Basic Principles of. Electricity. Basic Principles of Electricity. by Prof. Dr. Osman SEVAİOĞLU Electrical and Electronics Engineering Department
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1 Basic Principles of Electricity METU by Prof. Dr. Osman SEVAİOĞLU Electrical and Electronics Engineering Department EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 1
2 Course Syllabus EE 209 Fundamentals of Electrical and Electronics Engineering (3-0)3, Circuit Analysis, AC Circuits, AC Power, Phasors, Three Phase Systems, Transformers, Magnetic Circuits, Electrical Safety (Offered to non-ee students only) Prerequisite: PHYS 106 or consent of the department. EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 2
3 Book for the Course EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 3
4 Book for the Course Principles and Applications of Electrical Engineering, 4/e Giorgio Rizzoni The Ohio State University Mc. Graw Hill Book Company, ISBN: Copyright year: Pages Available in Reserve Division of the Middle East Technical University Central Library EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 4
5 Course Syllabus Chapters to be Covered, Circuit Analysis, AC Circuits, AC Power, Phasors, Three Phase Systems, Transformers, Magnetic Circuits, Electrical Safety EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 5
6 Basic Principles of the Course Examinations Two midterm examinations and a final exam Midterm Exam 1 (Three questions, equal credits, 90 min) 20 % Midterm Exam 2 (Three questions, equal credits, 90 min) 20 % Final Exam (Four questions, equal credits, 120 min) 30 % Attendance 30 % Total 100 % EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 6
7 Homework No homeworks will be assigned You are advise to examine; the homeworks in the book, examination questions that will be distributed EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 7
8 Basic Principles of the Course Examinations Midterm examinations will cover all the material taught until the examination date, Final Examination will cover the overall course material, Announced exam schedule can neither be changed nor discussed after it has been settled, Duration of the examination will never be extended, Questions will never be allowed during the examination EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 8
9 Basic Principles of the Course Make-up Examinations Will be given only to those students with valid documented excuse, Requests for make-up exam that does not include a valid documented excuse will be rejected, A single make-up exam will be given to all students with legitimate rights for the exam, Exam will be carried out in an officially settled date and hour, Exam will not be repeated, i.e. Make-up of make-up will not be performed, Students will be responsible for answering the questions only from the parts covered in the exam that they have missed EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 9
10 Questions and Solutions of the Previous Examinations Exam Questions and Solutions A file including all exam questions and solutions is available, A file including the questions and solutions of all the previous examinations will be submitted to a student who is elected by the class for photocopying and distributing this file to the class, This student will be responsible for the toll collection and distribution activity In case that there is no valunteer for the job, the task will be cancelled! EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 10
11 Yahoo Group for the Course Yahoo Group for the course is; This group is intended to be the main communication medium for information exchange and storage for the course Enrollment to this group is compulsory --- All students are obliged to subscribe to this group by using the procedure described in the next page EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 11
12 Group Enrollment To subscribe from the group, send an to: To unsubscribe from the group, send an to: EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 12
13 Group Nicknames Nicknames (User Ccodes) Please choose nicknames that reflect your personal identity, i.e. your surname and/or name and/or you name and surname augmented. Please do NOT choose improper or annoying nicknames, such as; Arizona Tigers, diabolic, best friend, miserable(68) etc. that does not reflect your personal identity EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 13
14 Group Communication All questions, suggestions, complaints, demands, requests and other communication concerning the course should be directed to the communication address of the group: The Course Instructor keeps the right of not answering some or all of the questions, suggestions, complaints, demands, requests forwarded in this mail group, in case that it is not necessary, or not relevant, or not possible EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 14
