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Chapter 32


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Chapter 32

  1. 1. start
  2. 2. <ul><li>Prepare for ASE Electrical/Electronic Systems (A6) certification test content area “A” (General Electrical/Electronic Systems Diagnosis). </li></ul><ul><li>State Ohm’s law. </li></ul><ul><li>Identify the parts of a complete circuit. </li></ul><ul><li>State Watt’s law. </li></ul><ul><li>Describe the characteristics of an open circuit, a short-to-ground, and a short-to-voltage. </li></ul>OBJECTIVES: After studying Chapter 32, the reader should be able to:
  3. 3. <ul><li>circuit • continuity ground (return) path • grounded insulated path kilowatt • load • Ohm’s law • open circuit power source • protection shorted • short-to-ground • short-to-voltage Watt’s law </li></ul>KEY TERMS:
  4. 4. CIRCUITS <ul><li>A circuit is a path that electrons travel from a power source (such as a battery) through a load (such as a light bulb) and back to the power source. It is called a circuit because the current must start and finish at the same place (power source). </li></ul>For any circuit to work, it must be continuous from the battery through all wires and components and back to the battery (ground). Figure 32–1 All complete circuits must have a power source, a power path, protection (fuse), an electrical load (light bulb in this case), and a return path back to the power source. A circuit that is continuous throughout is said to have continuity .
  5. 5. <ul><li>Parts of a Complete Circuit Every complete circuit contains: </li></ul>Continued <ul><li>A power source , such as a vehicle’s battery. </li></ul><ul><li>Fuses, circuit breakers, and fusible links, which are Protection from harmful overloads (excessive current flow). </li></ul><ul><li>A n insulated path for current flow from the power source to the resistance. This path from a power source to the load (a light bulb in this example) is usually an insulated copper wire. </li></ul><ul><li>The electrical load or resistance converts electrical energy into heat, light, or motion. </li></ul><ul><li>A ground (return) path for the electrical current from the load back to the power source so that there is a complete circuit. This ground path is usually the metal body, frame, and engine block of the vehicle. See Figure 32-2. </li></ul><ul><li>Switches and controls turn the circuit on and off. Figure 32-3. </li></ul>
  6. 6. Figure 32–2 The return path back to the battery can be any electrical conductor, such as the metal frame or body of the vehicle. Continued
  7. 7. Figure 32–3 An electrical switch opens the circuit and no current flows. The switch could also be on the return (ground) path wire. Continued
  8. 8. <ul><li>Open Circuits An open circuit is any circuit that is not complete, or that lacks continuity. See Figure 32–4. No current at all will flow through an incomplete circuit. An open circuit may be created by a break in the circuit or by a switch that opens (turns off) the circuit and prevents the flow of current. In any circuit containing a power load and ground, an opening anywhere in the circuit will cause the circuit to stop working. A light switch in a home and the headlight switch in a vehicle are examples of devices that open a circuit to control its operation. </li></ul>Continued
  9. 9. Figure 32–4 This figure shows examples of common causes of open circuits. Some of these causes are often difficult to find.
  10. 10. The wiring schematic is the “road map” of a circuit and shows all electrical paths. If an open occurs in a circuit, the current stops flowing and the electrical load device does not work. Trace the circuit by following the path from the battery through the power side component, load, and on the ground. Check for voltage at various points in the circuit to locate where the open is in the circuit. Use a Schematic as a Road Map
  11. 11. <ul><li>Short-to-Voltage If a wire (conductor) or component is shorted to voltage, it is commonly called shorted . </li></ul>Figure 32–5 A short circuit permits electrical current to bypass some or all of the resistance in the circuit. Continued
  12. 12. <ul><li>A short circuit : </li></ul>Figure 32–6 A fuse or circuit breaker opens the circuit to prevent possible overheating damage in the event of a short circuit. <ul><li>Is a complete circuit in which the current bypasses some or all of the resistance in the circuit. </li></ul><ul><li>Involves the power side of the circuit. </li></ul><ul><li>Involves a copper-to-copper connection. </li></ul><ul><li>Also called a short - to - voltage . </li></ul><ul><li>Usually affects more than one circuit. </li></ul><ul><li>May or may not blow a fuse. </li></ul>Continued
  13. 13. <ul><li>The technician tested all fuses using a conventional test light (not a low current test light) and found them to be okay. All body-to-engine block ground wires were clean and tight. All bulbs were of the correct trade number as specified in the owner’s manual. </li></ul>A technician was working on a Chevrolet pickup truck with unusual electrical problems including the following: The Short-To-Voltage Story - Part1 <ul><li>Whenever the brake pedal was depressed, the dash light and the side marker lights would light. </li></ul><ul><li>The turn signals caused all lights to blink and the fuel gauge needle to bounce up and down. </li></ul><ul><li>When the brake lights were on, the front parking lights also came on. </li></ul>NOTE: Using a single-filament bulb (ex: #1156) in place of a dual-filament bulb (ex: #1157) could also cause many of the same problems.
