ELECTRICAL ENGINEERING
LECTURE NOTES
I
CIRCUIT THEORY
Mr Ted Spooner & Dr B. Farah
THE UNIVERSITY OF NEW SOUTH WALES, SCHOOL OF ELECTRICAL ENGINEERING 2002
circuits4.doc
1
1.
INTRODUCTION:
Reasons for studying the “Circuit Theory”: To be able to understand the principles, specifications and performance of electrical and electronic devices (eg. filters, amplifiers), transformers, power supplies and machines (eg. DC and AC motors). To be able to communicate effectively with electrical engineers. Electrical systems are an integral part of many mechanical & mining systems. Mechanical Engineers need to have a broad understanding of important electrical issues. Need to know of electrical options for measurement and control of mechanical systems. 2.
The International System of Units “SI” 2.1.
Basic Units
Quantity Length Mass Time Electric Current Temperature
2.2.
Symbol l m t I or i T
Unit Metre Kilogram Second Ampere Kelvin
Mechanical Units Based on SI
Quantity Qty. Symb. Velocity u Acceleration a Force F (= ma) Weight W (= mg = 9.81m) Turning Moment or Torque T (=F.r) Work or Energy W.D. or E
Angular Velocity Frequency
circuits4.doc
Unit Symbol m kg s A K(=273 + deg.C)
ω f
1
Unit metre/second metre/(second)2 newton newton. newton.metre joule Megajoule watt second watt hour kilowatt hour radian/second hertz kilohertz Megahertz
Unit Symb. m/s m/s 2 N N N.m J (= 1 N.m) MJ(= 106 J) W.s(=1J) W.h(=3600J) kW.h(=3.6MJ) rad/s Hz kHz(=103 Hz) MHz(= 106Hz)
2.3.
Electrical Units Based on SI
Quantity Charge (or Qty. of Electricity) Current
QtY.Symb. Q, q I, i
Potential Difference or Voltage or Electromotive Force
V. v E
Resistance
R
Conductance
G(= 1/R)
Capacitance
C
Inductance Power
L P
3.
Unit coulomb ampere milliampere micro-ampere volt millivolt kilovolt ohm Micro-ohm Mega-ohm siemens or mhos farad microfarad nanofarad picofarad henry watt kilowatt
Symb. C A mA µA V mV kV Ω µΩ MΩ S Ω -1 F µ F(=10-6F) nF(=10-9 F) pF(=10-12 F) H W kW
FUNDAMENTAL CONCEPTS:
3.1.
Classification of Electrical Components:
All electrical components can be placed in one of two groups:
Electrical Components
Active elements: voltage source, current source, (which do supply energy to a circuit).
Passive elements: resistor, inductor, capacitor, diode, etc. (which do not supply energy to a circuit; rather they dissipate or store energy).
3.2.
Basic Components The three main passive components are:
3.2.1.
Resistor:
The current through a resistor is directly proportional to the voltage across it.
OHM’s LAW: V-I relationship for a resistor
V
Volts (V)
3.2.2.
=
Amps (A)
IR
Ohms (Ω)
Inductor
Wire wound coil. The magnitude of the induced voltage is proportional to the rate of change of the current flowing through the coil.
V vs I relationship for an inductor L: inductance (henrys) usually (microhenrys) or mH (millihenrys)
v
=
L
di dt
µH
3.2.3.
Capacitor
Two parallel plates. The charge on the plates is proportional to the voltage between the plates: q = Cv
However, as the current is defined as the rate of flow of charge, ie: i=
dq dt
=C
dv dt
C: Capacitance (farads)
Or
dv
=
1 idt C
V vs I relationship for a capacitor
v=
1 idt C
The farad is rather large, hence the units to be used practically are: Microfarad Nanofarad Picofarad
3.3.
= 10-6 farads = µF = 10-9 farads = nF = 10-12 farads = pF
Fundamental Terminology:
3.3.1.
Symbols: The symbol convention used in this course is: Capital letters for the time independent quantities eg. R, C Small letters for the time-varying quantities eg. v. i.
3.3.2.
Nodes: A node in an electrical circuit is any point where 2 or more components have a common connection viz.
NODE
3.3.3.
Branches: A branch is the part of a circuit which exists between 2 nodes, viz.
BRANCH
3.3.4.
Loops:
A loop is any physical loop, open or closed which can be described in a circuit:
Open Loop
Closed Loop
3.3.5.
Current & Voltage Sign Conventions
For the voltage sign convention (active or passive element) the terminal with algebraically highest potential is denoted by ' + ' and that with the algebraically lowest potential by ' - '. Thus the symbol representing a voltage generator is: +
+
More general as it can be used for both AC and DC sources.
