thevenins theorem
TRANSCRIPT
Thevenins Theorem Navigation
Tutorial: 7 of 10
Thevenins Theorem
In the previous 3 tutorials we have looked at solving complex electrical circuits using Kirchoff's Circuit Laws, Mesh Analysis and finally Nodal Analysis but there are many more "Circuit Analysis Theorems" available to calculate the currents and voltages at any point in a circuit. In this tutorial we will look at one of the more common circuit analysis theorems (next to Kirchoff´s) that has been developed, Thevenins Theorem.
Thevenins Theorem states that "Any linear circuit containing several voltages and resistances can be replaced by just a Single Voltage in series with a Single Resistor". In other words, it is possible to simplify any "Linear" circuit, no matter how complex, to an equivalent circuit with just a single voltage source in series with a resistance connected to a load as shown below. Thevenins Theorem is especially useful in analyzing power or battery systems and other interconnected circuits where it will have an effect on the adjoining part of the circuit.
Thevenins equivalent circuit.
As far as the load resistor RL is concerned, any "one-port" network consisting of resistive circuit elements and energy sources can be replaced by one single equivalent resistance Rs and 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 A and B with an open circuit and no load resistor Rs connected. Then, we get the following circuit.
Find the Equivalent Resistance (Rs)
Find the Equivalent Voltage (Vs)
We now need to reconnect the two voltages back into the circuit, and as VS = VAB the current flowing around the loop is calculated as:
so the voltage drop across the 20Ω resistor can be calculated as:
VAB = 20 - (20Ω x 0.33amps) = 13.33 volts.
Then the Thevenins Equivalent circuit is shown below with the 40Ω resistor connected.
and from this the current flowing in the circuit is given as:
which again, is the same value of 0.286 amps, we found using Kirchoff´s circuit law in the previous tutorial.
Thevenins theorem can be used as a circuit analysis method and is particularly useful if the load is to take a series of different values. It is not as powerful as Mesh or Nodal analysis in larger networks because the use of Mesh or Nodal analysis is usually necessary in any Thevenin exercise, so it might as well be used from the start. However, Thevenins equivalent circuits of Transistors, Voltage Sources such as batteries etc, are very useful in circuit design.
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 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 Navigation
Tutorial: 8 of 10
Nortons Theorem
In some ways Norton's Theorem can be thought of as the opposite to "Thevenins Theorem", in that Thevenin reduces his circuit down to a single resistance in series with a single voltage. Norton on the other hand reduces his circuit down to a single resistance in parallel with a constant current source. Nortons Theorem states that "Any linear circuit containing several energy sources and resistances can be replaced by a single Constant Current generator in parallel with a Single Resistor".
As far as the load resistance, RL is concerned this single resistance, RS is the value of the resistance looking back into the network with all the current sources open circuited and IS is the short circuit current at the output terminals as shown below.
Nortons equivalent circuit.
The value of this "constant current" is one which would flow if the two output terminals where shorted together while the source resistance would be measured looking back into the terminals, (the same as Thevenin).
For example, consider our now familiar circuit from the previous section.
To find the Nortons equivalent of the above circuit we firstly have to remove the centre 40Ω load resistor and short out the terminals A and B to give us the following circuit.
When the terminals A and B are shorted together the two resistors are connected in parallel across their two respective voltage sources and the currents flowing through each resistor as well as the total short circuit current can now be calculated as:
with A-B Shorted Out
If we short-out the two voltage sources and open circuit terminals A and B, the two resistors are now effectively connected together in parallel. The value of the internal resistor Rs is found by calculating the total resistance at the terminals A and B giving us the following circuit.
Find the Equivalent Resistance (Rs)
Having found both the short circuit current, Is and equivalent internal resistance, Rs this then gives us the following Nortons equivalent circuit.
Nortons equivalent circuit.
Ok, so far so good, but we now have to solve with the original 40Ω load resistor connected across terminals A and B as shown below.
Again, the two resistors are connected in parallel across the terminals A and B which gives us a total resistance of:
The voltage across the terminals A and B with the load resistor connected is given as:
Then the current flowing in the 40Ω load resistor can be found as:
which again, is the same value of 0.286 amps, we found using Kirchoff´s circuit law in the previous tutorials.