  14. 14. The Short-To-Voltage Story - Part 2 Because the trouble occurred when the brake pedal was depressed, the tech decided to trace all the wires in the brake light circuit. The problem was found near the exhaust system. A small hole in the tailpipe (after the muffler) directed hot exhaust gases to the wiring harness containing all of the wires for circuits at the rear of the truck. The heat melted the insulation causing most of the wires to touch. Whenever one circuit was activated (such as when the brake pedal was applied), the current had a complete path to several other circuits. A fuse did not blow because there was enough resistance in the circuits being energized that the current (in amps) was too low to blow any fuses.
  15. 15. <ul><li>Short-to-Ground A short - to - ground is a type of circuit failure wherein the current bypasses part of the normal circuit and flows directly to ground. Because the ground return circuit is metal (vehicle frame, engine, or body), this type of circuit is identified as having current flowing from copper to steel. See Figure 32–7. A defective component or circuit shorted to ground is commonly called grounded . For example, if a penny was inserted into a cigarette lighter socket, current would flow through the penny to ground. Because the penny has little resistance, an excessive amount of current flow causes the fuse to blow. </li></ul>Continued
  16. 16. Figure 32–7 A short-to-ground affects the power side of the circuit. Current flows directly to the ground return, bypassing some or all of the electrical loads in the circuit. There is no current in the circuit past the short.
  17. 17. <ul><li>High Resistance Another common fault is excessive resistance in the circuit. This can be caused by several circuit faults including: </li></ul><ul><li>Corrosion of wires on the terminals </li></ul><ul><li>Poor electrical connections at connectors </li></ul><ul><li>Loose ground connection </li></ul>Any of the above will cause current in amperes to decrease in the circuit. As a result, the electrical load device may operate, but with reduced speed or brightness. High-resistance faults can also be intermittent and cause problems just when conditions or temperatures cause a problem in the circuit.
  18. 18. A beginner technician cleaned the positive terminal of the battery to correct the problem of slow cranking. When questioned by the shop foreman as to why only the positive post had been cleaned, the technician responded that the negative terminal was “only a ground.” The foreman reminded the technician that the current, in amperes, is constant throughout a series circuit (such as the cranking motor circuit). If 200 amperes leaves the positive post of the battery, then 200 amperes must return to the battery through the negative post. The technician just could not understand how electricity can do work (crank an engine), yet return the same amount of current, in amperes, as left the battery. The shop foreman explained that even though the current is constant throughout the circuit, the voltage (electrical pressure or potential) is dropped to zero in the circuit. To explain further, the shop foreman drew a waterwheel. Think of a Waterwheel - Part 1
  19. 19. As water drops from a higher level to a lower level, high potential energy (or voltage) is used to turn the waterwheel and results in low potential energy (or lower voltage). The same amount of water (or amperes) reaches the pond under the waterwheel as started the fall above the waterwheel. As current (amperes) flows through a conductor, it performs work in the circuit (turns the waterwheel) while its voltage (potential) is dropped. Think of a Waterwheel - Part 2 Figure 32–8 Electrical flow through a circuit is similar to water flowing over a waterwheel. The more the water (amperes in electricity), the greater the amount of work (waterwheel). The amount of water remains constant, yet the pressure (voltage in electricity) drops as the current flows through the circuit.
  20. 20. OHM’S LAW <ul><li>German physicist, Georg Ohm, established that electric pressure (EMF) in volts, electrical resistance in ohms, and the amount of current in amperes flowing through any circuit are all related. Ohm’s law states: It requires 1 volt to push 1 ampere through 1 ohm of resistance . This means that if the voltage is doubled, then the number of amperes of current flowing through a circuit will also double if the resistance of the circuit remains the same. Ohm’s law can also be stated as a simple formula used to calculate one value of an electrical circuit if the other two are known. See Figure 32–9. </li></ul>Continued
  21. 21. Figure 32–9 To calculate one unit of electricity when the other two are known, simply use your finger and cover the unit you do not know. For example, if both voltage (E) and resistance (R) are known, cover the letter I (amperes). Notice that the letter E is above the letter R so divide the resistor’s value into the voltage to determine the current in the circuit. Continued
  22. 22. OHM’S LAW STATED <ul><li>I = E/R where: </li></ul><ul><li>I = Current in amperes ( A ) </li></ul><ul><li>E = Electromotive force (EMF) in volts ( V ) </li></ul><ul><li>R = Resistance in ohms ( Ω ) </li></ul>Continued <ul><li>Ohm’s law can determine the resistance if the volts and amperes are known: R = E/I . </li></ul><ul><li>Ohm’s law can determine the voltage if the resistance (ohms) and amperes are known: E = I  R . </li></ul><ul><li>Ohm’s law can determine the amperes if the resistance and voltage are known: I = E/R . </li></ul>
  23. 23. OHM’S LAW RELATIONSHIPS See the chart on Page 328 of your textbook.