15 Group In your s; Rules of Communication Be polite, Start your letter with; Dear Group Members or Dear Friends and end with; With best regards Do not use disturbing abbreviatons, such as slm for selam, Do not discuss your own personal, social or academic problems, Do not be aggressive to the Group members and to Course Instructor, Do not discuss subjects not relevant to the course, (such as last match of Fenerbahçe) People who violate the above rules will be deleted from the group EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 15
16 Group Course Instructor is the Main Moderator of the Group. Assistant Moderator Group Moderators An assistant moderator who is familiar with the management of yahoogroups activities, will be elected and appointed for managing the group from valunteer candidates in the class during the first hour. Moderators have identical authorities in group management in all respects EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 16
17 Problems Complaints and Expressions Complaints and expressions concerning your; personal, Social, Academic problems will never be listened, nor be appreciated nor be interested. Your personal, social and academic problems will never be an influencing factor in grading, Your personal, social and academic problems will not be taken into account at all This course is NOT a proper platform for expressing your own problems, negative or positive human feelings, such as, crying, complaining, hating, admiring, or any other physiologic, psychological expressions EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 17
18 Group Office Hours Unfortunately, there will not be any chance for office hour Please do not refer my office for any reason, and do not blame for that. EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 18
19 Group My GSM No: Telephone Calls Telephone calls for concerning your personal, social and academic problems will neither be listened, nor be appreciated nor be interested EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 19
20 Group Weekly Course Schedule (Three hours/week) Announced schedule can be discussed EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 20
21 Structure of atom Atom Helium Atom Electron is assumed to be negatively charged Proton is assumed to be positive charged Electron Proton Neutron EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 21
22 Definition Electrical Charge Unit of Electrical Charge Coulomb 6.3 x electrons = 1 Coulomb or Electrical charge / electron = 1/ (6.3 x ) Coulomb = x Coulomb EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 22
23 Basic Principle of Circuit Mechanical Example Inclined Surface h 2 h h 1 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 23
24 Water Circuit Water Current (I) Pump Water Current = Volume (m 3 ) / sec EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 24
25 Water Circuit Water Current (I) P 2 P = P 2 - P 1 P 1 Pump Gravitational force EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 25
26 Electromotive force METU Electrical Circuit Electrical Current (I) Consumer (Load) Electrical Current = No. of electrons / sec = 1 Coulomb / sec 6.3 x electrons / sec = 1 Amper EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 26
27 Electromotive force Voltage (V) METU Electrical Circuit Electrical Current (I) Current (I) Consumer (Load) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 27
28 Voltage Difference Pressure Difference Voltage Difference P 2 V 2 Current (I) P = P 2 - P 1 P 1 Pump Gravitational force V = V 2 - V 1 V 1 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 28
29 Ground Node (Earth Point) Definition Current (I) Ground Node is the point (junction) at which the voltage is assumed to be zero All other voltages takes their references with respect to this ground node V = V 2 - V 1 = V 2 V 2 Representation Ground Node V 1 V 1 = 0 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 29
30 Ground Node (Earth Point) Definition Ground Node is the point (junction) at which the voltage is assumed to be zero All other voltages takes their references with respect to this ground node Ground Node (Black Terminal ) Measured Node (Red Terminal ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 30
31 Electrical Current Current = no. of electrons transferred / time duration I = Q / t 1 Amp = 1 Coulomb / 1 Seconds Charge = Current x Time duration Q = I x t Current (I) Voltage difference ( V) Current (I) Consumer (Load) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 31
32 Traffic Current Cars Flowing in a Highway Traffic Current = Cars / minute EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 32
33 Water Current Birecik Dam (672 MW) Water Current = 500 m 3 / sec Water Current = Volume (m 3 ) / sec EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 33
34 1 meter METU Example: Electrical Current 2 mm diameter EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 34
35 1 meter METU Example: Electrical Current 2 mm diameter EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 35
36 1 meter METU Electrical Current 2 mm diameter EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 36
37 Electrical Current - Basic Principle Electromotive force Electrons Current V 2 V = V 2 - V 1 Electrons V 1 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 37