When a DC voltage generator (or a DC voltage source) is connected to a circuit, it drives current out of its positive terminal (and consequently into its negative terminal).
If an external circuit of a resistor is connected to the voltage source, then the current flows into the resistor’s positive terminal. Note that the polarity of its terminals is established by the voltage source. Thus:
For all passive elements, current flows in at the positive terminal.
Voltage Rise
+
+ Voltage Drop
-
3.3.6.
Open circuit Voltage & Short-Circuit Current
a) Consider any circuit with two free terminals A and B. The open-circuit voltage at terminals A and B is the voltage which appears at the terminals with nothing connected between them, ie. infinite resistance between them. Circuit
A Vo/c B
Open Circuit Voltage Vo/c = voltage across 2 terminals which are not connected in any way. (i.e. RAB=∞) b) if terminals A and B are connected together with a piece of wire i.e. zero resistance, then the short-circuit current is the current that flows along the wire.
CIRCUIT
ISC
Short Circuit Is/c = current which flows between 2 terminals when connected directly together. (i.e. RAB = 0)
3.3.7.
Electric Power
A considerable part of electrical technology is concerned with the transmission of power, so it is appropriate to take a quick look at the basic relations here. In an electric circuit: Power, P = v.i. watts
Power is the rate of doing work, or the rate of change of energy in a system, i.e.
P= Where: W:
dW dt
energy or work for an electrical system in joules:
W = Pdt = vidt As an example, consider the dissipation of power in a resistor in watts: PR
=
vi
=
i2R
v2/R
=
which gives the power dissipated in a resistor (due to a current flowing through the resistor) and lost as heat.
4. KIRCHHOFF'S LAWS 4.1.
KIRCHHOFF'S VOLTAGE LAW
It states that the algebraic sum of the voltage drops around any loop, open or closed, is zero.
Mathematically:
V =0 AroundLoop
Example
+
2Ω +
+ 10V
3Ω
I -
Going round the loop in the direction of the current, I, Kirchhoff's Voltage Law gives:
10- 2I - 3I = 0 - 2I and - 3I are negative, since they are voltage drops i.e. represent a decrease in potential on proceeding round the loop in the direction of I. For the same reason + 10V is positive as it is a voltage rise or increase in potential. Concluding: 5 I = 10 4.2.
Therefore, I = 2A
KIRCHOFF'S CURRENT LAW
It states that the algebraic sum of all currents entering a node is zero. Mathematically:
I =0 Into a node Currents are positive if entering a node Currents are negative if leaving a node. Example:
I2 = -3A I1= 5A
I3
I4 = 2A Applying Kirchhoff's current law: I1 + I2 + I3 + I4 = 0 (the negative sign in I2 indicates that I2 has a magnitude of 3A and is flowing in the direction opposite to that indicated by the arrow) Substituting: 5 - 3 + I3 + 2 = 0
Therefore, I3 = - 4A (ie 4A leaving node)
5. CIRCUITS A circuit is an interconnection of components, and as most electrical components are 2-terminal devices, such as a resistor, many of their interconnections can be resolved into series and parallel branches. 5.1.
SERIES CONNECTION OF RESISTORS:
Common Current = I = V1 / R1 = V2 / R2 = V3 / R3 = ( V1 + V2 + V3 )/ ( R1 + R2 + R3 ) = VT / RS As VT = V1 + V2 + V3
Therefore, RS = R1 + R2 + R3
RS is an equivalent resistor that has the same voltage across it, and the current through it as that through the 3 resistors in series. Example
Total Resistance = 6 + 8 = 14Ω 5.2. PARALLEL CONNECTION OF RESISTORS Common Voltage: V = I1 R1 = I2 R2 = I3 R3 = IT RP
As IT = I1 + I2 + I3
Then, V/RP = V/R1 + V/R2 + V/R3
Therefore, 1/RP = 1/R1 + 1/R2 + 1/R3 , or, GP = G1 + G2 + G3 Where, RP is the equivalent resistor, and GP the equivalent conductance (often useful in parallel circuits).
For 2 resistors, 1/RP = 1/R1 + 1/R2 = (R2 + R1 )/R1R2 ie. RP = (R1R2)/(R1 + R2) Example:
Equivalent resistance, RP = (6 x 8 )/(6 + 8) = 48 / 14 = 3.43Ω 6. CHARACTERISTICS The characteristic of a device, circuit, or system is a functional relationship between a dependent variable of interest and an independent variable. The description of a circuit, etc. by means of its characteristic allows the prediction of its operational conditions without the digression into perhaps unwanted detail as in the loop current method of analysis which produces every current and voltage in the circuit. 6.1.
