Chapter 33

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Chapter 33

  1. 1. start
  2. 2. <ul><li>Prepare for ASE Electrical/Electronic Systems (A6) certification test content area “A” (General Electrical/Electronic System Diagnosis). </li></ul><ul><li>Identify a series circuit. </li></ul><ul><li>State Kirchhoff’s voltage law. </li></ul><ul><li>Calculate voltage drops in a series circuit. </li></ul><ul><li>Explain series and parallel circuit laws. </li></ul>OBJECTIVES: After studying Chapter 33, the reader should be able to: Continued
  3. 3. <ul><li>State Kirchhoff’s current law. </li></ul><ul><li>Identify where faults in a series-parallel circuit can be detected or determined. </li></ul>OBJECTIVES: After studying Chapter 33, the reader should be able to:
  4. 4. <ul><li>branches • combination circuit • compound circuit Kirchhoff’s current law • Kirchhoff’s voltage law leg • parallel circuit series circuit • series-parallel circuits • shunt total circuit resistance • voltage drop </li></ul>KEY TERMS:
  5. 5. OHM’S LAW AND SERIES CIRCUITS <ul><li>A series circuit is a complete circuit that has more than one electrical load where all of the current has only one path to flow through all of the loads. Electrical components such as fuses and switches are generally not considered to be included in the determination of a series circuit. The circuit must be continuous or have continuity in order for current to flow through the circuit. </li></ul><ul><li>Ohm’s law can be used to calculate the value of one unknown (voltage, resistance, or amperes) if the other values are known. </li></ul>NOTE: An electrical load needs power and a ground to operate. A break (open) in a series circuit will cause the current in the circuit to stop.
  6. 6. <ul><li>Because all current flows through all resistances, total resistance is the sum (addition) of all resistances. </li></ul>Figure 33–1 A series circuit with three bulbs. All current flows through all resistances (bulbs). The total resistance of the circuit is the sum of the total resistance of the bulbs, and the bulbs will light dimly because of the increased resistance and the reduction of current flow (amperes) through the circuit. Total resistance of the circuit shown here is 6 ohms (1Ω = 2Ω + 3Ω). The formula for total resistance ( R T ) for a series circuit is: Using Ohm’s law to find current flow: Continued
  7. 7. <ul><li>With a total resistance of 6 ohms using a 12-volt battery in the series circuit shown, 2 amperes of current will flow through the entire circuit. If resistance is reduced, more current will flow. </li></ul>Figure 33–2 A series circuit with two bulbs. Continued Here, one resistance has been eliminated and now the total resistance is 3 ohms (1Ω + 2Ω). Using Ohm’s law to calculate current flow yields 4 amperes. Notice current flow was doubled (4 amps instead of 2 amps) when the resistance was cut in half (from 6 ohms to 3 ohms).
  8. 8. Electricity almost seems to act as if it “knows” what resistances are ahead on the long trip through a circuit. If the trip through the circuit has many high-resistance components, very few electrons (amperes) will choose to attempt to make the trip. If a circuit has little or no resistance (for example, a short circuit), then as many electrons (amperes) as possible attempt to flow through the complete circuit. If flow exceeds the capacity of the fuse or the circuit breaker, then the circuit is opened and all current flow stops. Farsighted Quality of Electricity
  9. 9. KIRCHOFF’S VOLTAGE LAW <ul><li>Voltage applied through a series circuit drops with each resistor. The greater the resistance, the greater the drop in voltage. German physicist, Gustav Robert Kirchhoff (1824–1887), developed laws about electrical circuits. His second law, Kirchhoff’s voltage law , concerns voltage drops. It states: </li></ul>Continued The voltage around any closed circuit is equal to the sum (total) of the voltage drops across the resistances .