38 Electrical Current DC (Direct Current) Sources V 2 V 2 V 1 V = V 2 - V 1 V = V 2 - V 1 V 1 = 0 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 38
39 Simple AC Circuit Current (I) Generator Customer EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 39
40 Kirchoff s Current Law (KCL) Basic Principle Cars entering = Cars leaving Balance Cars entering: 370 Cars leaving: EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 40
41 Kirchoff s Current Law (KCL) Charges entering Q 1 Q 2 Q n EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 41 Q in Charges leaving Q out = Q n Balance Q in = Q out or Q in Q out = 0 or Q = 0
42 Kirchoff s Current Law (KCL) Q n Coulombs/sec or Q = 0 or Q / t = 0 or i = n I i = 0 i =1 Kirchoff s First Law or Kirchoffs Current Law EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 42
43 Mechanical Force Definition F = m x a Force needed to accelarate 1 kg of mass to 1 meter / sec 2 is defined as 1 Newton 1 Newton = 1 kg x 1 meter / sec Newton = 1000 kg x 1 meter / sec 2 Accelaration = 1 m/sec 2 Wagon Mass = 1000 kg Force = 1000 Newton EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 43
44 Mechanical Energy Definition 1 Joule = 1 Newton x 1 Meter 1 Joule is the energy needed to move a mass 1 meter by using 1 Newton force 1 Meter 1 Newton Wagon Wagon EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 44
45 Power Definition Power is the work done within a certain unit of time, i.e. one second or one hour Power = Energy / Duration = 1 Joule / sec 1 sec 1 Meter 1 Newton Please note that force (and hence power) of the weak horse shown below is half of the first, but the work done (energy spent) is the same, i.e. Energy = 2 seconds x 0.5 Newton x 1 meter 2 seconds 1 Meter 0.5 Newton Wagon Wagon Wagon Wagon Energy = 1 Joule, Power = 1 Joule / sec. Energy = 1 Joule, Power = 1 Joule / 2 sec. EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 45
46 Mechanical Energy vs Electrical Energy Equivalance Mechanical Energy = Electrical Energy The same amount of energy may be spent out by using electricity Mechanical Work = Electrical Work 1 Meter 1 Newton Current (I) Wagon Wagon Voltage (V) 1 Joule Mechanical Energy (Work) = 1 Joule Electrical Energy (Work) = 1 Joule EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 46
47 Definition Electrical Power Similar to mechanical power, electrical power is the work done within a certain unit of time, i.e. one second or one hour Elecrical Power = Electrical Energy / Duration = 1 Joule / sec (1 sec) Current (I) 1 sec Wagon 1 Meter 1 Newton Wagon Voltage (V) Mechanical Power = 1 Joule / sec. Electrical Power = 1 Joule / sec. EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 47
48 Equivalence of Mechanical and Electrical Powers Equivalance Mechanical Power = Electrical Power 1 sec (1 sec) Current (I) 1 Meter 1 Newton Wagon Wagon Voltage (V) Mechanical Power = 1 Joule / sec. Electrical Power = 1 Joule / sec. EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 48
49 Electrical Power Definition 1 Joule / second = 1 Watt (1 Joule energy is spent within 1 second) 1 Joule = 1 Watt x second 1 Horse Power = 746 Watts = kwatt 1 sec (1 sec) Current (I) 1 Meter 1 Newton Wagon Wagon Voltage (V) 1 Joule / sec = 1 Watt Electrical Power = 1 Joule / sec. = 1 Watt EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 49
50 Electrical Power Definition P = V x I (Watt) = (Volt) x (Amp) Power = Voltage x Current Current (Amp) DC Voltage (Volt) Power (Watt) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 50
51 Voltage Definition Current Power = Voltage x Current or P = V x I Voltage = Power / Current or V = P / I DC Voltage Power EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 51
52 Voltage Definition Current (1 Amp) Power = Voltage x Current or Voltage = Power / Current DC Voltage (1 Volt) Power (1 Watt) or V = P / I 1 Volt = 1 Watt / 1 Amp EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 52
53 Electrical Energy Definition Current (I) Energy = Power x Time (Watt-sec) (Watt) (second) DC Voltage (V) Energy = Power x Time EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 53
54 Unit of Electrical Energy Definition Energy = Power x Time (Watt-sec) (Watt) (second) Energy = Power x Time (KiloWatt-hour) (KiloWatt) (hour) x 1000 x 1000 x KiloWatt = 1000 Watts 1 Hour = 3600 seconds DC Voltage (V) Current (I) 1 KiloWatt - hour = 1000 x 3600 Watt x seconds = Joules Energy = Power x Time EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 54
55 Example Calculate the monthly payment for the energy consumed by the lamp shown on the RHS Source voltage is 220 Volt Current drawn by the lamp is 5 Amp Price of electrical energy is 12 Cents / kwh Electrical Energy Power = Voltage x Current P = V x I P = 220 x 5 = 1100 Watts Energy = P x t = 1100 Watts x ( 24 hours /day x 30 days/month) = Watt hours = kwh Monthly payment = x 12 Cents / month = USD = YTL / month I = 5 Amp V = 220 V EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 55
56 Alternative Definition of Voltage 1 Volt = 1 Watt / 1 Amp = (1 Joule /sec) / 1 Amp = 1 Joule / (1 Amp x sec) = 1 Joule / 1 Coulomb (*) (*) Remember that 1 Amp = 1 Coulomb / 1 sec 1 Volt is the voltage needed; to move 1 Coulomb of electrical charge, to spend 1 Joule of energy for this movement in a conductor 1 Volt 1 Coulomb 1 Joule energy EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 56
