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Voltage, Current and Resistance Navigation
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Relationship Between Voltage, Current and Resistance
The fundamental relationship between voltage, current and resistance in an electrical circuit is called Ohm's Law. 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 by powerful 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. But if we separate them from each
other they want to reform and start to exert a potential of attraction called a potential difference.
Now if we create a closed circuit these loose electrons will start to move and drift back to the protons due to their attraction creating a flow of
electrons. This flow of electrons is called an electrical current. The electrons do not flow freely through the circuit as the material they move
through creates a restriction to the electron flow. This restriction is called resistance.
Then all basic electrical or electronic circuits consist of three separate but very much related electrical quantities called: Voltage, ( v ),
Current, ( i )and Resistance, ( Ω ).
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.
Then the difference in voltage between any two points, connections or junctions (called nodes) in a circuit is known as the Potential
Difference, ( p.d. ) sometimes called the Voltage Drop.
The Potential difference between two points is measured in Volts with the circuit symbol V, or lowercase "v", although Energy, Elowercase
"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 ampere through a
resistance of one Ohm. Voltages are generally expressed in Volts with prefixes used to denote sub-multiples of the voltage such as
microvolts( μV = 10-6
V ), millivolts( mV = 10-3
V ) or kilovolts( kV = 103 V ). 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.
General electronic circuits operate on low voltage DC battery supplies 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.
Voltage Symbols
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.
Electrical Current
Electrical Current, ( I ) is the movement or flow of electrical charge and is measured in Amperes, symbol i, for intensity). It is the
continuous and uniform flow (called a drift) of electrons (the 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.
Conventional Current Flow
Conventionally this is the flow of positive charge around a circuit, positive to negative. 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. To
continue with this line of thought, in all circuit diagrams and schematics, the arrows shown on symbols for components such as diodes and
transistors point in the direction of conventional current flow.
Then Conventional Current Flow gives the flow of electrical current from positive to negative and which is the opposite in direction to the
actual flow of electrons.
Electron Flow
The flow of electrons around the circuit is opposite to the direction of the conventional current flow being
negative to positive. The actual current flowing in an electrical circuit is composed of electrons that flow
from the negative pole of the battery (the cathode) and return back to the positive pole (the anode) of the
battery.
This is because the charge on an electron is negative by definition and so is attracted to the positive
terminal. This flow of electrons is called Electron Current Flow. Therefore, electrons actually flow around
a circuit from the negative terminal to the positive.
Both conventional current flow and electron flow are 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 micro amps ( μA = 10-6
A ) or milli amps( mA = 10-3
A
). Note that electrical current can be either positive in value or negative in value depending upon its direction of flow.
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, ( 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 ( Ω,Omega )
with prefixes used to denote Kilo-ohms ( kΩ = 103Ω )and Mega-ohms ( MΩ = 10
6Ω ). Note that Resistance cannot be negative in value
only positive.
Resistor Symbols
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 such as silicon or germanium, is a material whose resistance is half way between that of a good
conductor and a good insulator. Semiconductors are used to makeDiodes and Transistors etc.
Resistance can be linear in nature or non-linear in nature. Linear resistance obeysOhm'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 obeyOhm'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.
Remember that resistance is always positive, and never negative.
Resistance can also be classed as an attenuator as it has the ability to change the characteristics of a circuit by the effect of loading the
circuit 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 called Conductance, symbol ( G ) and represents the ability of a conductor
or device to conduct electricity. In other words the ease by which current flows. High values of conductance implies a good conductor such
as copper while low values of conductance implies a bad conductor such as wood. The standard unit of measurement given for 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.
The relationship between Voltage, ( v ) and Current, ( i ) in a circuit of constantResistance, ( R ).
Voltage, Current and Resistance Summary
Hopefully by now you should have some idea of how electrical Voltage,Current and Resistance are closely related together. 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. Then we can see that current flow around a circuit is
directly proportional ( ∝ ) to voltage, ( V↑ causes I↑ ) but inversely proportional ( 1/∝ ) to resistance as, ( R↑ causes I↓ ).
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 Ampere A
Resistance R Ohms Ω
In the next tutorial about DC Theory we will look atOhms 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 Navigation
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Ohms Law
The relationship between Voltage, Current andResistance in any DC electrical circuit was firstly discovered by the German
physicistGeorg 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
the Voltage,Current and Resistance forms the bases of Ohms Lawand is shown below.
