المحاضرة األولى. Electrical Circuits Analysis فهمى الخولى

Transcription

1 المحاضرة األولى Electrical Circuits Analysis أ.د. فهمى الخولى

2 Level 1 No. of Hrs/Week: Lecture 2 Tutorial 2 Time: Tusday 10:45-12:15 A.M. Location: L 316

3 Instructor Information Name: Prof. Fahmy El-Khouly Office Hours: Sunday to wensday.

4 - Course Objectives: 1 Understand the behavior of passive circuit elements. 2 Enrich the student knowledge about dc circuits' theories. 3 Develop the student ability to analyze and solve DC circuits.

5 2- Intended Learning Outcomes (ILO s) : a- Knowledge and understanding : - Define the different parameters of electrical circuits (voltage, current, power source, ). - Describe the voltage-current characteristics of different circuit elements. - Identify the different types of energy storage elements. - Define the different theories that can be applied to electric circuits. B- Intellectual Skills - Determine the characteristic curve of electrical circuit elements - Solve DC circuits using Mesh and Nodal Methods. - Apply the different theories to solve electrical circuits. - Create electrical circuits to verify linear circuit theorems. c- Professional skills: - Develop skills and practice in the design and fabrication of practical DC circuits used in electrical systems. - Design DC Circuits to meet certain specifications. - Use CAD tools to synthesis and simulate DC circuits.

6 D- General and Transferable Skills: - Present technical reports about basics of electrical circuits and different solving techniques. - Use IT effectively. - Work with limited or contradictory information.

7 Weighting of assessments: -Quizzes 20 (Degrees) 20 % Activities 20 (degrees) 20 % -Mid-Term Exam 20 (Degrees) 20 % - Final-Term Exam 40 (Degrees) 40 % Total 100 (Degrees) 100 % 6- List of References : Electric Circuits James Nilsson and Susan Riedel, Ninth Edition

8

9 CV Prof. Dr. Fahmy M. El-Khouly was born in Minoufiya Egypt in He received his B.Sc, M.Sc and ph.d. Degrees in Electrical Engineering from Minoufiya University in 1982, 1988, and 1995 respectively. - He joined the Department of Electrical Engineering Minoufiya University as a demonstrator in 1983, and became an assistant lecture in 1988 and a lecturer in He was promoted to an associate professor in 2001 and to full Professor in From 1992 to 1994 he was a visiting student in University of New Brunswick, Canada under academic channel system for ph.d. Study. - In 2005 he awarded the Minoufiya University prize in science engineering. - From July 2007 he joined Wuppertal University, Germany for three months as visitor researcher. - He was a department member of the engineering consulting center committee in the faculty of Engineering, Menoufiya University. - He was a member of Post graduate committee in the faculty of Engineering, Menoufiya University. - He was a member of Education committee in the faculty of Engineering, Menoufiya University. - He was preparing the study plan and table in the department of Elec. Eng., faculty of Engineering, Menoufiya University. - He was the manger of the consulting and research center in the faculty of Engineering, Menoufiya University. - Now, he is the vice dean of the student and education affair. - Professor El-Khouly published over 30 papers and has 25 years of teaching experience. He supervised over ten of M.sc. and ph.d. Students. His current interests are Power electronics, artificial intelligent control strategies, and Electrical Machines control.

10 Chapter 1 Circuit Elements [ chaprte 2 in the book ] Voltage and Current Sources Electrical Resistance ( Ohm s Law ) Kirchhoff s Laws Analysis of circuit containing dependent Sources

11 I: Current (A) Ampere. P : Power (W) Watt. V : Voltage (v) volt.

12 Electrical Resistance (Ohm s Law) : V = IR R= Resistance (Ω) ohm Condactanceعكس (G) = 1/R المقاومة S (Siemens) (ᵿ) maho الوحدات

13 Voltage and Current Source : ال يعتمد على شيء: -1 Independent Sources

14 Dependent Sources Constant = α, ᵦ, μ, ρ Figure 2.2 The circuit symbols for (a) an ideal dependent voltage-controlled voltage source, (b) an ideal dependent current- controlled voltage source, (c) an ideal dependent voltage-controlled current source, and (d) an ideal dependent currentcontrolled current source.