  10. 10. <ul><li>Applying Kirchhoff’s Voltage Law Kirchhoff states in his second law that voltage drops in proportion to resistance and the total of voltage drops will equal the applied voltage. </li></ul>Figure 33–3 As current flows through a circuit, the voltage drops in proportion to the amount of resistance in the circuit. Most, if not all, of the resistance should occur across the load such as the bulb in this circuit. All of the other components and wiring should produce little, if any, voltage drop. If a wire or connection did cause a voltage drop, less voltage would be available to light the bulb and the bulb would be dimmer than normal.circuit with two bulbs.
  11. 11. Figure 33–4 In a series circuit the voltage is dropped or lowered by each resistance in the circuit. The higher the resistance, the greater the drop in voltage. <ul><li>Total resistance of the circuit can be determined by adding the individual resistances (2Ω + 4Ω + 6Ω = 12Ω). </li></ul>Continued Resistance = 12Ω Voltage = 12V Current = 1A In the circuit shown, the following values are known: The current through the circuit is determined by using Ohm’s law, I = E / R = 12V/12Ω = 1A.
  12. 12. <ul><li>Everything is known except the voltage drop caused by resistance. This can be determined by using Ohm’s law and calculating for voltage ( E ) using the value of each resistance individually: </li></ul>According to Kirchhoff, the sum (addition) of the voltage drops should equal the applied voltage (battery voltage): Total of voltage drops = 2V + 4V + 6V = 12V = Battery voltage
  13. 13. Figure 33–5 A voltmeter reads the differences of voltage between the test leads. The voltage read across a resistance is the voltage drop that occurs when current flows through a resistance. A voltage drop is also called an “IR” drop because it is calculated by multiplying the current (I) through the resistance (electrical load) by the value of the resistance ( R). <ul><li>Another example of Kirchhoff’s second (voltage) law. </li></ul>NOTE: Notice that the voltage drop is proportional to the resistance. Higher resistance means greater voltage drop. A 6-ohm resistance dropped the voltage three times as much as the drop created by 2-ohm resistance. Continued
  14. 14. <ul><li>Use of Voltage Drops Due to built-in resistance, voltage drops are used in automotive electrical systems to drop the voltage in the following examples: </li></ul>Continued <ul><li>The Dash lights Most vehicles are equipped with a method of dimming the brightness of the dash lights by turning a variable resistor. This type of resistor can be changed and therefore varies the voltage to the dash light bulbs. A high voltage to the bulbs causes them to be bright, and a low voltage results in a dim light. </li></ul><ul><li>The Blower motor (heater or air-conditioning fan). Speeds are usually controlled by a fan switch sending current through high-, medium-, or low-resistance wire resistors. Highest resistance will drop voltage the most, causing the motor to run at the lowest speed. Highest speed of the motor will occur when no resistance is in the circuit. </li></ul>
  15. 15. <ul><li>A voltage drop test is also easier to perform because the resistance does not have to be known, only that unwanted loss of voltage in a circuit should be less than 3% or, about 0.14 volts for any 12-volt circuit. </li></ul>Imagine a wire with all strands cut except for one. An ohmmeter can be used to check resistance of this wire and the resistance would be low, indicating the wire was okay. But one strand cannot properly carry the current (amps) in the circuit. A voltage drop test is better for two reasons: Why Check Voltage Drop Instead of Resistance? <ul><li>An ohmmeter can only test a wire or component that has been disconnected from the circuit and is not carrying current. The resistance can, and does, change when current flows. </li></ul><ul><li>A voltage drop test is a dynamic test because as the current flows through a component, the conductor increases in temperature, which in turn increases resistance. This means that a voltage drop test is testing the circuit during normal operation and is therefore the most accurate way of determining circuit conditions. </li></ul>