Ohms Law Relationship
By knowing any two values of the Voltage,Current or Resistance quantities we can use Ohms Lawto 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 the Voltage, ( V )
[ V = I x R ] V (volts) = I (amps) x R (Ω)
To find the Current, ( I )
[ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)
To find the 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 of V, I and R have been
superimposed into a triangle (affectionately called the Ohms Law Triangle) giving voltage at the top with current and resistance at the
bottom. This arrangement represents the actual position of each quantity in the Ohms law formulas.
Ohms Law Triangle
and transposing the above Ohms Law equation gives us the following combinations of the same equation:
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 denotemilliwatts (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 the 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)
Again, the three quantities have been superimposed into a triangle this time called thePower 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
and again, transposing the basic Ohms Law equation above for power gives us the following combinations of the same equation to find the
various individual quantities:
One other point about Power, if the calculated power is positive in value for any formula the component absorbs the power, that is it is
consuming or using power. But if the calculated power is negative in value the component produces or generates 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 the old
measurement of "Horsepower" or hp. The relationship between horsepower and watts is given as: 1hp = 746W. So for example, a two-
horsepower motor has a rating of 1492W, (2 x 746) or 1.5kW.
Ohms Law Pie Chart
To help us understand the the relationship between the various values a little futher, we can take all of Ohm's Law equations from above for
finding Voltage, Current,Resistance and Power and condense them into a simpleOhms Law pie chart for use in AC and DC circuits and
calculations as shown.
Ohms Law Pie Chart
As well as using the Ohm's Law Pie Chart shown above, we can also put the individual Ohm's Law equations into a simple matrix table as
shown for easy reference when calculating an unknown value.
Ohms Law Matrix Table
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 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 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 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 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) or megajoules (MJ = 10
6J) 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.
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 theStandard 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.
Electrical Units of Measure Navigation
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Electrical Units of Measure
The standard SI units used for the measurement of voltage, current and resistance are theVolt [ 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 standard units
used in electrical formulas and component values.
Standard Electrical Units
Parameter Symbol Measuring
Unit Description
Voltage Volt V or E Unit of Electrical Potential
V = I × R
Current Ampere I or i Unit of Electrical Current
I = V ÷ R
Resistance Ohm R or Ω Unit of DC Resistance
R = V ÷ I
Conductance Siemen G or ℧ Reciprocal of Resistance
G = 1 ÷ R
Capacitance Farad C Unit of Capacitance
C = Q ÷ V
Charge Coulomb Q Unit of Electrical Charge
Q = C × V
Inductance Henry L or H Unit of Inductance
VL = -L(di/dt)
Power Watts W Unit of Power
P = V × I or I2 × R
Impedance Ohm Z Unit of AC Resistance
Z2 = R
2 + X
2
Frequency Hertz Hz Unit of Frequency
ƒ = 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 example, resistance can be lower than 0.01Ω's or higher than 1,000,000Ω'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 1 100
centi c 1/100 10-2
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 1,000 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 1,000 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, symbol ω.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 orπ 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.
In the next tutorial about DC Theory we will look atKirchoff'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 Navigation
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Kirchoffs Circuit Law
We saw in the Resistorstutorial that a single equivalent resistance, ( RT ) can be found when two or more resistors are connected together
in either series, parallel or combinations of both, and that these circuits obey Ohm's Law. However, sometimes in complex circuits such as
bridge or T networks, we can not simply use Ohm's Law alone to find the voltages or currents circulating within the circuit. For these types of
calculations we need certain rules which allow us to obtain the circuit equations and for this we can use Kirchoffs Circuit Law.
In 1845, a German physicist, Gustav Kirchoff developed a pair or set of rules or laws which deal with the conservation of current and
energy within electrical circuits. These two rules are commonly known as: Kirchoffs Circuit Laws with one of Kirchoffs laws dealing with the
current flowing around a closed circuit,Kirchoffs Current Law, (KCL) while the other law deals with the voltage sources present in a closed
circuit,Kirchoffs Voltage Law, (KVL).
Kirchoffs First Law - The Current Law, (KCL)
Kirchoffs Current Law or KCL, states that the "total current or charge entering a junction or node is exactly equal to the charge leaving the
node as it has no other place to go except to leave, as no charge is lost within the node". In other words the algebraic sum of ALL the
currents entering and leaving a node must be equal to zero, I(exiting) + I(entering) = 0. This idea by Kirchoff is commonly known as the
Conservation of Charge.