57 Alternative Definition of Voltage 1 Volt = 1 Joule / 1 Coulomb Please note that time parameter does not appear in the above equation, implying that it is arbitrary Case-1 Let t = 1 sec Then, I = 1 Coulomb / 1 sec = 1 Amp P = V x I = 1 Volt x 1 Amp = 1 Watt Energy = P x t = (1 Joule / sec) x sec = 1 Joule Case-2 Let now t = 2 sec Then, I = 1 Coulomb / 2 sec = 0.5 Amp P = V x I = 1 Volt x 0.5 Amp = 0.5 Watt = Energy / 2 = 0.5 Joule / sec Energy = P x t = 0.5 x 2 = 1 Joule again 1 Coulomb 1 Volt 1 Joule energy EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 57
58 Resistance Definition Resistance R 1 R 1 > R 2 Resistance R 2 Resistance is the reaction of a pipe against water flow Pump I 1 Pump I 2 Current I 1 I 2 > I 1 Current I 2 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 58
59 Resistance Definition Resistance R 1 Resistance R 2 Resistance is the reaction of a conductor against electrical current Current I 1 Current I 1 R 1 > R 2 I 1 < I 2 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 59
60 Ohm Law Basic Principles Current I (Amp) Current flowing in the circuit is; proportional to voltage, inversely proportional to resistance Voltage (Volt) Resistance (R) Hence Unit of resistance is Ohm 1 Ohm is the resistance that allows 1 Amper to pass at 1 Volts voltage; 1 Ohm = 1 Volt / 1 Amper or I = V / R (Amp) (Volt) (Ohm) V = R x I (Volt) (Ohm) (Amp) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 60
61 Ohm Law Two circuits with different Resistances, identical voltage sources Resistance R 1 R 1 > R 2 Resistance R 2 Current (I 1 ) Current (I 2 ) Voltage (V) Resistance (R 1 ) Voltage (V) Resistance (R 2 ) V / R 1 = I 1 I 1 < I 2 V / R 2 = I 2 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 61
62 Voltage (Volt) Voltage (Volts) METU Ohm Law V-I Characteristics Current I (Amp) 30,0 25,0 Resistance (R) 20,0 15,0 10,0 Slope = V / I = R V (Volt) = R x I (Ohm) (Amp) 5,0 0,0 V varies linearly with I 0,0 5,0 10,0 15,0 20,0 Current I (Amp) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 62
63 Ohm Law - Example Question Current I (Amp) Calculate the current flowing in the circuit shown on the RHS V= 220 V R = 5 Ohm V = R x I (Volt) = (Ohm) x (Amp) I = V / R = 220 / 5 = 44 Amps EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 63
64 Voltage (Volts) METU V = R x I (Volt) (Ohm) (Amp) R = R 0 (1 t) t = T 23 0 C R 0 = Resistance at 23 0 C = The temperature coeff. of the metal Ohm Law Nonlinear V-I Characteristics Note that resistance increases with temperature, hence current is reduced V = 220 V Slope = V / I = R I = 5 Amp Load Current I (Amp) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 64
65 Resistance Formula Resistance Formula R = l / A where, R is the resistance of conductor, is the resistivity coefficient, = 1 / 56 Ohm-mm 2 /m (Copper) 1 / 32 Ohm-mm 2 /m (Aluminum) l (m) is the length of the conductor A (mm 2 ) is the cross sectional area of the conductor ACSR Conductor (Aluminum Conductor Steel Reinforced) l (meter) A (mm 2 ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 65
66 Resistance Formula Resistance Formula Resistance of a cable is proportional to the length and inversely proportional to the cross sectional area of the cable R = l / A where, R is the resistance of conductor, is the resistivity coefficient, = 1 / 56 Ohm-mm 2 /m (Copper) 1 / 32 Ohm-mm 2 /m (Aluminum) l (m) is the length of the conductor A (mm 2 ) is the cross sectional area of the conductor EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 66
67 Resistance Formula Resistance Formula Example Calculate the resistance of a copper cable with length 3200 meters and cross section 240 mm 2 Solution R = (1 / 56) 3200 / 240 = Ohms Aluminum Conductors EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 67
68 Resistance Formula Resistance Formula Example Calculate the resistance of a copper cable with length 3200 meters and cross section 240 mm 2 Solution R = (1 / 56) 3200 / 240 = Ohms ACSR Conductor (Aluminum Conductor Steel Reinforced) l 3200 (m) A = 240 (mm 2 ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 68
69 Resistivity Coefficients of Various Metals R = l / A Formula = 1 / 56 Ohms/meter (Copper) = Ohm-mm 2 /m where, R is the resistance of conductor, is the resistivity coefficient, = 1 / 56 Ohm-mm 2 /m (Copper) 1 / 32 Ohm-mm 2 /m (Aluminum) l (m) is the length of the conductor A (mm 2 ) is the cross sectional area of the conductor Material Resistivity Coefficients Resistivity Coefficient Ohm-mm 2 /m Resistance Ohms/feet Silver Copper Gold Aluminum Mercury Brass Nickel Iron Platinium Steel Lead EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 69
70 Color Codes for Resistances Rule EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 70