Nortons Theorem Summary
The basic procedure for solving a circuit using Nortons Theorem is as follows:
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.
Maximum Power Transfer Theorem Navigation
Tutorial: 9 of 10
Maximum Power Transfer
We have seen in the previous tutorials that any complex circuit or network can be replaced by a single energy source in series with a single internal source resistance, RS. Generally, this source resistance or even impedance if inductors or capacitors are involved is of a fixed value in Ohm´s. However, when we connect a load resistance, RL across the output terminals of the power source, the impedance of the load will vary from an open-circuit state to a short-circuit state resulting in the power being absorbed by the load becoming dependent on the impedance of the actual power source. Then for the load resistance to absorb the maximum power possible it has to be "Matched" to the impedance of the power source and this forms the basis of Maximum Power Transfer.
The Maximum Power Transfer Theorem is another useful analysis method to ensure that the maximum amount of power will be dissipated in the load resistance when the value of the load resistance is exactly equal to the resistance of the power source. The relationship between the load impedance and the internal impedance of the energy source will give the power in the load. Consider the circuit below.
Thevenins Equivalent Circuit.
In our Thevenin equivalent circuit above, the maximum power transfer theorem states that "the maximum amount of power will be dissipated in the load resistance if it is equal in value to the Thevenin or Norton source resistance of the network supplying the power".
In other words, the load resistance resulting in greatest power dissipation must be equal in value to the equivalent Thevenin source resistance, then RL = RS but if the load resistance is lower or higher in value than the Thevenin source resistance of the network, its dissipated power will be less than maximum. For example, find the value of the load resistance, RL that will give the maximum power transfer in the following circuit.
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 I P0 0 05 3.3 5510 2.8 7815 2.5 9320 2.2 97
RL I P25 2.0 10030 1.8 9740 1.5 9460 1.2 83100 0.8 64
Using the data from the table above, we can plot a graph of load resistance, RL against power, P for different values of load resistance. Also notice that power is zero for an open-circuit (zero current condition) and also for a short-circuit (zero voltage condition).
Graph of Power against Load Resistance
From the above table and graph we can see that the Maximum Power Transfer occurs in the load when the load resistance, RL is equal in value to the source resistance, RS that is: RS = RL = 25Ω. This is called a "matched condition" and as a general rule, maximum power is transferred from an active device such as a power supply or battery to an external device when the impedance of the external device exactly matches the impedance of the source. One good example of impedance matching is between an audio amplifier and a loudspeaker. Improper impedance matching can lead to excessive power loss and heat dissipation.
Transformer Impedance Matching
One very useful application of impedance matching in order to provide maximum power transfer between the source and the load is in the output stages of amplifier circuits. Signal transformers are used to match the loudspeakers higher or lower impedance value to the amplifiers output impedance to obtain maximum sound power output. These audio signal transformers are called "matching transformers" and couple the load to the amplifiers output as shown below.
Transformer Impedance Matching
The maximum power transfer can be obtained even if the output impedance is not the same as the load impedance. This can be done using a suitable "turns ratio" on the transformer with the corresponding ratio of load impedance, ZLOAD to output impedance, ZOUT matches that of the ratio of the transformers primary turns to secondary turns as a resistance on one side of the transformer becomes a different value on the other. If the load impedance, ZLOAD is purely resistive and the source impedance is purely resistive, ZOUT then the equation for finding the maximum power transfer is given as:
Where: NP is the number of primary turns and NS the number of secondary turns on the transformer. Then by varying the value of the transformers turns ratio the output impedance can be "matched" to the source impedance to achieve maximum power transfer. For example,
Example No2.
If an 8Ω loudspeaker is to be connected to an amplifier with an output impedance of 1000Ω, calculate the turns ratio of the matching transformer required to provide maximum power transfer of the audio signal. Assume the amplifier source impedance is Z1, the load impedance is Z2 with the turns ratio given as N.
Generally, small transformers used in low power audio amplifiers are usually regarded as ideal so any losses can be ignored.
In the next tutorial about DC Theory we will look at Star Delta Transformation which allows us to convert balanced star connected circuits into equivalent delta and vice versa.
