Chapter 33

[object Object],[object Object],[object Object],[object Object],[object Object],OBJECTIVES: After studying Chapter 33, the reader should be able to: Continued

[object Object],[object Object],OBJECTIVES: After studying Chapter 33, the reader should be able to:

OHM’S LAW AND SERIES CIRCUITS ,[object Object],[object Object],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.

[object Object],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

[object Object],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).

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

KIRCHOFF’S VOLTAGE LAW ,[object Object],Continued The voltage around any closed circuit is equal to the sum (total) of the voltage drops across the resistances .

[object Object],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.

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. ,[object Object],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.

[object Object],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

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). ,[object Object],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

[object Object],Continued ,[object Object],[object Object]

[object Object],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? ,[object Object],[object Object]

SERIES CIRCUIT LAWS ,[object Object],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

SERIES CIRCUIT EXAMPLES ,[object Object],Continued ,[object Object],[object Object],[object Object]

Figure 33–7 Example 1. ,[object Object],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

Figure 33–8 Example 2. ,[object Object],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

Figure 33–9 Example 3. ,[object Object],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

Figure 33–10 Example 4. ,[object Object],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

PARALLEL CIRCUITS ,[object Object],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.

KIRCHOFFS CURRENT LAW ,[object Object],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.

PARALLEL CIRCUIT LAWS ,[object Object],Continued NOTE: A parallel circuit drops the voltage from source voltage to zero (ground) across the resistance in each leg of the circuit.

Figure 33–12 The current in a parallel circuit splits (divides) according to the resistance in each branch. ,[object Object],Continued

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

[object Object],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

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. ,[object Object],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.

[object Object],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.

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

[object Object],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.

[object Object],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

Figure 33–16 Using an electronic calculator to determine the total resistance of a parallel circuit. ,[object Object],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

[object Object],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.

[object Object],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 ).

PARALLEL CIRCUIT EXAMPLES ,[object Object],Continued ,[object Object],[object Object],[object Object]

Figure 33–19 Example 1. ,[object Object],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.

Figure 33–20 Example 2. ,[object Object],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)

Figure 33–21 Example 3. ,[object Object],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

Figure 33–22 Example 4. ,[object Object],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

SERIES-PARALLEL CIRCUITS ,[object Object],Continued

[object Object],Figure 33–23 A series-parallel circuit. Continued

[object Object],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.

[object Object],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.

SERIES-PARALLEL CIRCUIT EXAMPLES ,[object Object],Continued ,[object Object],[object Object],[object Object]

Figure 33–26 Example 1. ,[object Object],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

Figure 33–27 Example 2. ,[object Object],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

Figure 33–28 Example 3. ,[object Object],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:

Figure 33–29 Example 4. ,[object Object],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.

SUMMARY ,[object Object],Continued ,[object Object],[object Object],[object Object],[object Object]

SUMMARY ,[object Object],Continued ( cont. ) ,[object Object],[object Object],[object Object],[object Object]

SUMMARY ,[object Object],( cont. ) ,[object Object],[object Object],[object Object],[object Object]

Chapter 33

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Chapter 33