71 Insulator Insulator HV side Insulator is a material with almost infinite resistance Insulators are used to support HV lines and conductors In practice, all materials have resistances. Hence, they conduct a certain amount of current when a voltage is applied to the terminals. Insulator are materials that conduct only a very small amount of current, even when an extremely high voltage is applied to the terminals. Ground side EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 71
72 Power dissipation in a Resistance V = R x I (Volt) = (Ohm) (Amp) Current, I On the other hand, it was shown in this lecture that; Voltage, V R Power = Voltage x Current or P = V x I Power Hence, power dissipation in resistance R is Power = R x I x I = R x I 2 Watt EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 72
73 Series Connected Resistances Equivalent Resistance Formula R 1 = l 1 / A 1 R 2 = l 2 / A 2 Let A 1 = A 2 Hence; l total = l 1 l 2 l 3200 (m) A = 240 (mm 2 ) l 3200 (m) A = 240 (mm 2 ) R total = l total / A = (l 1 l 2 ) / A = l 1 / A l 2 / A = R 1 R 2 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 73
74 Series Connected Resistances Equivalent Resistance Formula R total = R 1 R 2 Series connected resistances are added R 1 R 2 R k R total = R 1 R 2... R k l 3200 (m) A = 240 (mm 2 ) l 3200 (m) A = 240 (mm 2 ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 74
75 Voltage (Volt) Voltage (Volt) METU Ohm Law for Series Resistances V = R total x I (Volt) (Ohm) (Amp) Current I (Amp) V = R 1 x I R 2 x I (Volt) (Volt) (Volt) Current I (Amp) Resistance (R total ) Resistance (R 1 ) Resistance (R 2 ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 75
76 METU Ohm Law for Series Resistances Current I (Amp) Aluminum Cables and Conductors V (Volt) V 1 (Volt) V 2 (Volt) V = R 1 x I R 2 x I (Volt) (Volt) (Volt) = V 1 V 2 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 76
77 Admittance Definition Inverse of resistance is called Admittance g (Siemens) = 1 / R (1 / Ohm) Unit of Admittance is Siemens Example Calculate the admittance of 10 k resistance shown on the RHS g = 1 / 10 4 = 10-4 Siemens EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 77
78 METU Shunt Connected Resistances Equivalent Resistance Formula V = V T V = V T I1 R 1 R 1 I 1 I 2 I 1 = V T / R 1 = V T x g 1 R 2 I Total R 2 I 2 I Total I 2 = V T / R 2 = V T x g 2 R k I k I k I k = V T / R k = V T x g k R k EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 78
79 METU Shunt Connected Resistances Equivalent Resistance Formula V = V T R 1 I 1 I 1 = V T / R 1 I 2 R 2 I Total I 2 = V T / R 2 R k I k I k = V T / R k V = V T I total = V T (1 / R 1 1 / R / R k ) = V T / R equivalent Hence, 1 R equiv = / R 1 1 / R / R k EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 79
80 Shunt Connected Resistances Example Find the equivalent resistance of the following connection R 1 = 1 Ohm V = V T I 1 1 R equiv = / R 1 1 / R / R k 1 R equiv = / 1 1 / 2 1 / 4 = 1 / (7/4) = 4 / 7 = Ohm R 2 = 2 Ohms I 2 I Total R k = 4 Ohms I k EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 80
81 Shunt Connected Resistances Example Find the equivalent admittance of the following connection 1 R equiv = / R 1 1 / R / R k V = V T g 1 = 1/1 Siemens I 1 or 1 1 / g equiv = g 1 g 2... g k g 2 = 1/2 Siemens g k = 1/4 Siemens I 2 I k I Total g equiv = g 1 g 2 g 3 = 1 / 1 1 / 2 1 / 4 = 7 / 4 = 1.75 Siemens EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 81
82 Voltages on Series Connected Elements Voltages on series connected elements are added V = V 1 V 2 (Volt) (Volt) (Volt) or generalizing V = V 1... V n-1 (Volt) (Volt) (Volt) Current I (Amp) V 1 (Volt) Current I (Amp) V 1 (Volt) V s (Volt) V (Volt)... V 2 (Volt) V n-1 (Volt) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 82
83 Voltages on Series Connected Elements Voltages on series connected elements are added V = V 1... V n-1 (Volt) (Volt) (Volt) Current I (Amp) i = n-1 V 1 (Volt) V = V i i =1 i = n-1 V n - V i = 0 i = 1 V s (Volt) i = n V i = 0 i = 1 V 2 (Volt) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 83
84 Kirchoff s Voltage Law (KVL) Statement Current I (Amp) The above result may be expressed as; Sum of voltages in a closed loop is zero V s (Volt) V 1 (Volt) or i = n V i = 0 i = 1 V 2 (Volt) Kirchoff s Second Law or Kirchoff s Voltage Law EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 84
85 Kirchoff s Voltage Law (KVL) Example i = n V i = 0 i = 1 V s = 220 Volts V = 220 V Current I (Amp) V 1 = -100 (Volt) V s V 1 V 2 = = 0 V 2 = -120 (Volt) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 85
86 Kirchoff s Voltage Law (KVL) Simple Rules Head (pinpoint) of the arrow is negative, Tail of the arrow is positive V 1 (Volt) This current is assigned such a direction that it always enters from the side of the resistance V s (Volt) V 2 (Volt) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 86
87 A Simple Rule for applying Kirchoff s Voltage Law (KVL) A Simple Rule Choose a ground node, Assume that current I flows clockwise, Starting from the ground node, assign and - signs to those passive elements (i.e. those elements other than source) in such a direction that the current enters to side and the leaves from the - side, Assign sign to the that side of the source from which current is leaving V s - Current I Ground node V 1 V k EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 87
