V I

Conductance Siemen G orReciprocal of Resistance

G = 1 R

Capacitance Farad CUnit of Capacitance

C = Q V

Charge Coulomb QUnit of Electrical Charge

Q = C V

Inductance Henry L or HUnit of Inductance

VL = -L(di/dt)

Power Watts WUnit of Power

P = V I

Impedance Ohm ZUnit of AC Resistance

Z2 = R2 + X2

Frequency Hertz Hz Unit of Frequency = 1 T

Multiples and Sub-multiples

There is a huge range of values encountered in electrical and electronic engineering

between a maximum value and a minimum value of a standard electrical unit. For


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example, resistance can be lower than 0.01's or higher than 1,000,000's. By using

multiples and submultiple's of the standard unit we can avoid having to write too many

zero's to define the position of the decimal point. The table below gives their names and

abbreviations.

Prefix Symbol Multiplier Power of Ten

Terra T 1,000,000,000,000 1012

Giga G 1,000,000,000 109

Mega M 1,000,000 106

kilo k 1,000 103

none none 1 10

0

centi c 1/100 10-2

milli m 1/1,000 10-3

micro 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.

47k = 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.


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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 andquicker 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 ofkWh (Kilowatt-hour) which is

1,000 watt-hours orMWh (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.

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.

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, symbol . This is a rotational unit of

angular frequency 2 with units in radians per second, rads/s. The

complete revolution of one cycle is 360 degrees or 2, therefore, half


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a revolution is given as 180 degrees or rad.

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 StepResponse input. It is a measure of reaction time.

In the next tutorial about DC Theory we will look at Kirchoff's Circuit Law which along

with Ohms Law allows us to calculate the different voltages and currents circulating

around a complex circuit.

Kirchoffs Circuit Law

We saw in the Resistors tutorial that a single equivalent resistance, ( RT ) can be found

when two or more resistors are connected together in either series, parallel or

combinations of both, and that these circuits obey Ohm's Law. However, sometimes in

complex circuits such as bridge or T networks, we can not simply use Ohm's Law alone

to find the voltages or currents circulating within the circuit. For these types of

calculations we need certain rules which allow us to obtain the circuit equations and for

this we can use Kirchoffs Circuit Law.

In 1845, a German physicist, Gustav Kirchoffdeveloped a pair or set of rules or laws

which deal with the conservation of current and energy within electrical circuits. The

rules are commonly known as: Kirchoffs Circuit Laws with one of these laws dealing

with current flow around a closed circuit, Kirchoffs Current Law, (KCL) and the other

which deals with the voltage around a closed circuit, Kirchoffs Voltage Law, (KVL).

Kirchoffs First Law - The Current Law, (KCL)

Kirchoffs Current Law or KCL, states that the "total current or charge entering ajunction or node is exactly equal to the charge leaving the node as it has no other place

to go except to leave, as no charge is lost within the node". In other words the algebraic

sum of ALL the currents entering and leaving a node must be equal to zero,

I(exiting) + I(entering) = 0. This idea by Kirchoff is known as the Conservation of Charge.

Kirchoffs Current Law
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Here, the 3 currents entering the node, I1, I2, I3 are all positive in value and the 2

currents leaving the node, I4 and I5 are negative in value. Then this means we can also

rewrite the equation as;

I1 + I2 + I3 - I4 - I5 = 0

The term Node in an electrical circuit generally refers to a connection or junction of two

or more current carrying paths or elements such as cables and components. Also for

current to flow either in or out of a node a closed circuit path must exist. We can use

Kirchoff's current law when analysing parallel circuits.

Kirchoffs Second Law - The Voltage Law, (KVL)

Kirchoffs Voltage Law or KVL, states that "in any closed loop network, the total

voltage around the loop is equal to the sum of all the voltage drops within the same

loop" which is also equal to zero. In other words the algebraic sum of all voltages withinthe loop must be equal to zero. This idea by Kirchoff is known as the Conservation of

Energy.

Kirchoffs Voltage Law


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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 startingpoint. It is important to maintain the same direction either clockwise or anti-clockwise or

the final voltage sum will not be equal to zero. We can use Kirchoff's voltage law when

analysing series circuits.