Star Delta Transformation Navigation
Tutorial: 10 of 10
Star Delta Transformation
We can now solve simple series, parallel or bridge type resistive networks using Kirchoff´s Circuit Laws, mesh current analysis or nodal voltage analysis techniques but in a balanced 3-phase circuit we can use different mathematical techniques to simplify the analysis of the circuit and thereby reduce the amount of math's involved which in itself is a good thing.
Standard 3-phase circuits or networks take on two major forms with names that represent the way in which the resistances are connected, a Star connected network which has the symbol of the letter, Υ (wye) and a Delta connected network which has the symbol of a triangle, Δ (delta). If a 3-phase, 3-wire supply or even a 3-phase load is connected in one type of configuration, it
can be easily transformed or changed it into an equivalent configuration of the other type by using either the Star Delta Transformation or Delta Star Transformation process.
A resistive network consisting of three impedances can be connected together to form a T or "Tee" configuration but the network can also be redrawn to form a Star or Υ type network as shown below.
T-connected and Equivalent Star Network
As we have already seen, we can redraw the T resistor network to produce an equivalent Star or Υ type network. But we can also convert a Pi or π type resistor network into an equivalent Delta or Δ type network as shown below.
Pi-connected and Equivalent Delta Network.
Having now defined exactly what is a Star and Delta connected network it is possible to transform the Υ into an equivalent Δ circuit and also to convert a Δ into an equivalent Υ circuit using a the transformation process. This process allows us to produce a mathematical
relationship between the various resistors giving us a Star Delta Transformation as well as a Delta Star Transformation.
These 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 P in a Star network can be found as Equation 1 plus (Equation 3 minus Equation 2) or Eq1 + (Eq3 - Eq2).
Similarly, to find resistor Q in a star network, is equation 2 plus the result of equation 1 minus equation 3 or Eq2 + (Eq1 - Eq3) and this gives us the transformation of Q as:
and again, to find resistor R in a Star network, is equation 3 plus the result of equation 2 minus equation 1 or Eq3 + (Eq2 - Eq1) and this gives us the transformation of R as:
When converting a delta network into a star network the denominators of all of the transformation formulas are the same: A + B + C, and which is the sum of ALL the delta resistances. Then to convert any delta connected network to an equivalent star network we can summarized the above transformation equations as:
Delta to Star Transformations Equations
If the three resistors in the delta network are all equal in value then the resultant resistors in the equivalent star network will be equal to one third the value of the delta resistors, giving each branch in the star network as: RSTAR = 1/3RDELTA
Example No1
Convert the following Delta Resistive Network into an equivalent Star Network.
Star Delta Transformation
We have seen above that when converting from a delta network to an equivalent star network that the resistor connected to one terminal is the product of the two delta resistances connected to the same terminal, for example resistor P is the product of resistors A and B connected to terminal 1.
By rewriting the previous formulas a little we can also find the transformation formulas for converting a resistive star network to an equivalent delta network giving us a way of producing a star delta transformation as shown below.
Star to Delta Network.
The value of the resistor on any one side of the delta, Δ network is the sum of all the two-product combinations of resistors in the star network divide by the star resistor located "directly opposite" the delta resistor being found. For example, resistor A is given as:
with respect to terminal 3 and resistor B is given as:
with respect to terminal 2 with resistor C given as:
with respect to terminal 1.
By dividing out each equation by the value of the denominator we end up with three separate transformation formulas that can be used to convert any Delta resistive network into an equivalent star network as given below.
Star Delta Transformation Equations
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
Nodal Voltage Analysis Navigation
Tutorial: 6 of 10
Nodal Voltage Analysis
As well as using Mesh Analysis to solve the currents flowing around complex circuits it is also possible to use nodal analysis methods too. Nodal Voltage Analysis complements the previous mesh analysis in that it is equally powerful and based on the same concepts of matrix analysis. As its name implies, Nodal Voltage Analysis uses the "Nodal" equations of Kirchoff's first law to find the voltage potentials around the circuit.
So by adding together all these nodal voltages the net result will be equal to zero. Then, if there are "n" nodes in the circuit there will be "n-1" independent nodal equations and these alone are sufficient to describe and hence solve the circuit.