Kirchoffs Current Law
Here, the 3 currents entering the node, I1, I2, I3are all positive in value and the 2 currents leaving the node, I4 and I5 are negative in value.
Then this means we can also rewrite the equation as;
I1 + I2 + I3 - I4 - I5 = 0
The term Node in an electrical circuit generally refers to a connection or junction of two or more current carrying paths or elements such as
cables and components. Also for current to flow either in or out of a node a closed circuit path must exist. We can use Kirchoff's current law
when analysing parallel circuits.
Kirchoffs Second Law - The Voltage Law, (KVL)
Kirchoffs Voltage Law or KVL, states that "in any closed loop network, the total voltage around the loop is equal to the sum of all the
voltage drops within the same loop" which is also equal to zero. In other words the algebraic sum of all voltages within the loop must be
equal to zero. This idea by Kirchoff is known as theConservation of Energy.
Kirchoffs Voltage Law
Starting at any point in the loop continue in the same direction noting the direction of all the voltage drops, either positive or negative, and
returning back to the same starting point. It is important to maintain the same direction either clockwise or anti-clockwise or the final voltage
sum will not be equal to zero. We can use Kirchoff's voltage law when analysing series circuits.
When analysing either DC circuits or AC circuits using Kirchoffs Circuit Laws a number of definitions and terminologies are used to
describe the parts of the circuit being analysed such as: node, paths, branches, loops and meshes. These terms are used frequently in
circuit analysis so it is important to understand them.
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.
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 40Ω Resistor, R3
The circuit has 3 branches, 2 nodes (A and B) and 2 independent loops.
Using Kirchoffs Current Law, KCL the equations are given as;
At node A : I1 + I2 = I3
At node B : I3 = I1 + I2
Using Kirchoffs Voltage Law, KVL the equations are given as;
Loop 1 is given as : 10 = R1 x I1 + R3 x I3 = 10I1 + 40I3
Loop 2 is given as : 20 = R2 x I2 + R3 x I3 = 20I2 + 40I3
Loop 3 is given as : 10 - 20 = 10I1 - 20I2
As I3 is the sum of I1 + I2 we can rewrite the equations as;
Eq. No 1 : 10 = 10I1 + 40(I1 + I2) = 50I1 + 40I2
Eq. No 2 : 20 = 20I2 + 40(I1 + I2) = 40I1 + 60I2
We now have two "Simultaneous Equations" that can be reduced to give us the value of bothI1 and I2
Substitution of I1 in terms of I2gives us the value of I1 as -0.143 Amps
Substitution of I2 in terms of I1gives us the value of I2 as +0.429 Amps
As : I3 = I1 + I2
The current flowing in resistor R3 is given as : -0.143 + 0.429 = 0.286 Amps
and the voltage across the resistor R3 is given as : 0.286 x 40 = 11.44 volts
The negative sign for I1 means that the direction of current flow initially chosen was wrong, but never the less still valid. In fact, the 20v
battery is charging the 10v battery.
Application of Kirchoffs Circuit Laws
These two laws enable the Currents and Voltages in a circuit to be found, ie, the circuit is said to be "Analysed", and the basic procedure
for using Kirchoff's Circuit Laws is as follows:
1. Assume all voltages and resistances are given. ( If not label them V1, V2,... R1, R2, etc. )
2. Label each branch with a branch current. ( I1, I2, I3 etc. )
3. Find Kirchoff's first law equations for each node.
4. Find Kirchoff's second law equations for each of the independent loops of the circuit.
5. Use Linear simultaneous equations as required to find the unknown currents.
As well as using Kirchoffs Circuit Law to calculate the various voltages and currents circulating around a linear circuit, we can also use
loop analysis to calculate the currents in each independent loop which helps to reduce the amount of mathematics required by using just
Kirchoff's laws. In the next tutorial aboutDC Theory we will look atMesh Current Analysis to do just that.