  16. 16. SERIES CIRCUIT LAWS <ul><li>Law 1 Total resistance in a series circuit is the sum total of the individual resistances. The resistance values of each electrical load are simply added together. Law 2 The current is constant throughout the entire circuit. If 2 amperes of current leave the battery, 2 amperes of current return to the battery. </li></ul>Figure 33–6 In this series circuit with a 2-ohm resistor and a 4-ohm resistor, current (2 amperes) is the same throughout even though the voltage drops across each resistor. Continued
  17. 17. <ul><li>Law 3 Although current (in amps) is constant, voltage drops across each resistance in the circuit. The voltage drop across each load is proportional to the value of the resistance compared to the total resistance. For example, if the resistance is one-half of the total resistance, the voltage drop across that resistance will be one-half of the applied voltage. The sum total of all individual voltage drops equals the applied source voltage. </li></ul>Continued
  18. 18. SERIES CIRCUIT EXAMPLES <ul><li>Each of the four examples includes solving for the following: </li></ul>Continued <ul><li>Total resistance in the circuit </li></ul><ul><li>Current flow (amperes) through the circuit </li></ul><ul><li>Voltage drop across each resistance </li></ul>
  19. 19. Figure 33–7 Example 1. <ul><li>The unknown in this problem is the value of R 2 . The total resistance, however, can be calculated using Ohm’s law. </li></ul>Continued Because R 1 is 3 ohms and the total resistance is 4 ohms, the value of R 2 is 1 ohm. R Total = E/I = 12 volts/3A = 4Ω
  20. 20. Figure 33–8 Example 2. <ul><li>The unknown in this problem is the value of R 3 . The total resistance, however, can be calculated using Ohm’s law. </li></ul>Continued Total resistance of R 1 (3 ohms) and R 2 (1 ohm) equals 4 ohms so the value of R 3 is the difference between the total resistance (6 ohms) and the value of known resistance (4 ohms). R Total = E/I = 12 volts/2 A = 6Ω 6 – 4 = 2 ohms = R 3
  21. 21. Figure 33–9 Example 3. <ul><li>The unknown in this problem is voltage of the battery. To solve for voltage, use Ohm’s law ( E = I × R ). The “ R” in this problem refers to the total resistance (RT). The total resistance of a series circuit is determined by adding the values of the individual resistors. </li></ul>Continued Placing total resistance (3Ω) into the equation results in battery voltage of 12 volts. R T = 1Ω + 1Ω + 1Ω R T = 3Ω E = 4A × 3Ω E = 12 volts
  22. 22. Figure 33–10 Example 4. <ul><li>The unknown in this example is the current (amperes) in the circuit. To solve for current, use Ohm’s law. </li></ul>Notice that the total resistance in the circuit (6 ohms) was used in this example, which is the total of the three individual resistors: ( 2Ω + 2Ω + 2Ω = 6Ω ) The current through the circuit is two amperes. I = E/R = 12 volts/6 ohms = 2A
  23. 23. PARALLEL CIRCUITS <ul><li>A parallel circuit is a complete circuit that has more than one path for the current. The separate paths which split and meet at junction points are called branches , legs , or shunts . </li></ul>The current flow through each branch or leg varies depending on the resistance in that branch. A break or open in one leg or section of a parallel circuit does not stop current flow through the remaining legs of the circuit.
  24. 24. KIRCHOFFS CURRENT LAW <ul><li>Kirchhoff’s current law (first law) states: </li></ul>Figure 33–11 The amount of current flowing into junction point A equals the total amount of current flowing out of the junction. The current flowing into any junction of an electrical circuit is equal to the current flowing out of that junction. Illustrated using Ohm’s law, seen here: The 6-ohm leg requires 2 amps and the 3-ohm resistance leg requires 4 amps. The wire from the battery to junction A must be capable of handling 6 amps. The sum of current flowing out is equal to current flowing into the junction, proving Kirchhoff’s current law.
  25. 25. PARALLEL CIRCUIT LAWS <ul><li>Law 1 Total resistance of a parallel circuit is always less than that of the smallest-resistance leg. This occurs because not all current flows through each leg or branch. With many branches, more current can flow from the battery. Law 2 The voltage is the same for each leg of a parallel circuit. </li></ul>Continued NOTE: A parallel circuit drops the voltage from source voltage to zero (ground) across the resistance in each leg of the circuit.