88 A Simple Rule for applying Kirchoff s Voltage Law (KVL) A Simple Rule Then write down the voltages on each element by using Ohm Law on a path in a clockwise direction, Assign sign to those voltage terms in the equation that you pass from - to, Assign - sign to those voltage terms in the equation that you pass from to -, Stop and equate it to zero when you come again to the ground node that you have started Example; V s V 1 V 2 = 0 V s = V 1 V 2 V s - Current I Ground node V 1 V k EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 88
89 V = 220 V METU Kirchoff s Current Law (KCL) Summary of Kirchoff s Laws Kirchoff s Voltage Law (KVL) Algebraic sum of currents entering a junction is zero i = n I i = 0 i = 1 Node (Junction) Algebraic sum of voltages in a closed loop is zero i = n V i = 0 i = 1 Current I (Amp) Current I n Current I 1 Current I 2... V 1 = -100 (Volt) Current I n-1 V n-1 = -120 (Volt) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 89
90 Voltage Division Principle V 1 = R 1 x I V 2 = R 2 x I... V k = R k x I V s = V 1 V 2... V k = ( R 1... R k ) x I V k / V s = R k / ( R 1... R k ) Vs Input Voltage Current I... R 1 R 2 V 1 V 2 R k Voltage Division Ratio = R 1... R k R k V k Output Voltage = V s R k / (R 1..R k ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 90
91 Potentiometer (Voltage Divider) Principle Input voltage is divided and a certain fraction is given to the output R k V k = V s R 1... R k Current I R 1 R k Division Ratio = R 1... R k V 1 Vs R 2 V 2... R k V k A B C EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 91
92 Potentiometer (Voltage Divider) Circuit Arrangement Rotary Potentiometer R k Division Ratio = R 1... R k Rotary Potentiometer EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 92
93 Current Division Principle V T x g 1 = I 1 V T x g 2 = I 2... V T x g k = I k V T (g 1... g k ) = I 1... I k g k I k = I s g 1... g k g 1 g k Division Ratio = g 1... g k V = V T I 1 = V T x g 1 or Source Current I s g 2 I 2 = V T x g 2 I s V T (g 1... g k ) = I s I k / I s = g k / ( g 1... g k ) g k I k = V T x g k = I s g k / (g 1...g k ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 93
94 Voltage Sources Definition Voltage source is an element which creates a voltage difference at its terminals V = 24 Volts A simple Rule: Current is assigned such a direction that it always leaves the side of the voltage or current source. Voltage Source DC Voltage Source V = 24 V V = 24 V EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 94
95 Terminal Voltage (Volts) METU Definition Ideal Voltage Source An ideal voltage source is the one that the terminal voltage does not change with the current drawn An ideal voltage source has zero internal resistance 30,0 V = 24 Volts Battery Load Current I 25,0 20,0 15,0 10,0 5,0 V T = constant Internal resistance V = R x I = 0 V s = 24 V V T = V s 0,0 0,0 5,0 10,0 15,0 20,0 25,0 Load Current I (Amp) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 95
96 Non-Ideal (Real) Voltage Sources Definition V = 24 Volts A voltage source always has an internal resistance R connected in series with the source Writing down KVL for the above cct; V s V V T = 0 or Battery Load Current I V T = V s V where, V = R x I is called internal voltage drop Internal resistance V = R x I V s = 24 V V T = 24 - V Terminal voltage V T is reduced by V EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 96
97 Terminal Voltage V T (Volts) METU Non-Ideal (Real) Voltage Sources Definition Writing down KVL for the above cct; V T = V s V = Vs R x I 30,0 25,0 20,0 V T depends on I i.e. it is reduced as I is increased Battery Load Current I 15,0 Slope = - V T / I = - R Internal resistance V = R x I V s = 24 V V T = 24 - V 10,0 5,0 0,0 0,0 5,0 10,0 15,0 20,0 25,0 Load Current I (Amp) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 97
98 Ideal Current Source Definition Ideal Current Source An ideal current source is an element providing a constant current from its terminals Source Current I s Terminal Current I T I T = I s Ideal Current Source Source Current I s EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 98
99 METU Non-Ideal (Real) Current Source Definition Non-Ideal Current Source A non ideal current source is an element with a current depending on terminal voltage Terminal Current I T I s Ideal Current I Source I s g =1 / R V T I T = I s I I T = I s g x V T Current Source I s EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 99
100 Terminal Current I s (Amp) METU Non-Ideal (Real) Current Source Definition: Non-Ideal Current Source A non ideal current source is an element with a current depending on terminal voltage Non-Ideal Current Source Terminal Current I T 30,0 25,0 I T = I s I I T = I s g x V T I T depends on V i.e. it is reduced as V is increased 20,0 Ideal Current Source I s I g =1 / R V T 15,0 10,0 Slope = - I T / V T = - g 5,0 0,0 0,0 5,0 10,0 15,0 20,0 Terminal Voltage (Volts) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page ,0
