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BEKG1113 PRINCIPLE OF ELECTRICAL AND ELECTRONICS CHAPTER 2 (WEEK 3) FACULTY OF ELECTRONIC AND COMPUTER ENGINEERING 2 Chapter 2 • The topics for this chapter: – Voltage and current sources – Resistor – Resistor Color code – Ohm’s Law – Types of circuit - series, parallel and seriesparallel circuit – Circuit ground, KVL, KCL, Power dissipation 3 Voltage source – Ideal voltage source Can provide a constant voltage for any current required by a circuit [1]. – Voltage source can either be AC (Alternating current) or DC (Direct current) [2]. 4 Voltage source • The symbols for voltage sources are represented here [1]: VI characteristics 6 5 V (Volts) • VI characteristics for an ideal voltage source is shown by the graph [1] 4 3 V 2 1 0 1 2 3 4 I (Ampere) 5 6 7 5 Voltage source • Six categories of voltage sources [1]: – Batteries – Solar cells – Generator – The electronic power supply – Thermocouples – Piezoelectric sonsors 6 • Thermocouples – It is a thermoelectric type of voltage source that is commonly used to sense temperature. The operation is based on Seebeck effect. – The types of thermocouples are categorized by letters (J,K,E,N, B and R) that depends on the range of temperature. 7 • The graph of thermocouple characteristics. • Note that K is the most common thermocouple used. 8 • A few types of thermocouples which depend on their specification requirements. 9 • Piezoelectric sensors – They act as voltage sources and are based on the piezoelectric effect where a voltage is generated when a piezoelectric material is mechanically deformed by an external force. – Types of piezoelectric material; quartz & ceramic – Applications: pressure sensor, force sensor 10 Chapter 2: Direct Current (DC) Circuit Chapter 2: Direct Current (DC) • The current source – Definition: The ideal current source can provide a constant voltage for any load. – The symbol for a current source is shown below 11 Current Source • Constant current sources – a type of power supply 12 Current Source • Constant current sources from most transistor circuits • Constant current battery chargers 13 Basic DC Circuit • It consists of a source of electrical energy, some sort of load to make use of that energy, and electrical conductors connecting the source and the load. + SOURCE - LOAD 14 Electrical Circuit Requirements • Control Device: Allows the user control to turn the circuit on or off. Switches are the devices commonly used for controlling the oppening or closing of circuits. • Protection Device: Current must be monitored and not allowed to exceed a safe level as to protect users from shock, to protect the equipment from damage, and to prevent fire hazards. 15 Protective Devices • Fuses: Fuses use a metallic element that melts when current exceeds a preset value. It will blow if more current passes through it. • Circuit breaker: Circuit breaker works on the different principle. When the current exceeds the rated value of breaker, the magnetic field produced by excessive current operates a mechanism that trips open a switch. After the fault or overload condition has been cleared, the breaker can reset and used again. 16 17 Current Direction 18 Circuit Ground • Grounding is achieved in an electrical system when one of the conductive wires serving as part of the circuit path is intentionally given a direct path to the earth. This method of grounding is called earth ground. • In most electronic equipment, a large conductive area on printed circuit board or the metal housing is used as the circuit ground or chassis ground. 19 • Ground is the reference point in electric circuits and has a potential of 0 V with respect to other points in the circuit. All of the ground points in a circuit are electrically the same and therefore common points. 