  26. 26. Figure 33–12 The current in a parallel circuit splits (divides) according to the resistance in each branch. <ul><li>Law 3 The sum of the individual currents in each leg will equal the total current. The amount of current flow through a parallel circuit may vary for each leg depending on the resistance of that leg. The current flowing through each leg results in the same voltage drop (from the power side to the ground side) as for every other leg of the circuit. </li></ul>Continued
  27. 27. There is an old saying that electricity will always take the path of least resistance. This is true, especially if there is a fault such as in the secondary (high-voltage) section of the ignition system. If there is a path to ground that is lower than the path to the spark plug, the high-voltage spark will take the path of least resistance. In a parallel circuit where there is more than one path for the current to flow, most of the current will flow through the branch with the lower resistance. This does not mean that all of the current will flow through the lowest resistance, because the other path does provide a path to ground, and the amount of current flow through the other branches is determined by the resistance and the applied voltage according to Ohm’s law. Therefore, the only place where electricity takes the path of least resistance is in a series circuit where there are not other paths for the current to flow . The Path of Least Resistance
  28. 28. <ul><li>Five common methods used to determine total resistance in a parallel circuit: </li></ul>DETERMINING TOTAL RESISTANCE IN A PARALLEL CIRCUIT NOTE: Determining the total resistance of a parallel circuit is very important in automotive service. Electronic fuel-injector and diesel engine glow plug circuits are two of the most commonly tested circuits where parallel circuit knowledge is required. Also, when installing extra lighting, the tech must determine the proper gauge wire and protection device. Method 1 The total current (in amperes) can be calculated first by treating each leg of the parallel circuit as a simple circuit. Each leg has its own power and ground (–), and therefore, the current through each leg is independent of the current through any other leg. Continued
  29. 29. Figure 33–13 In a typical parallel circuit, each resistance has power and ground and each leg operates independently of the other legs of the circuit. <ul><li>Total current is the sum total of the individual currents for each leg. Total current from the battery is 9 amps: ( 4A +3A + 2A = 9A ) </li></ul>Continued If total circuit resistance ( R T ) is needed, Ohm’s law can be used to calculate it because voltage ( E ) and current ( I ) are now known.
  30. 30. <ul><li>Total resistance (1.33Ω) is smaller than the smallest-resistance leg of the circuit. This characteristic of a parallel circuit holds true as not all current flows through all resistances as in a series circuit. </li></ul>Continued Because the current has alternative paths to ground through the various legs of a parallel circuit, as additional resistances (legs) are added to a parallel circuit, the total current from the battery (power source) increases . Additional current can flow when resistances are added in parallel, because each leg of a parallel circuit has its own power and ground and the current flowing through each leg is strictly dependent on the resistance of that leg.
  31. 31. Figure 33–14 A schematic showing two resistors in parallel connected to a 12-volt battery. Method 2 If only two resistors are connected in parallel, the total resistance ( R T ) can be found using R T = ( R 1 × R 2 ) / ( R 1 + R 2 ) Using this circuit, substituting 3 ohms for R 1 and 4 amperes for R 2 , R T = (3 × 4)/(3 × 4) = 12/7 = 1.7Ω Note that the total resistance (1.7Ω) is smaller than that of the smallest-resistance leg of the circuit. Continued
  32. 32. <ul><li>This formula can be used for more than two resistances in parallel, but only two resistances can be calculated at a time. After solving for R T for two resistors, use the value of R T as R 1 and the additional resistance in parallel as R 2 . Then solve for another R T . Continue the process for all resistance legs of the parallel circuit. Note that it might be easier to solve for R T when there are more than two resistances in parallel by using Method 3 or 4. </li></ul>Continued NOTE: Which resistor is R 1 and which is R 2 is not important. Position in the formula makes no difference in multiplication and addition of values.
