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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. 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

9. Figure 32–4 This figure shows examples of common causes of open circuits. Some of these causes are often difficult to find.

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

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.

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.

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. 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.

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

23. OHM’S LAW RELATIONSHIPS See the chart on Page 328 of your textbook.

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. 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. 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?

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