15 * Active Elements أجزاء تعطي طاقة Give power أجزاء ال ت عطي طاقة * Passage Elements Can not generate power Loadمثل (resistance,capastance)

16 *Power dissipated in resistance : كل األحمال تستهلك طاقة و لكن المصادر ت عطي طاقة في األساس و لكن قد تستهلك طاقة

17 يستهلك طاقة ي عطي طاقة P = V I P = - V I

18 Kirchoff s Law : 1- Current Law :σ ί = 0 at any node 2- Voltage Law :σ v = 0 For any close loop Ia+Ib-Ic = 0 E I *R1 - I *R2 = 0

19 Example a) Use Kirchhoff's laws and Ohm's law to find i0 in the circuit shown in Fig. b) Test the solution for i0 by verifying that the total power generated equals the total power dissipated.

20

21

22 5 Ia A Example: Find V 500v V - + V - 20 Ib 5Ia 5Ic

23 Example: Find Vo

24 المحاضرة الثانية Electric Circuits Analysis أ.د. فهمى الخولى

25 1. Resistors in series Chapter 2 Simple Resistive Circuits Two resistances is said to be in series when: they connected at a single node and carry the same current as shown in fig 2.1 R 1 R 2 + i s i 1 i 2 Vs - R 5 R 4 i 3 R 3 Vs + i s R eq - i 5 i 4 Fig. 2.1

26 i s = i 1 = i 2 = i 3 = i 4 = i 5 Kirchhoff's voltage law in fig 2.1 V s = i 1 R 1 + i 2 R 2 + i 3 R 3 + i 4 R 4 + i 5 R 5 = i s R 1 + R 2 + R 3 + R 4 + R 5 = i s R eq Then, the equivalent resistance for series resistances is: R eq = R 1 + R 2 + R 3 + R 4 + R 5

27 1. Resistors in Parallel Two resistances is said to be in parallel when: 1. An one terminal of each is connected together in a node and the other terminal of each is connected together in another node. 2. The total current input to them is the same which output from both. 3. The total current is the summation of their individual currents

28 + Vs - i 1 i 2 i 3 i 4 + i s R 1 R 2 R 3 R 4 Vs - i s R eq Fig. 2.2 As shown in fig. 2.2 Applying Kirchhoff's current law i s = i 1 + i 2 + i 3 + i 4 Also, V s = i 1 R 1 = i 2 R 2 = i 3 R 3 = i 4 R 4

29 Then i 1 = v s R 1 i 2 = v s R 2 i 3 = v s R 3 i 4 = v s R 4 i s = v s ( 1 R R R R 4 ) i s v s = 1 R eq = ( 1 R R R R 4 )

30 For k numbers of resistances 1 R eq k 1 = = ( ) R i R 1 R 2 R k i=1 Using the conductance G =1/R G eq = σk i=1 G i For only two resistances 1 R eq = = G 1 + G G k 1 R R 2 = R 1 + R 2 R 1 R 2 R eq = R 1 R 2 R 1 + R 2

31 Voltage divider and current divider 3-1 Voltage divider V s = i s R 1 + i s R 2 i s = v s R 1 + R 2 Vs -+ i s V1 - V2 - R 1 R 2 v 1 = i s R 1 = v s R 1 R 1 + R 2 v 2 = i s R 2 = v s R 2 R 1 + R 2

32 + + + If a load is connected to R2 : v o = i s R eq = v s R eq R 1 + R eq Vs -+ i s V1 - V2 - R 1 R 2 Vo - R L Where R eq = R L R 2 R L + R 2