  33. 33. <ul><li>Method 3 A formula to find total resistance for any number of resistances in parallel is: </li></ul>Figure 33–15 A parallel circuit with three resistors connected to a 12-volt battery. Continued To solve for R T for the three resistance legs shown, substitute the values of the resistances for R 1 , R 2 , and R 3 : 1/ R T = 1/3 + 1/4 + 1/6
  34. 34. <ul><li>The fractions cannot be added together unless they all have the same denominator. The lowest common denominator here is 12. 1/3 becomes 4/12 , 1/4 becomes 3/12 , and 1/6 becomes 2/12 . 1/ R T = 4/12 + 3/12 + 2/12 or 9/12 Cross multiplying R T = 12/9 = 1.33Ω Note the result (1.33Ω) is the same regardless of method used. (see Method 1). The most difficult part is determining lowest common denominator, especially for circuits containing a wide range of ohmic values for various legs. For an easier method using a calculator, see Method 4. </li></ul>Continued
  35. 35. Figure 33–16 Using an electronic calculator to determine the total resistance of a parallel circuit. <ul><li>Method 4 This method uses an electronic calculator, available at very low cost. Instead of determining the lowest common denominator as in Method 3, one can use the calculator to convert the fractions to decimal equivalents. </li></ul>Memory buttons on most calculators can be used to keep a running total of the fractional values. Calculate the total resistance ( R T ) by pushing the indicated buttons on the calculator. Continued See also Figure 33-17
  36. 36. <ul><li>The memory recall (MRC) and equals (=) buttons invert the answer to give the correct value for total resistance (1.33Ω). The inverse (1/X or X -1 ) button can be used with the sum (SUM) button on scientific calculators without using the memory button. </li></ul>Figure 33–17 Another example of how to use an electronic calculator to determine the total resistance of a parallel circuit. The answer is 13.45 ohms. Notice that the effective resistance of this circuit is less than the resistance of the lowest branch (20 ohms). This method can be used to find the total of any number of resistances in parallel.
  37. 37. <ul><li>Method 5 Used when two or more resistances connected in parallel are the same value. Total resistance ( R T ) of equal-value resistors: divide the number of equal resistors into value of resistance. R T = Value equal resistance/Number equal resistances = 12W/4 = 3W </li></ul>Figure 33–18 A parallel circuit containing four 12-ohm resistors. When a circuit has more than one resistor of equal value, total resistance can be determined by simply dividing the value of resistance (12 ohms here) by the number of equal-value resistors (4 in this example), to get 3 ohms. NOTE: As most automotive and light-truck electrical circuits involve multiple use of the same resistance, this method is most useful. If six additional 12-ohm lights were added to a vehicle, the additional lights would represent just 2 ohms of resistance ( 12 Ω /6 lights = 2 ). 6 amperes of additional current would be drawn ( I = E/R = 12V/2 Ω = 6A ).
  38. 38. PARALLEL CIRCUIT EXAMPLES <ul><li>Each of the four examples includes solving for the following: </li></ul>Continued <ul><li>Total resistance </li></ul><ul><li>Current flow (amperes) through each branch as well as total current flow </li></ul><ul><li>Voltage drop across each resistance </li></ul>
  39. 39. Figure 33–19 Example 1. <ul><li>In this example, the voltage of the battery is unknown and the equation to be used is E = I × R — where R represents the total resistance of the circuit. Using the equation for two resistors in parallel, the total resistance is 6 ohms. </li></ul>E = I × R E = 2A × 6Ω E = 12 volts Continued Placing total value of resistors in the equation results in a value for the battery voltage of 12 volts.