  24. 24. <ul><li>Ohm’s Law Applied to Simple Circuits If a battery with 12 volts is connected to a resistor of 4 ohms, as shown here, how many amperes will flow through the circuit? Using Ohm’s law, we can calculate the number of amperes that will flow through the wires and the resistor. </li></ul>Figure 32–10 This closed circuit includes a power source, power-side wire, circuit protection (fuse), resistance (bulb), and return path wire. If two factors are known (volts and ohms in this example), the remaining factor (amperes) can be calculated using Ohm’s law. Continued
  25. 25. <ul><li>I = E/R = 12 V/4Ω The values for voltage (12) and resistance (4) were substituted for the variables E and R , making I = 3 amperes (12/4 = 3). </li></ul><ul><li>If we want to connect a resistor to a 12-volt battery, we now know that this simple circuit requires 3 amperes to operate. This may help us for two reasons. </li></ul><ul><li>We can now determine the wire diameter that we will need based on the number of amperes flowing through the circuit. </li></ul><ul><li>T he correct fuse rating can be selected to protect the circuit. </li></ul>
  26. 26. WATT’S LAW <ul><li>James Watt (1736–1819), a Scottish inventor, first determined the power of a typical horse while measuring the amount of coal being lifted out of a mine. The power of one horse was determined to be 33,000 foot-pounds per minute. </li></ul>Continued Electricity can also be expressed in a unit of power called a watt and the relationship is known as Watt’s law , which states: A watt is a unit of electrical power represented by a current of 1 ampere through a circuit with a potential difference of 1 volt .
  27. 27. <ul><li>The symbol for a watt is the capital letter W . The formula for watts is: </li></ul><ul><li>W = I  E </li></ul><ul><li>Another way to express this formula uses the letter P to represent the unit of power. The formula then becomes: P = I  E </li></ul>Continued HINT: An easy way to remember this equation is that it spells “pie.”
  28. 28. <ul><li>To calculate watts, both the current in amperes and the voltage in the circuit must be known. If two of these factors are known, the other remaining factor can be determined by the following equation: P = I  E ( watts equal amperes times voltage ) </li></ul><ul><li>I = P/E ( amperes equal watts divided by voltage ) </li></ul><ul><li>E = P/I ( voltage equals watts divided by amperes ) </li></ul><ul><li>A Watt’s circle can be drawn and used like the Ohm’s law circle diagram. See Figure 32–11. </li></ul><ul><li>Magic Circle The formulas for calculating any combination of electrical units are shown in Figure 32–12. </li></ul>Continued
  29. 29. Figure 32–11 To calculate one unit when the other two are known, simply cover the unknown unit to see what unit needs to be divided or multiplied to arrive at the solution. Figure 32–12 “Magic circle” of most of the formulas for problems involving Ohm’s law. Each quarter of the “pie” has formulas used to solve for a particular unknown value: current (amperes), in the upper right segment; resistance (ohms), in the lower right; voltage ( E ), in the lower left; and power (watts), in the upper left.
  30. 30. The brightness of a light bulb, such as an automotive headlight or courtesy light, depends on the number of watts available. The watt is the unit by which electrical power is measured. If the battery voltage drops, even slightly, the light becomes noticeably dimmer. The formula for calculating power ( P ) in watts is P × I = E , also be expressed Watts = Amps × Volts . According to Ohm’s law, I = E/R . Therefore, E/R can be substituted for I in the previous formula resulting in P = E/R × E or P = E 2 / R . E 2 means E multiplied by itself. A small change in the voltage ( E ) has a big effect on the total brightness of the bulb. (Remember, household light bulbs are sold according to wattage.) Thus, if voltage to an automotive bulb is reduced, such as by a poor electrical connection, brightness of the bulb is greatly affected. A poor electrical ground causes a voltage drop. The voltage at the bulb is reduced and the bulb’s brightness is reduced. Wattage Increases by the Square of the Voltage
  31. 31. Engine power is commonly rated in watts or kilowatts (1,000 watts equal 1 kilowatt) because 1 horsepower is equal to 746 watts. For example, a 200-horsepower engine can be rated as having the power equal to 149,200 watts or 149.2 kilowatts (kW). Why are Vehicle Engines for Europe Rated in Kilowatts?
  32. 32. SUMMARY <ul><li>All complete electrical circuits have a power source (such as a battery), a circuit protection device (such as a fuse), a power-side wire or path, an electrical load, a ground return path, and a switch or a control device. </li></ul><ul><li>A short-to-voltage involves a copper-to-copper connection and often affects more than one circuit. </li></ul><ul><li>A short-to-ground involves a copper-to-steel connection and usually causes the fuse to blow. </li></ul><ul><li>An open is a break in the circuit resulting in absolutely no current flow through the circuit. </li></ul>
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