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Circuit Analysis
In the previous tutorial we saw that complex circuits such as bridge or T-networks can be solved usingKirchoff's Circuit Laws. While
Kirchoff´s Laws give us the basic method for analysing any complex electrical circuit, there are different ways of improving upon this method
by using Mesh Current Analysis 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, I1 and I2flowing in the two resistors. Then there is no need to calculate the current I3 as its just the sum of I1 and I2. So
Kirchoff's second voltage law simply becomes:
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 Analysis which is also sometimes called
Maxwell´s Circulating Currents method. Instead of labelling the branch currents we need to label each "closed loop" with a circulating
current. As a general rule of thumb, only label inside loops in a clockwise direction with circulating currents as the aim is to cover all the
elements of the circuit at least once. Any required branch current may be found from the appropriate loop or mesh currents as before using
Kirchoff´s method.
For example: : i1 = I1 , i2 = -I2 and I3 = I1 - I2
We now write Kirchoff's voltage law equation in the same way as before to solve them but the advantage of this method is that it ensures
that the information obtained from the circuit equations is the minimum required to solve the circuit as the information is more general and
can easily be put into a matrix form.
For example, consider the circuit from the previous section.
These equations can be solved quite quickly by using a single mesh impedance matrix Z. Each element ON the principal diagonal will be
"positive" and is the total impedance of each mesh. Where as, each element OFF the principal diagonal will either be "zero" or "negative"
and represents the circuit element connecting all the appropriate meshes. 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 andI2 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 usingKirchoff´s circuit law in the previous tutorial.
Mesh Current Analysis Summary.
This "look-see" method of circuit analysis is probably the best of all the circuit analysis methods with the basic procedure for solving Mesh
Current Analysis equations is as follows:
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
o R11 = the total resistance in the first loop.
o
o Rnn = the total resistance in the Nth loop.
o
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.
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 Theorywe will look at Nodal Voltage Analysis
to do just that.
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Nodal Voltage Analysis
As well as using Mesh Analysis to solve the currents flowing around complex circuits it is also possible to use nodal analysis methods
too.Nodal Voltage Analysis complements the previous mesh analysis in that it is equally powerful and based on the same concepts of
matrix analysis. As its name implies, Nodal Voltage Analysis uses the "Nodal" equations of Kirchoff's first law to find the voltage potentials
around the circuit.
So by adding together all these nodal voltages the net result will be equal to zero. Then, if there are "n" nodes in the circuit there will be "n-1"
independent nodal equations and these alone are sufficient to describe and hence solve the circuit.
At each node point write down Kirchoff's first law equation, that is: "the currents entering a node are exactly equal in value to the currents
leaving the node" then express each current in terms of the voltage across the branch. For "n" nodes, one node will be used as the
reference node and all the other voltages will be referenced or measured with respect to this common node.
For example, consider the circuit from the previous section.
Nodal Voltage Analysis Circuit
In the above circuit, node D is chosen as the reference node and the other three nodes are assumed to have voltages, Va, Vb and Vc with
respect to node D. For example;
As Va = 10v andVc = 20v , Vb can be easily found by:
again is the same value of 0.286 amps, we found usingKirchoff'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)
matrices for "N" independent nodes.
2. Write the admittance matrix [Y] of the network where:
o
o Y11 = the total admittance of the first node.
o
o Y22 = the total admittance of the second node.
o
o RJK = the total admittance joining node Jto 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.
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.
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Thevenins Theorem
In the previous 3 tutorials we have looked at solving complex electrical circuits usingKirchoff's Circuit Laws,Mesh Analysis and
finallyNodal 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 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.
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 resistor Rs 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 Aand 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 usingKirchoff´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 asMesh 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 of Transistors, 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 resistor RL or component concerned.
2. Find RS by shorting all voltage sources or by open circuiting all the current sources.
3. Find VS by the usual circuit analysis methods.
4. Find the current flowing through the load resistor RL.
In the next tutorial we will look at Nortons Theoremwhich 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.
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Nortons Theorem
In some ways Norton's Theorem can be thought of as the opposite to "Thevenins Theorem", in that Thevenin reduces his circuit down to a
single resistance in series with a single voltage. Norton on the other hand reduces his circuit down to a single resistance in parallel with a
constant current source. Nortons Theorem states that "Any linear circuit containing several energy sources and resistances can be
replaced by a single Constant Current generator in parallel with a Single Resistor".
As far as the load resistance, RL is concerned this single resistance,RS is the value of the resistance looking back into the network with all
the current sources open circuited and IS is the short circuit current at the output terminals as shown below.