RESISTOR
The V-I relationship for a simple resistor (load line) is given by Ohm's Law which may be plotted thus:
The resistor can only have values of voltage across it, and current flowing though it, which are consistent with its characteristic. This is true for any device or system. 6.2.
IDEAL VOLTAGE SOURCES
An ideal voltage source is represented in the diagram. For this ideal source V is constant for all I, and its open circuit voltage: Vo/c = E.
We could also make I the dependent variable and plot the characteristic as:
6.3.
NON-IDEAL VOLTAGE SOURCE
Non-ideal voltage source, for example a torch battery, can be modelled or represented by an ideal voltage source in series with a resistor.
V Internal Resistance E
I
+
Ideal Voltage Source
VOC = E +
Slope = dV/dI =d(E-IRi)/dI = -Ri
V V = E-IRi Non-Ideal Voltage Source
ISC
I
Here, taking KVL around the loop : - IRi - V + E = 0
Therefore, V = E - IRi
The V-I relationship for the non-ideal source shows that as the current drawn from the source, I, increases, the voltage drop across the internal resistance, IRi , increases. Since E remains constant this reduces the voltage V, available at the output terminals, Therefore, V is not constant for all I The V-I characteristic can be plotted from the equation, V = E – IRi , as shown in the above figure. From this equation: If I = 0 (open circuit), then Vo/c = E, and if V = 0 (short circuit), then Is/c = E / Ri
Comparing the characteristics for the two types of voltage sources we can comment on the internal resistance of both: The internal resistance of ideal V-source = 0Ω , and of non-ideal V-source = Ri Ω.
I Internal Resistance E
+
Ideal Voltage Source
I
ISC = E/RiI +
Slope = dI/dV = -1/Ri = -G
V
Non-Ideal Voltage Source
VOC = E
V
We could of course rearrange our equation to make I the dependent variable and V the independent variable as shown below: As, V = E - I Ri , then, I = ( E/Ri ) - ( V / Ri ).
Example Kirchhoff’s Voltage Law (KVL) around the loop:
- 2 I - V + 10 = 0
Therefore, V = 10 - 2 I.
Slope, dI/dV = - 1 / Ri
7.
VOLTAGE REGULATION:
By comparing the V-I characteristics for the ideal and the non-ideal cases, we can see how good a representation of constant voltage source is the non-ideal source. We would prefer to have a voltage source whose terminal voltage remained constant over a wide range of currents. However, in real life we are usually dealing with nonideal sources where, as a result of the internal resistance of the source, the terminal voltage decreases as the current increases. The usual method of characterising the change in the terminal voltage V. with change in current drawn from the voltage generator is to specify the regulation of the generator. This is a measure of how constant V remains as the source goes from the 'no load' condition to the 'full load' condition as explained below. It is particularly important when considering transformers and electrical machinery.
V Ri E
B I
+
VOC = E
V = E-IRi
+ V
A
I
No-load condition means that no load resistor is connected between A and B
i.e. , I = 0 ,
.'. Vno load = Vo/c = E (for this circuit).
Connecting a load resistor:
V Ri E
+
B I
Operating Point
Vno load VFull Load
+ V RL
Load Line
A
IMax
I
Full load condition: The load resistor is connected between A and B such that I has its maximum possible value. (This means that if I were to exceed Imax the generator would burn out.) Therefore, I = E / ( Ri + RL ) , and Vfull load = I RL = E RL / ( Ri + RL ) < Vno load . The 'regulation' of a voltage source is a measure of how constant the terminal voltage remains as the circuit goes from the 'no load' to the 'full load' condition.
Voltage Regulation =
V no − load − V full −load V full − load
Example: With the load resistor (4Ω ) disconnected, RAB = ∞ ,
∴
×100%
I = 0,
V = Vno-load = 10V. With the load resistor now connected,
I = 10 / ( 1 + 4) = 2A.
∴ V = Vfull-load = 2A x 4Ω = 8V (Alternatively, V = 10 - 2A x 1Ω = 10 - 2 = 8V)
∴ Voltage Regulation = 10 - 8 x 100 % = 25 % 8
hence
8. THEVENIN'S THEOREM Quite often, particularly in a complex circuit we need more information than just the operating conditions which we could obtain from the characteristics, but at the same time we are not interested in every voltage and current in the circuit as would result from a loop current analysis. For example, consider the circuit of a DC transmission system such as that found in a car, we may be confronted by this multi-loop circuit.