20 Resistors • Definition: – A component that is specifically designed to have a certain amount of resistance is called resistor. Color bands Resistance material (carbon composition) Insulation coating Leads 21 Resistors • Fixed resistors: provides a specific constant value of resistance – – – – Carbon-composition Chip resistor Film resistor Wirewound resistor • Variable resistor: values can be changed easily with manual or an automatic adjustment – Potentiometer – Rheostat 22 Carbon-composition • Mixture of finely ground carbon, insulating filler and resin binder. • The ratio of carbon to insulating filler sets the resistance value. • Commonly use fixed resistor. 23 Chip Resistors • SMT (Surface Mount Technology) component • Very small in size ,suitable for compact assemblies. 24 Film Resistors • The resistive material could be carbon film or metal film. • The desired resistance value is obtained by removing part of resistive material in a spiral pattern along the rod 25 • Another type of film type resistor 26 Wirewound Resistors • Constructed with resistive wire wound around an insulating rod and then sealed. • Used in application that require higher power rating resistance 27 • Several types of wirewound resistors 28 Potentiometer • used to divide voltage • Has 3 terminal; terminal 1&2 have a fixed resistance between them, which is the total resistance. Terminal 3 is connected to moving contact. 29 Rheostat Chapter 2: Direct Current (DC) • Used to control current • Potentiometer can be used as a rheostat by connecting terminal 3 to either terminal 1 or terminal 2 30 Resistors • Resistor color code 31 32 Ohm’s Law • Ohm’s law states that current is directly proportional to voltage and inversely proportional to resistance. The formula given are: V = IR where: I = current in amperes (A) V = voltage in volts (V) R = resistance in ohms (Ω) 33 • Voltage - is not affected by either current or resistance. It is either too low, normal, or too high. If it is too low, current will be low. If it is normal, current will be high if resistance is low, or current will be low if resistance is high. If voltage is too high, current will be high. • Current - is affected by either voltage or resistance. If the voltage is high or the resistance is low, current will be high. If the voltage is low or the resistance is high, current will be low. • Resistance - is not affected by either voltage or current. It is either too low, okay, or too high. If resistance is too low, current will be high at any voltage. If resistance is too high, current will be low if the voltage is fix. 34 The linear relationship of current and voltage. • The relationship can be portrayed by the graph below: 35 Example 1 Show that if the voltage in the circuit is increased to three times its present value, the current will triple in value. 10V R 4.7kΩ 36 Example 2 Assume that you are measuring the current in a circuit that is operating with 25V. The ammeter reads 50mA. Later, you notice that the current has dropped to 40mA. Assuming that the resistance did not change, you must conclude that the voltage source has changed. How much has the voltage changed, and what is its new value? 37 The inverse relationship of current and resistance. • As you have seen, current varies inversely with resistance as expressed by Ohm’s law, I = V/R. When the resistance is reduced, the current goes up; when the resistance is increased, the current goes down. For example, if the source voltage is held constant and the resistance is halved, the current doubles in value; when the resistance is doubled, the current is reduced by half. 38 The Inversely Proportional Relationship Chapter 2: Direct Current (DC) Circuit • The inversely proportional can be explained by this graph 39 Example 3 How many amperes of current are in the circuit of figure below? 100V R 22Ω If the resistance in above figure is changed to 47 Ω and the voltage to 50 V, what is the new value of current? 