Nortons equivalent circuit.
The value of this "constant current" is one which would flow if the two output terminals where shorted together while the source resistance
would be measured looking back into the terminals, (the same as Thevenin).
For example, consider our now familiar circuit from the previous section.
To find the Nortons equivalent of the above circuit we firstly have to remove the centre40Ω load resistor and short out the terminals A andB
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 Aand 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 40Ωload 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 usingKirchoff´s 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 resistor RL or component concerned.
2. Find RS by shorting all voltage sources or by open circuiting all the current sources.
3. Find IS by placing a shorting link on the output terminals A and B.
4. Find the current flowing through the load resistor RL.
In a circuit, power supplied to the load is at its maximum when the load resistance is equal to the source resistance. In the next tutorial we
will look at Maximum Power Transfer. The application of the maximum power transfer theorem can be applied to either simple and
complicated linear circuits having a variable load and is used to find the load resistance that leads to transfer of maximum power to the load.
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Maximum Power Transfer
We have seen in the previous tutorials that any complex circuit or network can be replaced by a single energy source in series with a single
internal source resistance, RS. Generally, this source resistance or even impedance if inductors or capacitors are involved is of a fixed value
in Ohm´s. However, when we connect a load resistance, RL across the output terminals of the power source, the impedance of the load will
vary from an open-circuit state to a short-circuit state resulting in the power being absorbed by the load becoming dependent on the
impedance of the actual power source. Then for the load resistance to absorb the maximum power possible it has to be "Matched" to the
impedance of the power source and this forms the basis 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.
Thevenins Equivalent Circuit.
In our Thevenin equivalent circuit above, the maximum power transfer theorem states that "the maximum amount of power will be dissipated
in the load resistance if it is equal in value to the Thevenin or Norton source resistance of the network supplying the power".
In other words, the load resistance resulting in greatest power dissipation must be equal in value to the equivalent Thevenin source
resistance, then RL = RS but if the load resistance is lower or higher in value than the Thevenin source resistance of the network, its
dissipated power will be less than maximum. For example, find the value of the load resistance, RLthat will give the maximum power transfer
in the following circuit.
Example No1.
Where:
RS = 25Ω
RL is variable between 0 - 100Ω
VS = 100v
Then by using the following Ohm's Law equations:
We can now complete the following table to determine the current and power in the circuit for different values of load resistance.
Table of Current against Power
RL (Ω) I (amps) P (watts)
0 0 0
5 3.3 55
10 2.8 78
15 2.5 93
20 2.2 97
RL (Ω) I (amps) P (watts)
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, RLagainst power, P for different values of load resistance. Also
notice that power is zero for an open-circuit (zero current condition) and also for a short-circuit (zero voltage condition).
Graph of Power against Load Resistance
From the above table and graph we can see that the Maximum Power Transfer occurs in the load when the load resistance, RL is equal in
value to the source resistance,RS that is: RS = RL = 25Ω.This is called a "matched condition" and as a general rule, maximum power is
transferred from an active device such as a power supply or battery to an external device when the impedance of the external device exactly
matches the impedance of the source.
One good example of impedance matching is between an audio amplifier and a loudspeaker. The output impedance, ZOUT of the amplifier
may be given as between 4Ωand 8Ω, while the nominal input impedance, ZIN of the loudspeaker may be given as 8Ω only. Then if the 8Ω
speaker is attached to the amplifiers output, the amplifier will see the speaker as an 8Ω load. Connecting two 8Ω speakers in parallel is
equivalent to the amplifier driving one 4Ωspeaker and both configurations are within the output specifications of the amplifier.
Improper impedance matching can lead to excessive power loss and heat dissipation. But how could you impedance match an amplifier and
loudspeaker which have very different impedances. Well, there are loudspeaker impedance matching transformers available that can
change impedances from 4Ω to 8Ω,or to 16Ω's to allow impedance matching of many loudspeakers connected together in various
combinations such as in PA (public address) systems.
Transformer Impedance Matching
One very useful application of impedance matching in order to provide maximum power transfer between the source and the load is in the
output stages of amplifier circuits. Signal transformers are used to match the loudspeakers higher or lower impedance value to the amplifiers
output impedance to obtain maximum sound power output. These audio signal transformers are called "matching transformers" and couple
the load to the amplifiers output as shown below.