When analysing either DC circuits or AC circuits using Kirchoffs Circuit Laws a

number of definitions and terminologies are used to describe the parts of the circuit

being analysed such as: node, paths, branches, loops and meshes. These terms are

used frequently in circuit analysis so it is important to understand them.

Circuit - a circuit is a closed loop conducting path in which an electrical current

flows.

Path - a line of connecting elements or sources with no elements or sources

included more than once.

Node - a node is a junction, connection or terminal within a circuit were two or

more circuit elements are connected or joined together giving a connection point

between two or more branches. A node is indicated by a dot.

Branch - a branch is a single or group of components such as resistors or a

source which are connected between two nodes.

Loop - a loop is a simple closed path in a circuit in which no circuit element or

node is encountered more than once.

Mesh - a mesh is a single open loop that does not have a closed path. No

components are inside a mesh.


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Components are connected in series if they carry the same current.

Components are connected in parallel if the same voltage is across them.

Example No1

Find the current flowing in the 40 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 nodeA : I1 + I2 = I3


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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 ofI1 + I2 we can rewrite the equations as;

Eq. No 1 : 10 = 10I1 + 40(I1 + I2) = 50I1 + 40I2

Eq. No 2 : 20 = 20I2 + 40(I1 + I2) = 40I1 + 60I2

We now have two "Simultaneous Equations" that can be reduced to give us the value

of both I1 and I2

Substitution ofI1 in terms ofI2 gives us the value ofI1 as -0.143 Amps

Substitution ofI2 in terms ofI1 gives us the value ofI2 as +0.429 Amps

As : I3 = I1 + I2

The current flowing in resistorR3 is given as : -0.143 + 0.429 = 0.286 Amps

and the voltage across the resistorR3 is given as : 0.286 x 40 = 11.44 volts

The negative sign forI1 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 circuitis said to be "Analysed", and the basic procedure for using Kirchoff's 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)


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3. Find Kirchoff's first law equations for each node.

4. Find Kirchoff's second law equations for each of the independent loops of the

circuit.

5. Use Linear simultaneous equations as required to find the unknown currents.

As well as using Kirchoffs Circuit Law to calculate the voltages and currents

circulating around a linear circuit, we can also use loop analysis to calculate the

currents in each independent loop helping to reduce the amount of mathematics

required using just Kirchoffs laws. In the next tutorial about DC Theory we will look at

Mesh Current Analysis to do just that.

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 Kirchoffs Laws give us the basic

method for analysing any complex electrical circuit, there are different ways of

improving upon this method by using Mesh Current Analysis orNodal 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 analyse the circuit

using Kirchoff's Current Law equations to determine the currents, I1and I2 flowing in the

two resistors. Then there is no need to calculate the current I3 as its just the sum of

I1and I2. So Kirchoff's second voltage law simply becomes:

Equation No 1 : 10 = 50I1 + 40I2


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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

orLoop Analysis which is also sometimes called Maxwells 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 Kirchoffs method.

For example: : i1 = I1 , i2 = -I2 and I3 = I1 - I2

We now write Kirchoff's voltage law equation in the same way as before to solve them

but the advantage of this method is that it ensures that the information obtained from

the circuit equations is the minimum required to solve the circuit as the information is

more general and can easily be put into a matrix form.


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:


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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 Kirchoffs circuit law in the

previous tutorial.

Mesh Current Analysis Summary.

This "look-see" method of circuit analysis is probably the best of all the circuit analysis

methods with the basic procedure for solving Mesh Current Analysis equations is as

follows:

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.


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As well as using Mesh Current Analysis, we can also use node analysis to calculate

the voltages around the loops, again reducing the amount of mathematics required

using just Kirchoff's laws. In the next tutorial about DC Theory we will look at Nodal

Voltage Analysis to do just that.

Nodal Voltage Analysis

As well as using Mesh Analysis to solve the currents flowing around complex circuits it

is also possible to use nodal analysis methods too. Nodal Voltage Analysis

complements the previous mesh analysis in that it is equally powerful and based on the

same concepts of matrix analysis. As its name implies, Nodal Voltage Analysis uses

the "Nodal" equations of Kirchoff's first law to find the voltage potentials around the

circuit. By adding together all these nodal voltages the net result will be equal to zero.