101 Controlled (Dependent) Sources Definition: Controlled Sources Voltage Controlled Current Source A controlled source is an element with a current or voltage depending on any other voltage or current in the circuit Controlled Source: Current I s = A V x A = Amplification coefficient Voltage Controlled Current Source I s = A V x Is I s Terminal Current: I T V x EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 101
102 Controlled (Dependent) Sources Definition: Controlled Sources A controlled source is an element with a current or voltage depending on any other voltage or current in the circuit Current Controlled Current Source I s Current Controlled Current Source I s = A I x Terminal Current: I T I t - I s Controlled Source: Current I s = A I x I s I x EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 102
103 Controlled (Dependent) Sources Definition: Controlled Sources A controlled source is an element with a current or voltage depending on any other voltage or current in the circuit Voltage Controlled Voltage Source Voltage Controlled Voltage Source V s = A V x Controlled Source: Voltage V s = A V x V s V x EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 103
104 Controlled (Dependent) Sources Definition: Controlled Sources Current Controlled Voltage Source A controlled source is an element with a current or voltage depending on any other voltage or current in the circuit Controlled Source: Voltage V s = A I x I s Current Controlled Voltage Source V s = A I x Terminal Current: I T I t - I s V s I x EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 104
105 Example Question Current Controlled Voltage Source Solve the circuit on the RHS for current I x Write down KVL; Solution I s Current Controlled Voltage Source V s = 10 I x Terminal Current: I T 10 V V s 10 2 I x = 0 10 I x 10 2 I x = 0 8 I x = 10 I x = 10 / 8 = 1.25 Amp V s I x 2 Ohm EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 105
106 Measuring Devices - Ammeter An ammeter is a measuring instrument used to measure the flow of electric current in a circuit. Electric currents are measured in amperes, hence the name The word "ammeter" is commonly misspelled or mispronounced as "ampmeter" by some The earliest design is the D'Arsonval galvanometer. It uses magnetic deflection, where current passing through a coil causes the coil to move in a magnetic field The voltage drop across the coil is kept to a minimum to minimize resistance in any circuit into which the meter is inserted Coil Magnet Spring forcing arm back to zero position EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 106
107 Measuring Devices - Ammeter Ampere - Volt - Ohm (AVO)Meter An ammeter is always series connected in the circuit measured Ammeter I load R L EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 107
108 Measuring Devices - Ammeter An ammeter is always series connected in the circuit measured Internal Resistance of Ammeter Ammeter R amp I load R L EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 108
109 Ideal Ammeter Definition An ideal ammeter is the one with zero internal resistance (Short Circuit) An ideal ammeter behaves as a short circuit, i.e. R amp 0. An ideal ammeter has zero resistance so that the measured current is not influenced V s Ammeter R amp 0 I load R L No ammeter can ever be ideal, and hence all ammeters have some internal resistance EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 109
110 An ammeter should not influence the current measured I = V s / ( R R amp ) R amp 0 Hence, I = V s / ( R R amp ) V s / R Ideal Ammeter Ammeter R amp 0 I load R L EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 110
111 Non-Ideal (Real) Ammeter Definition No ammeter can ever be ideal, and hence all ammeters have some internal resistance Non-ideal Ammeter Ammeter R amp I load A real (non-ideal) ammeter has always an internal resistance in series V s A non ideal ammeter behaves as a series resistance with: R amp 0 Hence the the measured current is influenced (reduced) R L I load = V s / ( R R amp ) I load = V s / R } I load < I ideal EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 111
112 Measuring Devices Clamp Ammeter The Need for Clamp Ammeter Sometimes the electrical service carried out by the circuit may be so vital that it can not be interrupted by breaking the line for a series connection of the ammeter Ammeter shown on the RHS is a particular design for such circuits to measure current flowing in the circuit as well as resistance without braeaking the circuit EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 112
113 Measuring Devices - Voltmeter A voltmeter has a high internal resistance so that it passes only a small current An ideal voltmeter has a very large resistance so that the the circuit in which it has been placed is not disturbed An ideal voltmeter is an open circuit However, no voltmeter can ever be ideal, and therefore all voltmeters draw some small current Voltmeter is always parallel connected to the terminals measured EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 113