At each node point write down Kirchoff's first law equation, that is: "the currents entering a node are exactly equal in value to the currents leaving the node" then express each current in terms of the voltage across the branch. For "n" nodes, one node will be used as the reference node and all the other voltages will be referenced or measured with respect to this common node.
For example, consider the circuit from the previous section.
Nodal Voltage Analysis Circuit
In the above circuit, node D is chosen as the reference node and the other three nodes are assumed to have voltages, Va, Vb and Vc with respect to node D. For example;
As Va = 10v and Vc = 20v , Vb can be easily found by:
again is the same value of 0.286 amps, we found using Kirchoff's Circuit Law in the previous tutorial.
From both Mesh and Nodal Analysis methods we have looked at so far, this is the simplest method of solving this particular circuit. Generally, nodal voltage analysis is more appropriate when there are a larger number of current sources around. The network is then defined as: [ I ] = [ Y ] [ V ] where [ I ] are the driving current sources, [ V ] are the nodal voltages to be found and [ Y ] is the admittance matrix of the network which operates on [ V ] to give [ I ].
Nodal Voltage Analysis Summary.
The basic procedure for solving Nodal Analysis equations is as follows:
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 J to node K.
3. For a network with "N" independent nodes, [Y] will be an (N x N) matrix and that Ynn will be positive and Yjk will be negative or zero value.
4. The voltage vector will be (N x L) and will list the "N" voltages to be found.
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.
Mesh Current Analysis Navigation
Tutorial: 5 of 10
Circuit Analysis
In the previous tutorial we saw that complex circuits such as bridge or T-networks can be solved using Kirchoff's Circuit Laws. While Kirchoff´s Laws give us the basic method for analysing any complex electrical circuit, there are different ways of improving upon this method by using Mesh Current Analysis or Nodal Voltage Analysis that results in a lessening of the math's involved and when large networks are involved this reduction in maths can be a big advantage.
For example, consider the 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 I2 flowing in the two resistors. Then there is no need to calculate the current I3 as its just the sum of I1 and I2. So Kirchoff's second voltage law simply becomes:
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 and I2 as -0.429 Amps
As : I3 = I1 - I2
The current I3 is therefore given as : -0.143 - (-0.429) = 0.286 Amps
which is the same value of 0.286 amps, we found using Kirchoff´s circuit law in the previous tutorial.
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 Theory we will look at Nodal Voltage Analysis to do just that.
Kirchoffs Circuit Law Navigation
Tutorial: 4 of 10
Kirchoffs Circuit Law
We saw in the Resistors tutorial that a single equivalent resistance, ( RT ) can be found when two or more resistors are connected together in either series, parallel or combinations of both, and that these circuits obey Ohm's Law. However, sometimes in complex circuits such as bridge or T networks, we can not simply use Ohm's Law alone to find the voltages or currents circulating within the circuit. For these types of calculations we need certain rules which allow us to obtain the circuit equations and for this we can use Kirchoffs Circuit Law.
In 1845, a German physicist, Gustav Kirchoff developed a pair or set of rules or laws which deal with the conservation of current and energy within electrical circuits. These two rules are commonly known as: Kirchoffs Circuit Laws with one of Kirchoffs laws dealing with the current flowing around a closed circuit, Kirchoffs Current Law, (KCL) while the other law deals with the voltage sources present in a closed circuit, Kirchoffs Voltage Law, (KVL).
Kirchoffs First Law - The Current Law, (KCL)
Kirchoffs Current Law or KCL, states that the "total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node". In other words the algebraic sum of ALL the currents entering and leaving a node must be equal to zero, I(exiting) + I(entering) = 0. This idea by Kirchoff is commonly known as the Conservation of Charge.
Kirchoffs Current Law
Here, the 3 currents entering the node, I1, I2, I3 are all positive in value and the 2 currents leaving the node, I4 and I5 are negative in value. Then this means we can also rewrite the equation as;
I1 + I2 + I3 - I4 - I5 = 0
The term Node in an electrical circuit generally refers to a connection or junction of two or more current carrying paths or elements such as cables and components. Also for current to flow either in or out of a node a closed circuit path must exist. We can use Kirchoff's current law when analysing parallel circuits.