Transformer Impedance Matching
The maximum power transfer can be obtained even if the output impedance is not the same as the load impedance. This can be done using
a suitable "turns ratio" on the transformer with the corresponding ratio of load impedance, ZLOAD to output impedance, ZOUTmatches 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, ZLOADis 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 andNS 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 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 atStar Delta Transformation which allows us to convert balanced star connected
circuits into equivalent delta and vice versa.
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Star Delta Transformation
We can now solve simple series, parallel or bridge type resistive networks usingKirchoff´s Circuit Laws, mesh current analysis or nodal
voltage analysis techniques but in a balanced 3-phase circuit we can use different mathematical techniques to simplify the analysis of the
circuit and thereby reduce the amount of math's involved which in itself is a good thing.
Standard 3-phase circuits or networks take on two major forms with names that represent the way in which the resistances are connected, a
Star connected network which has the symbol of the letter, Υ (wye) and a Delta connected network which hasthe symbol of a triangle, Δ
(delta). If a 3-phase, 3-wire supply or even a 3-phase load is connected in one type of configuration, it can be easily transformed or changed
it into an equivalent configuration of the other type by using either the Star Delta Transformation orDelta Star Transformation process.
A resistive network consisting of three impedances can be connected together to form aT or "Tee" configuration but the network can also be
redrawn to form aStar or Υ type network as shown below.
T-connected and Equivalent Star Network
As we have already seen, we can redraw the T resistor network to produce an equivalent Star or Υ type network. But we can also convert a
Pi or π type resistor network into an equivalent Delta or Δ type network as shown below.
Pi-connected and Equivalent Delta Network.
Having now defined exactly what is a Star and Delta connected network it is possible to transform the Υ into an equivalent Δcircuit and also
to convert a Δ into an equivalent Υcircuit 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 aDelta Star Transformation.
These transformations allow us to change the three connected resistances by their equivalents measured between the terminals 1-2, 1-3 or
2-3 for either a star or delta connected circuit. However, the resulting networks are only equivalent for voltages and currents external to the
star or delta networks, as internally the voltages and currents are different but each network will consume the same amount of power and
have the same power factor to each other.
Delta Star Transformation
To convert a delta network to an equivalent star network we need to derive a transformation formula for equating the various resistors to
each other between the various terminals. Consider the circuit below.
Delta to Star Network.
Compare the resistances between terminals 1 and 2.
Resistance between the terminals 2 and 3.
Resistance between the terminals 1 and 3.
This now gives us three equations and taking equation 3 from equation 2 gives:
Then, re-writing Equation 1 will give us:
Adding together equation 1 and the result above of equation 3 minus equation 2 gives:
From which gives us the final equation for resistor P as:
Then to summarize a little the above maths, we can now say that resistor Pin a Star network can be found as Equation 1 plus (Equation 3
minus Equation 2) or Eq1 + (Eq3 - Eq2).
Similarly, to find resistor Q in a star network, is equation 2 plus the result of equation 1 minus equation 3 or Eq2 + (Eq1 - Eq3) and this
gives us the transformation of Q as:
and again, to find resistor R in a Star network, is equation 3 plus the result of equation 2 minus equation 1 or Eq3 + (Eq2 - Eq1) and this
gives us the transformation of R as:
When converting a delta network into a star network the denominators of all of the transformation formulas are the same: A + B + C, and
which is the sum of ALL the delta resistances. Then to convert any delta connected network to an equivalent star network we can
summarized the above transformation equations as:
Delta to Star Transformations Equations
If the three resistors in the delta network are all equal in value then the resultant resistors in the equivalent star network will be equal to one
third the value of the delta resistors, giving each branch in the star network as: RSTAR = 1/3RDELTA
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 andB connected
to terminal 1.
By rewriting the previous formulas a little we can also find the transformation formulas for converting a resistive star network to an equivalent
delta network giving us a way of producing a star delta transformation as shown below.
Star to Delta Network.
The value of the resistor on any one side of the delta, Δnetwork 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 Ais given as:
with respect to terminal 3 and resistor B is given as:
with respect to terminal 2 with resistor C given as:
with respect to terminal 1.
By dividing out each equation by the value of the denominator we end up with three separate transformation formulas that can be used to
convert any Delta resistive network into an equivalent star network as given below.
Star Delta Transformation Equations
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