33 3-2 Current divider + i 1 i 2 Vs - i s R 1 R 2

34 1. Voltage Division and Current Division It is a general case of the previous section as following 4-1 Voltage division R 1 R 2 Vs + i i + R i V o - R n R n-1 - i = v s R eq i V O = i R i = V S R i R eq

35 4-2 Current division Vs + - i s R 1 R 2 R n i n + V -

36 5-Measuring resistance The Wheatstone Bridge G is a galvanometer to detect the current in µa. when its current is zero then Vs + i s a R 1 G R 2 b - R 3 R X Then

37 And Then

38 6- Delta to Wye (pi- to- tee) Equivalent Circuits 6-1 deta or pi a R c b a R c b R b R a R b R a c 6-2 Wye or Tee ( Y or T ) c a b a R 1 R 2 R 1 R 2 b R 3 R 3 c c

39 Converting from Y to Delta and vies versa Delta to Y a b R 1 R 2 R 3 a R c b c R b R a c

40 Solving the above three equations give

41 Also, Y to Delta

42 Example V + i s a b Required : is Solution and power supplied by the source a b

43 V i s a b

44 + Quiz 2 i o 10 V -+ i s 6 3 is 3 Vo - Find: Vo, power generated, and power absorbed

45 المحاضرة الرابعة Electrical Circuits Analysis فهمى الخولى أ.د.

46 Chapter 3 Techniques of circuit Analysis 1. Definitions Node : a point where two or more circuit elements join Essential node : a point where three or more circuit elements join Branch : a path that connect two nodes Essential a branch: a path that connect two essential nodes without passing through an essential node Loop : a path whose last node is the same as the starting node Mesh : a loop that does not enclose any other loops

47 1. Introduction to the node-voltage method - Mark clearly the essential nodes - Select one as essential node, usually it is the node with the most branches and flag it by and use it as reference node - Define the node voltage with respect to the reference voltage - Write the node kirchhoff's current law as function of node voltage assuming that the summation of all currents go out from the node equal zero

48 Example: V A - Node 1 Node 2

49 3- The node voltage method and dependent sources Node ix + 10 V ix A -

50 4-The node voltage method: some special cases In case when there is a source between a two nodes 1 1 V V + 5 i 10 2 A -

51 By adding the two equations supernode 10 V A -

52 5- Introduction to the Mesh-Current method 8 (i a - i b ) +2 i a = =4 i c + 7 (i c - i b )

53 6- Mesh-Current method with dependent sources -15 i x = 4 i c + 7 (i c - i b ) -15 (i a - i b ) =4 i c + 7 (i c - i b ) (1)

54 8 (i b i a ) + 6 i b +7 (i b i c ) =0

55 7- Mesh-Current method with a branch includes a current source 8 (i a - i b ) +2 i a = 40 8 (i b - i a ) +6 i b +4 i c = - 20

56 8- Source transformations Voltage source Current source Suppose RL is connected between a and b, then For Voltage source for Current source

57 Equating il in the two cases then i s = V s R

58 Thevenen and Norton Equivalent A general circuit Thevenen equivalent circuit At no load At short circuit v Th =v ab v Th =i R sc Th R Th = v Th i sc

59 Example Find RTh and VTh between a and b v v = 0 v 1 = 32 = v Th

60 v v 2 = 16 + v v 2 4 = 0 i sc = 16 4 = 4 A R Th = v Th i sc = 32 4 = 8 Ω

61 Superposition theorem + i = i - i i 1 = i 1 + i 1 i 2 = i 2 + i 2

62 R eq = =8.33 i = = 3 A i 1 = = 2.5 A i 2 = = 0.5 A

63 R eq1 = = 4 i 1= 3 i 2= = 1.5 A = 0.3 A i = = 1.2 A i = i - i = 3-1.2= 1.8 A i 1 = i 1 + i 1= 2.8 A i 2 = i 2 + i 2= 0.8 A

64 Example Find Vth and Rth between a and b

65 Example Find Vo using superposition

66