40 Example 4 • Calculate the current in figure below. 50V R 1.0kΩ 41 Example 5 – How many milliamperes are in the circuit below? 30V R 5.6kΩ 42 Example 6 – In the circuit below, how much voltage is needed to produce 5 A of current? 5A V 5A R 100Ω 43 Example 7 – How much voltage will be measured across the resistor below? V 5 mA R 56Ω V 44 Example 8 – Suppose that there is a current of 8 µA through a 10 Ω resistor. How much voltage is across the resistor? 45 Example 9 – In the circuit of figure below, how much resistance is needed to draw 3.08 A of current from battery? 12 V 3.08 A R 46 Example 10 – Suppose that the ammeter in figure below indicates 4.55 mA of current and the voltmeter across the supply reads 150V. What is the value of R? 47 Types of Circuit • Series • Parallel • Series-parallel circuit 48 Series Circuit • Resistor in series A R1 1k R2 1k R3 1k R4 1k B • A series circuit provides only one path for current between two points so that the current is the same through each series resistor [1]. 49 • The equation of a series resistors circuit RT R1 R2 R3 .... RN • The current flow is the same through each element of the series circuit. IT I1 I 2 I 3 ... I N • The voltage across the source or power supply is equal to the sum of the voltage drops across the separate resistors in series. VT V1 V2 V3 .... VN 50 Power Distribution • Power distribution in series circuit – It is represented by this equation [2]: PE PR1 PR2 PR3 – The power delivered by the supply can be determined using the equation [2]: PE EI s 51 Voltage Sources • Voltage sources in series [2]: 52 Example 11 • In figures below, find out the total resistance between point A and point B A R1 20 R2 220 R3 1.2k RT B R4 5.6k A A R2 3.3k R1 3.3k R3 3.3k RT R1 1k R2 1k RT R3 1k R4 3.3k B RT B B A R5 1k R6 1k R7 1k R8 1k R9 1k R10 1k R4 1k 53 54 Voltage Divider • The voltage across a resistor in a series circuit is equal to the value of that resistor times the total applied voltage divided by the total resistance of the series configuration [2]. VR X RX VT RT 55 Example 12 a. Find the total resistance RT. b. Calculate the resulting source current Is. c. Determine the voltage across each resistor. 2 2 Is E R1 + R2 20V 5 5 R3 1 1 56 Example 13 E + 36V 1k 3k R1 R2 R3 2k 57 Example 13 a. b. c. d. e. Determine the total resistance RT. Calculate the source current IS. Determine the voltage across each resistor. Find the power supplied by the battery. Determine the power dissipated by each resistor. f. Comment on whether the total power supplied equals the total power dissipated. 58 Parallel Circuit • The parallel resistors circuit + V1 10V R1 1k R2 1k R3 1k R4 1k R5 1k 59 • The equation of a parallel resistors circuit 1 1 1 1 1 ... Req R1 R2 R3 RN • Voltage across each resistor is the same as the voltage across the parallel combination. VT V1 V2 V3 .... VV • The current flowing through the parallel combination is the sum of the current in the separate branches. IT I1 I 2 I 3 ..... I N 60 Power in Parallel Circuits BENG 1113: CHAPTER 2 WEEK 5 • Total power in a parallel circuit is found by adding up the powers of all the individual resistors, the same as for the series circuits. PT P1 P2 P3 .... PN 61 Exercise 14 • What is the total resistance between point A and point B B A R5 1k R6 1k R7 1k R8 1k R9 1k R10 1k 62 Current Divider • A parallel circuit acts as a current divider because the current entering the junction of parallel branches “divides” up into several individual branch currents. RT IX IT RX 63 Series-Parallel 64 KVL • KVL is the abbreviation of Kirchoff’s Voltage Law • The sum of the voltage drops around a closed loop is equal to the sum of the voltage sources of that loop • Total voltage = 0 65 E VR1 VR2 VR3 2 R1 E + R2 20V 5 R3 1 66 KCL • Kirchoff’s Current Law • The current arriving at any junction point in a circuit is equal to the current leaving that junction • Current in = current out 67 I T I R1 I R2 I R3 I R4 I R5 BENG 1113: CHAPTER 2 WEEK 5 + V1 10V R1 1k R2 1k R3 1k R4 1k R5 1k 68 The Branch Current Method • The branch current method is a circuit analysis method using KVL and KCL to find the current in each branch of a circuit by generating simultaneous equation. 