Then, if there are "N" nodes in the circuit there will be "N-1" independent nodal

equations and these alone are sufficient to describe and hence solve the circuit.

At each node point write down Kirchoff's first law equation, that is: "the currents entering

a node are exactly equal in value to the currents leaving the node " then express each

current in terms of the voltage across the branch. For "N" nodes, one node will be used

as the reference node and all the other voltages will be referenced or measured with

respect to this common node.


Nodal Voltage Analysis Circuit

In the above circuit, node D is chosen as the reference node and the other three nodes

are assumed to have voltages, Va, Vb and Vc with respect to node D. For example;
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As Va = 10v and Vc = 20v , Vb can be easily found by:

again is the same value of0.286 amps, we found using Kirchoff's Circuit Law in the

previous tutorial.

From both Mesh and Nodal Analysis methods we have looked at so far, this is the

simplest method of solving this particular circuit. Generally, nodal voltage analysis is

more appropriate when there are a larger number of current sources around. The

network is then defined as: [ I ] = [ Y ] [ V ] where [ I ] are the driving current sources, [ V

] are the nodal voltages to be found and [ Y ] is the admittance matrix of the network

which operates on [ V ] to give [ I ].

Nodal Voltage Analysis Summary.

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

1. Write down the current vectors, assuming currents into a node are positive. ie,

a (N x 1) matricesfor "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.


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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.

We have now seen that a number of theorems exist that simplify the analysis of linear

circuits. In the next tutorial we will look at Thevenins Theorem which allows a network

consisting of linear resistors and sources to be represented by an equivalent circuit with

a single voltage source and a series resistance.

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 Kirchoffs) 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.
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As far as the load resistorRL is concerned, any "one-port" network consisting of

resistive circuit elements and energy sources can be replaced by one single equivalent

resistance Rs and equivalent voltage Vs, where Rs is the source resistance value

looking back into the circuit and Vs is the open circuit voltage at the terminals.


Firstly, we have to remove the centre 40 resistor and short out (not physically as this

would be dangerous) all the emfs connected to the circuit, or open circuit any current

sources. The value of resistorRs is found by calculating the total resistance at the

terminalsA and B with all the emfs removed, and the value of the voltage required Vs

is the total voltage across terminalsA and B with an open circuit and no load resistorRs

connected. Then, we get the following circuit.


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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 20 resistor can be calculated as:


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

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

and from this the current flowing in the circuit is given as:

which again, is the same value of0.286 amps, we found usingKirchoffs 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 orNodal

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 ofTransistors, Voltage Sources such as

batteries etc, are very useful in circuit design.

Thevenins Theorem Summary

The basic procedure for solving a circuit using Thevenins Theorem is as follows:

1. Remove the load resistorRL 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 resistorRL.
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In the next tutorial we will look at Nortons Theorem which allows a network consisting

of linear resistors and sources to be represented by an equivalent circuit with a single

current source in parallel with a single source resistance.

Nortons Theorem

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 resistancelooking 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.


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To find the Nortons equivalent of the above circuit we firstly have to remove the centre

40 load resistor and short out the terminalsA and B to give us the following circuit.

When the terminalsA 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 terminalsA and B, the two

resistors are now effectively connected together in parallel. The value of the internal


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resistorRs is found by calculating the total resistance at the terminalsA 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 40 load resistor

connected across terminalsA and B as shown below.


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Again, the two resistors are connected in parallel across the terminalsA and B which

gives us a total resistance of:

The voltage across the terminalsA and B with the load resistor connected is given as:

Then the current flowing in the 40 load resistor can be found as:

which again, is the same value of0.286 amps, we found usingKirchoffs circuit law in

the previous tutorials.

Nortons Theorem Summary

The basic procedure for solving a circuit using Nortons Theorem is as follows:

1. Remove the load resistorRL 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 terminalsA and B.
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4. Find the current flowing through the load resistorRL.

In a circuit, power supplied to the load is at its maximum when the load resistance is

equal to the source resistance. In the next tutorial we will look at Maximum Power

Transfer. The application of the maximum power transfer theorem can be applied toeither simple and complicated linear circuits having a variable load and is used to find

the load resistance that leads to transfer of maximum power to the load.