Kirchoffs Second Law - The Voltage Law, (KVL)
Kirchoffs Voltage Law or KVL, states that "in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop" which is also equal to zero. In other words the algebraic sum of all voltages within the loop must be equal to zero. This idea by Kirchoff is known as the Conservation of Energy.
Kirchoffs Voltage Law
Starting at any point in the loop continue in the same direction noting the direction of all the voltage drops, either positive or negative, and returning back to the same starting point. It is important to maintain the same direction either clockwise or anti-clockwise or the final voltage sum will not be equal to zero. We can use Kirchoff's voltage law when analysing series circuits.
When analysing either DC circuits or AC circuits using Kirchoffs Circuit Laws a number of definitions and terminologies are used to describe the parts of the circuit being analysed such as: node, paths, branches, loops and meshes. These terms are used frequently in circuit analysis so it is important to understand them.
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 both I1 and I2
Substitution of I1 in terms of I2 gives us the value of I1 as -0.143 Amps
Substitution of I2 in terms of I1 gives us the value of I2 as +0.429 Amps
As : I3 = I1 + I2
The current flowing in resistor R3 is given as : -0.143 + 0.429 = 0.286 Amps
and the voltage across the resistor R3 is given as : 0.286 x 40 = 11.44 volts
The negative sign for I1 means that the direction of current flow initially chosen was wrong, but never the less still valid. In fact, the 20v battery is charging the 10v battery.
Application of Kirchoffs Circuit Laws
These two laws enable the Currents and Voltages in a circuit to be found, ie, the circuit is said to be "Analysed", and the basic procedure for using 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 about DC Theory we will look at Mesh Current Analysis to do just that.
Electrical Units of Measure Navigation
Tutorial: 3 of 10
Electrical Units of Measure
The standard SI units used for the measurement of voltage, current and resistance are the Volt [ V ], Ampere [ A ] and Ohms [ Ω ] respectively. Sometimes in electrical or electronic circuits and systems it is necessary to use multiples or sub-multiples (fractions) of these standard units
when the quantities being measured are very large or very small. The following table gives a list of some of the standard units used in electrical formulas and component values.
Standard Electrical Units
Parameter SymbolMeasuring
UnitDescription
Voltage Volt V or EUnit of Electrical Potential
V = I × R
Current Ampere I or iUnit of Electrical Current
I = V ÷ R
Resistance Ohm R or ΩUnit of DC Resistance
R = V ÷ I
Conductance Siemen G or ℧ Reciprocal of ResistanceG = 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 or I2 × R
Impedance Ohm ZUnit of AC Resistance
Z2 = R2 + X2
Frequency Hertz HzUnit 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 or π rad.
• τ − Time Constant, The Time Constant of an impedance circuit or linear first-order system is the time it takes for the output to reach 63.7% of its maximum or minimum output value when subjected to a Step Response input. It is a measure of reaction time.
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.
Ohms Law Navigation
Tutorial: 2 of 10
Ohms Law
The relationship between Voltage, Current and Resistance in any DC electrical circuit was firstly discovered by the German physicist Georg Ohm, (1787 - 1854). Georg Ohm found that, at a constant temperature, the electrical current flowing through a fixed linear resistance is directly proportional to the voltage applied across it, and also inversely proportional to the resistance.
This relationship between the Voltage, Current and Resistance forms the bases of Ohms Law and is shown below.
Ohms Law Relationship
By knowing any two values of the Voltage, Current or Resistance quantities we can use Ohms Law to find the third missing value. Ohms Law is used extensively in electronics formulas and calculations so it is "very important to understand and accurately remember these formulas".
To find 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 denote milliwatts (mW = 10-3W) or kilowatts (kW = 103W). By using Ohm's law and substituting for V, I and R the formula for electrical power can be found as:
To find 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 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
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 simple Ohms 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 = 106J) are used instead. If the electrical power is measured in "kilowatts" and the time is given in hours then the unit of energy is in kilowatt-hours or kWh which is commonly called a "Unit of Electricity" and is what consumers purchase from their electricity suppliers.
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.