69 Example: The Branch Current Method + 5Ω - + 2Ω - a I3 + 10Ω I1 20V - b I2 8V 70 • Currents I1, I2 and I3 are assigned to the branches a as shown. • Applying KCL at node a I1 I 2 I 3 • Appling KVL at both loops resulting; 20 5I1 10 I 3 0 2 I 2 10 I 3 8 0 71 • The following equations can be written 5I 2 15I 3 20 2 I 2 10 I 3 8 • Solving all three equations will give us I1 2 A I 2 1A I 3 1A 72 The Mesh Current Method BENG 1113: CHAPTER 2 WEEK 5 • In the loop current method, you will work with loop current instead of branch current. 73 The Mesh Current Method 5Ω 20V 10Ω I1 b 2Ω a I2 8V 74 • Appling KVL at both loops resulting; 20 5I1 10I1 I 2 0 2 I 2 10I 2 I1 8 0 • Solving both equations will give us; I1 2 A I 2 1A 75 The Node Voltage Method BENG 1113: CHAPTER 2 WEEK 5 • Another method of analysis is node voltage method. • It is based on finding the voltages at each node in the circuit using KCL. 76 The Node Voltage Method + 5Ω - + 2Ω - a I3 + 10Ω I1 20V - b I2 8V 77 • Currents I1, I2 and I3 are assigned to the branches a as shown. • Applying KCL at node a I1 I 2 I 3 • Express the currents in terms of circuit voltages using Ohms’s law 20 Va I1 5 Va 8 I2 2 Va I3 10 78 • Substituting these terms into the current equation yields; BENG 1113: CHAPTER 2 WEEK 5 20 Va Va 8 Va 5 2 10 Va 10V 79 Superposition • A linear network which contains two or more independent sources can be analyzed to obtain the various voltages and branch currents by allowing the sources to act one at a time, then superposing the results. • This principle applies because the linear relationship between current and voltage. • Voltage sources are replaced by short circuits; current sources are replaced by open circuits. • Superposition can be applied directly to the computation of power. 80 R1 4ohm R2 47ohm Is1 20A R3 27ohm + Vs1 200V - R4 23ohm 81 R1 4ohm R2 47ohm o.c R3 27ohm + Vs1 200V - R4 23ohm 82 • By applying superposition principle, with the 200V acting alone, the 20A current source is replaced by an open circuit. RT 60.5 I T 3.31A I 23 1.65 A 83 R1 4ohm R2 47ohm Is1 20A R3 27ohm s.c R4 23ohm 84 • When the 20-A source acts alone, the 200V source is replaced by a short circuit. RT 11.02 I 23 11.02 20 9.58 A 23 • The total current in the 23Ω resistor is I 23 I 23old I 23new 11.23 A 85 Source Transformation Norton’s equivalent Thevenin’s equivalent a 5ohm a + 10V 2A 5ohm b b 86 Example: The Node Voltage Method 4 VA RA 1 RC RB 2 RD 3 5 RE VB 87 • Assign all currents are going out from node 1 &2. • From node 1; V1 VA V1 V1 V2 0 RA RB RC • From node 2; V2 V1 V2 V2 VB 0 RC RD RE 88 Example: Mesh Analysis 10Ω 2Ω 5Ω 2Ω I1 I2 25V 4Ω I3 50V 89 • The matrix that we’ll get 2 I1 5I1 I 2 25 0 25 5I 2 I1 10 I 2 4I 2 I 3 0 4I 3 I 2 2 I 3 50 0 • Solving I1 1.31A I 2 3.17 A I 3 10.45 A -ve sign in I1 represents that the assigned current direction opposed the actual current direction. 90 • Solve the same circuit using node voltage method 1 10Ω 2 2Ω 5Ω 2Ω I1 I2 25V 3 4Ω I3 50V 91 Solution • at node 1 V1 V1 25 V1 V2 0 2 5 10 • At node 2 V2 V1 V2 V2 50 0 10 4 2 92 • From these V1 2.61V V2 29.1V • The currents can be determined V1 I1 1.31A 2 V1 V2 I2 3.17 A 10 V2 50 I3 10.45 A 2 93 Test 1 Week 7 • Scientific notation and engineering notation, Resistor color code • Ohm’s Law, KVL and KCL in Mesh or Node Analysis • Basic Circuit measurement and measurement equipment 94 References [1] Thomas L.Floyd; Principle of Electric Circuits 8th Ed; Pearson Education; 2007 [2] Robert L.Boylestad; Introductory Circuit Analysis11th Ed; Pearson Education; 2007