  40. 40. Figure 33–20 Example 2. <ul><li>R 3 is unknown. Because voltage (12 volts) and current (12 A) are known, it is easier to solve for the unknown resistance by treating each branch or leg as a separate circuit. Using Kirchhoff’s law, total current equals total current flow through each branch. </li></ul>Resistance must be 4Ω ( I = E / R = 12V/4Ω = 3A) Continued The current flow through R 1 is 3A ( I = E / R = 12 V/4Ω = 3A) and the current flow through R 2 is 6A ( I = E / R = 12V/2Ω = 6A) Current through the two known branches: 9A (3A + 6A = 9A) Because there are 12A leaving and returning to the battery, the current flow through R 3 must be 3A (12A – 9A = 3A)
  41. 41. Figure 33–21 Example 3. <ul><li>In this example, the voltage of the battery is unknown. The equation to solve for voltage according to Ohm’s law is: </li></ul>Continued The R refers to the total resistance. As there are four resistors of equal value, the total can be determined by the equation: R Total = Value Resistors/Number Equal Resistors = 12Ω/4 = 3Ω Inserting value of total resistors of the parallel circuit (3Ω) into Ohm’s law results in a battery voltage of 12V. E = 4A × 3Ω E = I × R E = 12 V
  42. 42. Figure 33–22 Example 4. <ul><li>The unknown is the amount of current in the circuit. The Ohm’s law equation for determining current is: </li></ul>Continued I = E/R R represents total resistance. Because there are two equal resistances (8Ω), they can be replaced by one resistance of 4Ω ( R Total = Value/Number = 8Ω/2 = 4Ω) Total resistance of circuit containing two 8-ohm and one 4-ohm resistor is 2 ohms. Battery current flow is then calculated to be 6A. I = E/R = 12V/2Ω = 6A
  43. 43. SERIES-PARALLEL CIRCUITS <ul><li>Series-parallel circuits are a combination of series and parallel segments in one complex circuit. A series-parallel circuit is called a compound or a combination circuit . Many automotive circuits include sections that are in parallel and in series. A series-parallel circuit includes both parallel loads or resistances, plus additional loads or resistances that are electrically connected in series. There are two basic types of series-parallel circuits. </li></ul>Continued
  44. 44. <ul><li>A circuit where the load is in series with other loads is parallel. An example of this type of circuit is a dash light dimming circuit. The variable resistor is used to limit current flow to the dash light bulbs, which are wired in parallel. </li></ul>Figure 33–23 A series-parallel circuit. Continued
  45. 45. <ul><li>A circuit where a parallel circuit contains resistors or loads is in series in one or more branches. A headlight and starter circuit is an example of this type. </li></ul>Figure 33–24 This complete headlight circuit with all bulbs and switches is a series- parallel circuit. A headlight switch is usually connected in series with a dimmer switch and in parallel with the dash light dimmer resistors. The headlights are also connected in parallel along with the taillights and side marker lights.
  46. 46. <ul><li>Series-Parallel Circuit Faults If a conventional parallel circuit, such as a taillight circuit, had a fault that increased the resistance in one branch of the circuit, the current flow through that one branch will be reduced. Added resistance, due to corrosion or similar cause, would create a voltage drop. As a result of this drop, a lower voltage would be applied and the bulb in the taillight would be dimmer than normal. Because brightness of the bulb depends on voltage and current applied, lower voltage and current would cause the bulb to be dimmer. If added resistance occurred in part of the circuit that fed both taillights, then both taillights would be dimmer than normal. In this case, the added resistance created a series-parallel circuit that was originally just a simple parallel circuit. </li></ul>Continued
  47. 47. <ul><li>The key to solving series-parallel circuit problems is to combine or simplify as much as possible. </li></ul>SOLVING SERIES-PARALLEL CIRCUIT PROBLEMS Figure 33–25 Solving a series-parallel circuit problem. If there are two loads or resistances in series in a parallel branch or leg, then the circuit can be made simpler if the two are first added together before attempting to solve the parallel section.