114 Measuring Devices - Voltmeter A voltmeter has a high internal resistance so that it passes only a small current A voltmeter is always shunt (parallel) connected in the circuit that it measures Measured voltage; Red lines are not part of the circuit I Source R 1 I Load I Voltmeter (negligible) R m Voltmeter R L V o = V s R 1 R L V s R L V o Voltmeter is always parallel connected to the terminals measured EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 114
115 Ideal Voltmeter Definition An ideal voltmeter is the one with infinite internal resistance (Open circuit) An ideal voltmeter has a very large resistance, R m. i.e. it behaves as an open circuit, so that the the measured circuit is not influenced However, no voltmeter can ever be ideal, and therefore all voltmeters draw some current A real voltmeter has a certain internal resistance so that it passes a certain current V s I Source R 1 I Load R L V o I m (negligible) R m Voltmeter EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 115
116 Ideal Voltmeter No voltmeter can ever be ideal, and therefore all voltmeters draw some current. I Source R 1 I Load I m (negligible) R m Voltmeter R m i.e. R m >> R L I m << I Load V s R L V o I Source = I Load I m I Load V o = R L ( I source I m ) = R L I Source R L I m R L I Source } Negligible EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 116
117 METU Example Problem Ideal Voltmeter I m 0 Calculate the internal admittance g m of a voltmeter, if it reads Volts when connected to a 0.48 ma current source with an internal admittance of g s = 4 x 10-5 Siemens I g V read = V Non-ideal Voltmeter I g I m (negligible) Voltmeter Siemens = 1/ I S =0.48 ma g s = 4 x 10-5 S g m = 1/R m EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 117
118 METU Problem Example Ideal Voltmeter I m 0 R s = 1 / g s = 1 / (4 x10-5 ) Siemens = 10 5 / 4 = 25 k I g Non-ideal Voltmeter I s x R eq = V read = Volts Hence, R eq = V read / I s = / (0.48 x 10-3 ) = I S =0.48 ma V read = V g s = 4 x 10-5 S I g g m = 1/R m I m (negligible) Voltmeter R eq = R s // R m Hence, R eq =(R s x R m ) / (R s R m ) = Rm = M EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 118
119 Advanced Measuring Devices Power Quality Analyzer EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 119
120 Wheatstone Bridge The Wheatstone Bridge is an electrical circuit used to determine an unknown resistance R x by adjusting the values of known resistances, so that the current measured in the line connecting the terminals C and D is zero R b R x R 1 R 2 A x C I = 0 D B EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 120
121 Wheatstone Bridge Principle Adjust the resistances R 1, R 2 and R b such that the ammeter connected between the terminals C and D reads zero current Hence, the voltage difference between the terminals C and D is zero V CD = 0 or V C = V D EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 121
122 Wheatstone Bridge Principle A V C = V D V C = V s R b / ( R x R b ) V D = V s R 2 / ( R 1 R 2 ) V s C x I = 0 D V s R b / ( R x R b ) = V s R 2 / ( R 1 R 2 ) or R b / ( R x R b ) = R 2 / ( R 1 R 2 ) R b ( R 1 R 2 ) = R 2 ( R x R b ) R b R 1 R b R 2 = R 2 R x R 2 R b or R x = R b x R 1 / R 2 V s A B A R x R 1 C D R R b 2 B EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 122
123 Wheatstone Bridge Basic Rule Cross multiplication branch resistances must be equal at balance condition x A R x x R 2 = R b x R 1 Please note that voltage V s is neither used, nor needed in the above equation, i.e. its value is arbitrary V s C I = 0 D B EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 123
124 Wheatstone Bridge Example Calculate the value of unknown resistance R x in the balanced Wheatstone Bridge shown on the RHS Cross multiplication of branch resistances must be equal at balance condition: R x x R 2 = R b x R 1 R x = R b x R 1 / R 2 = 100 x 100 / 20 = 500 Ohm EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 124
125 Switch or Circuit Breaker Switch - Circuit Breaker Switch or circuit breaker is a device used to open an electrical circuit manually or automatically by an electronic relay system Switch I Load R 1 V s R 1 Open Off Closed On EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 125
126 Meaning of Open and Closed (Highly Important) Breaker Open Circuit or Switch Closed Circuit or Switch Open (On), Closed (Off) Open Circuit Open (Off) Switch Closed Circuit Closed (On) Switch I Load = 0 I Load R 1 V s R 1 R 1 V s R 1 Closed Switch (On) does NOT mean that there is no voltage (current) in the circuit! EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 126
127 Thermal-Magnetic Circuit Breaker 220 Volt, 63 Amp. Thermal-Magnetic (Molded-Case) Breaker Open Circuit Open (Off) Switch Closed Circuit Closed (On) Switch I Load = 0 I Load R 1 R 1 V s R 1 V s R 1 Closed Switch (On) does NOT mean that there is no voltage (current) in the circuit! EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 127
128 Medium Voltage (36 kv) Vacuum Circuit Breaker Open Circuit R 1 I Load = 0 Vacuum tube enclosing breaker poles V s R 1 Tripping buttons Spring handle (taken off) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 128
129 Did everbody understand the? EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 129
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