Appliance of interest say radio
Lead Resistance
I V
+
A
V
B
Appliances, Lights, Motors etc
If we are interested only in what current is drawn by the load (radio), clearly we have quite a task ahead of us, if we set up, say 4 loop equations, and find every loop current. It would be a decided advantage if we could replace everything to the left of line AB by a much simpler, yet EQUIVALENT CIRCUIT. By equivalent we mean having the same V-I characteristics at the terminals A and B. Thevenin's Theorem gives us just such an equivalent circuit. In fact it is a method of modelling or simulating the behaviour of a complex circuit whose actual configuration may be unknown, but whose characteristics at its output terminals are well known. 8.1. DEFINITION OF THEVENIN'S THEOREM It states that any complex, linear, active or passive 2 terminal network may be modelled by an ideal voltage source in series with a resistor.
RT Linear Network
I V
ET
+
I
V
V VOC = ET
V-I Relationship: V = ET - I RT
Slope = dV/dI = -RT
ET :
Vo/c at terminals (A and B) of the linear network RT : Resistance of the Linear network at terminals (A and B) with all energy sources replaced by their internal resistances.
ISC =ET/RT Example: Consider the following linear network, determine the V- I characteristic at its output terminals, then calculate the Thevenin's equivalent of the circuit, plot its V-I characteristic and see if there is any connection:
I1 50V
2Ω
I
4Ω
A +
I2
V 2Ω B
Derivation of the V- I characteristic at terminals A and B using Kirchoff's laws: Using Kirchoff's current law. I1 − I 2 − I = 0 Using Kirchoff's voltage law around first loop: 50 - 2I1 - 2I2 =0 Using Kirchoff's voltage law around second loop: 2I2 - 4I - V = 0 Substituting current equn in first loop eqn: 50 - 2(I2+I) - 2I2 =0 50 - 4I2 - 2I =0 From 2 nd loop eqn 2I2 = 4I + V Substituting in above 50 - 2(4I + V) - 2I =0 50 - 10I - 2V = 0 V = 25 -5I
I
V V OC = 2 5V S lo p e = d V /d I = -5 Ω
IS C = 5 A Plotting the V-I relationship: if A - B short-circuited V = 0 and I = 5A If A - B open circuited then I = 0 and V = 25V
I
Superposition
Y
Y1 Y1 x1
x2 (x1+x2)
X
If an output is a linear function of an input parameter y = f(x) and y=0 when x=0. then f(x1+x2) = f(x1) + f(x2) = y1 + y2 Consider a linear deflecting beam:
F1 D1 F2 D2 F1+F2 D1+D2 If the deflection of the beam is completely linear or that the only region we are studying is the linear region then if: we apply a force of F1 and get a deflection D1 we apply a force of F2 and get a deflection D2 Then if we apply a force of F1+F2 we will get a deflection D1+D2.
Another similar example using the same beam is shown below only now the forces are applied at different positions. F1 D1
F2
D2 F2
F1 D1+D2
The same superposition applies.... If: we apply a force of F1 and get a deflection D1 we apply a force of F2 and get a deflection D2 Then if we apply a force of F1 and F2 we will get a deflection D1+D2 So we can for such a system do tests or analysis of the system and apply forces separately and look at the systems reaction. The result of applying both forces simultaneously is the sum of the individual reactions. Superposition in circuits The same principle may be used in analysing circuits. ie if cause and effect are linearly related then the total effect of several causes is the sum of the effects of the individual causes . Consider the circuit below.
V2
+
R1 +
I3
V1 R2
R3
We could analyse this circuit by doing loop analysis or writing down all the node equations but it can be simpler to solve two separate simpler circuits and find the currents due to each voltage source individually and then to sum the results. sum of effects of each voltage source taken individually.
R1
R1
+
V2
+
I3 =
I31
V1 R2
R3
I32 R2
R3
I3 = I31 + I32 Note: When a source is removed it is replaced by its internal resistance... for an ideal voltage source this is 0Ω or a short circuit. for an ideal current source this is ∞Ω or an open circuit. Example of circuit with Current Source
R1 +
I3
V1 R2
I2
R3
This is analysed in two parts as before 1) by open circuiting the current source I2 as shown below and calculate I31.
R1 +
I31
V1 R2
R3
2) put current source back and replace the voltage source V1 by a short circuit and calculate I32.
R1 I32 R2
I2
R3
Again I3 = sum of effects of each source taken individually. I3 = I31 + I32