Maximum Power Transfer

We have seen in the previous tutorials that any complex circuit or network can be

replaced by a single energy source in series with a single internal source resistance, RS.

Generally, this source resistance or even impedance if inductors or capacitors areinvolved is of a fixed value in Ohms. 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 ofMaximum Power

Transfer.

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.


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Thevenins Equivalent Circuit.

In our Thevenin equivalent circuit above, the maximum power transfer theorem states

that "the maximum amount of power will be dissipated in the load resistance if it isequal 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 = 25RL is variable between 0 - 100VS = 100v

Then by using the following Ohm's Law equations:


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We can now complete the following table to determine the current and power in the

circuit for different values of load resistance.

Table of Current against Power

RL I P

0 0 0

5 3.3 55

10 2.8 78

15 2.5 93

20 2.2 97

RL I P

25 2.0 100

30 1.8 97

40 1.5 94

60 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 Transferoccurs

in the load when the load resistance, RL is equal in value to the source resistance, RS sothen: RS = RL = 25. 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.


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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 isachieved by using a matching transformerto couple the load to the amplifiers output as

shown below.

Transformer Coupling

The maximum power transfer can be obtained even if the output impedance is not the

same as the load impedance. This can be done using a suitable "turns ratio" on the

transformer with the corresponding ratio of load impedance, ZLOAD to output impedance,

ZOUT matches that of the ratio of the transformers primary turns to secondary turns as a

resistance on one side of the transformer becomes a different value on the other. If the

load impedance, ZLOAD is purely resistive and the source impedance is purely resistive,

ZOUT then the equation for finding the maximum power transfer is given as:

Where: NP is the number of primary turns and NS the number of secondary turns on the

transformer. Then by varying the value of the transformers turns ratio the output

impedance can be "matched" to the source impedance to achieve maximum power

transfer. For example,

Example No2.

If an 8 loudspeaker is to be connected to an amplifier with an output impedance of

1000, calculate the turns ratio of the matching transformer required to provide


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maximum power transfer of the audio signal. Assume the amplifier source impedance is

Z1, the load impedance is Z2 with the turns ratio given as N.

Generally, small transformers used in low power audio amplifiers are usually regarded

as ideal so any losses can be ignored.

In the next tutorial about DC Theory we will look at Star Delta Transformation which

allows us to convert balanced star connected circuits into equivalent delta and vice

versa

Star Delta Transformation

We can now solve simple series, parallel or bridge type resistive networks using

Kirchoffs 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, (wye) and a Delta connected

network which has the symbol of a triangle, (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


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transformed or changed it into an equivalent configuration of the other type by using

either the Star Delta Transformation orDelta 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 Staror typenetwork 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 Staror type network. But we can also convert a Pi or type resistor

network into an equivalent Delta or type network as shown below.

Pi-connected and Equivalent Delta Network.

Having now defined exactly what is a Starand Delta connected network it is possible to

transform the into an equivalent circuit and also to convert a into an equivalent


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circuit using a the transformation process. This process allows us to produce a

mathematical relationship between the various resistors giving us a Star Delta

Transformation as well as a Delta Star Transformation.

These transformations allow us to change the three connected resistances by theirequivalents measured between the terminals 1-2, 1-3 or2-3 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.


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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:


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From which gives us the final equation for resistorP as:

Then to summarize a little the above maths, we can now say that resistorP in a Star

network can be found as Equation 1 plus (Equation 3 minus Equation 2) or Eq1 + (Eq3

- Eq2).

Similarly, to find resistorQ 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 ofQ as:

and again, to find resistorR 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 ofR 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


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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 resistorP is the product of

resistorsA 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.


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Star to Delta Network.

The value of the resistor on any one side of the delta, 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, resistorA isgiven as:

with respect to terminal 3 and resistorB is given as:

with respect to terminal 2 with resistorC 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.


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Star Delta Transformation Equations

Star Delta Transformations allow us to convert one circuit type of circuit connection to

another in order for us to easily analyise a circuit and one final point about converting a

star resistive network to an equivalent delta network. If all the resistors in the star

network are all equal in value then the resultant resistors in the equivalent delta network

will be three times the value of the star resistors and equal, giving: RDELTA = 3RSTAR

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