  48. 48. SERIES-PARALLEL CIRCUIT EXAMPLES <ul><li>Each of the four examples includes solving for the following: </li></ul>Continued <ul><li>Total resistance </li></ul><ul><li>Current flow (amperes) through each branch, as well as total current flow </li></ul><ul><li>Voltage drop across each resistance </li></ul>
  49. 49. Figure 33–26 Example 1. <ul><li>The unknown resistor is in series with the other two resistances, which are connected in parallel. The Ohm’s law equation to determine resistance is: </li></ul>Continued R = E/I = 12V/3A = 4Ω The total resistance of the circuit is 4 ohms, and the value of the unknown can be determined by subtracting the value of the two resistors that are connected in parallel. The parallel branch resistance is 2Ω . The value of the unknown resistance is 2Ω Total R = 4Ω – 2Ω = 2Ω
  50. 50. Figure 33–27 Example 2. <ul><li>The unknown unit in this circuit is the voltage of the battery. The Ohm’s law equation is: E = I × R </li></ul>Continued Because each branch contains two 4-ohm resistors in series, values in each can be added to help simplify the circuit. By adding the resistors in together, the circuit now consists of two 8-ohm resistors. Inserting the value for total resistance into the Ohm’s law equation results in a value of 12 volts for the battery voltage. E = 3A × 4Ω E = 12 volts E = I × R
  51. 51. Figure 33–28 Example 3. <ul><li>In this example, the total current through the circuit is unknown. The Ohm’s law equation to solve for it is: I = E/R </li></ul>Continued To solve for total resistance, the circuit can be simplified by adding R 3 and R 4 together because these two resistors are in series in the same branch of the parallel circuit, similar to Example 2. With the branches reduced to just one 4-ohm resistor, this can be added to the 2-ohm ( R 1 ) resistor because it is in series, creating a total circuit resistance of 6 ohms. I = E/R = 12V/6Ω = 2A Current flow can now be determined from Ohm’s law:
  52. 52. Figure 33–29 Example 4. <ul><li>In this example, the value of resistor R 1 is unknown. Total resistance of the circuit is 3 ohms: R = E/I = 12V/4A = 3Ω </li></ul>Continued Knowing total resistance is not enough to determine value of R 1 . To simplify, R 2 and R 5 combine to create a parallel branch resistance value of 8 ohms because they are in series. With the circuit simplified to one resistor in series ( R 1 ) with 4 ohm branches with 4 ohms, the branches can be reduced to the equal of one 2-ohm resistor. 3Ω – 2Ω = 1Ω The circuit includes one 2- ohm resistor plus unknown R 1 . Total resistance is 3 ohms, so R 1 must be 1 ohm.
  53. 53. SUMMARY <ul><li>Series circuits: </li></ul>Continued <ul><li>In a simple series circuit, the current remains constant throughout, but the voltage drops as current flows through the resistances of the circuit. </li></ul><ul><li>The voltage drop across each resistance or load is directly proportional to the value of the resistance compared to the total resistance in the circuit. </li></ul><ul><li>The sum (total) of the voltage drops equals the applied voltage (Kirchhoff’s voltage law). </li></ul><ul><li>An open or a break anywhere in a series circuit stops all current from flowing. </li></ul>
  54. 54. SUMMARY <ul><li>Parallel circuits: </li></ul>Continued ( cont. ) <ul><li>A parallel circuit, such as is used for all automotive lighting, has the same voltage available to each resistance (bulb). </li></ul><ul><li>The total resistance of a parallel circuit is always lower than the smallest resistance. </li></ul><ul><li>The separate paths that split and meet at junction points are called branches, legs, or shunts. </li></ul><ul><li>Kirchhoff’s current law states: “The current flowing into a junction of an electrical circuit is equal to current flowing out of that junction.” </li></ul>
  55. 55. SUMMARY <ul><li>Series-Parallel circuits: </li></ul>( cont. ) <ul><li>A series-parallel circuit is also called a compound circuit or a combination circuit. </li></ul><ul><li>A series-parallel circuit is a combination of a series and a parallel circuit, which does not include fuses or switches. </li></ul><ul><li>A fault in a series portion of the circuit would affect the operation if the series part was in the power or the ground side of the parallel portion of the circuit. </li></ul><ul><li>A fault in one leg of a series-parallel circuit will affect just the component(s) in that one leg. </li></ul>
  56. 56. end

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