AC/DC ELECTRICAL SYSTEMS

Transcription

1 AC/DC ELECTRICAL SYSTEMS LEARNING ACTIVITY PACKET COMBINATION CIRCUITS BB227-BC05UEN

2 LEARNING ACTIVITY PACKET 5 COMBINATION CIRCUITS INTRODUCTION This LAP will continue to build on series circuits and parallel circuits discussed in previous LAPs by combining them into combination circuits. Combination circuits combine the desirable characteristics of series and parallel circuits. In fact, many of the circuits found in industry, home, and commercial electrical systems are actually combination circuits. ITEMS NEEDED Amatrol Supplied 1 T7017 AC/DC Electrical Learning System FIRST EDITION, LAP 5, REV. A Amatrol, AMNET, CIMSOFT, MCL, MINI-CIM, IST, ITC, VEST and Technovate are trademarks or registered trademarks of Amatrol, Inc. All other brand and product names are trademarks or registered trademarks of their respective companies. Copyright 2012 by AMATROL, INC. All rights Reserved. No part of this publication may be reproduced, translated, or transmitted in any form or by any means, electronic, optical, mechanical, or magnetic, including but not limited to photographing, photocopying, recording or any information storage and retrieval system, without written permission of the copyright owner. Amatrol,Inc., 2400 Centennial Blvd., Jeffersonville, IN USA, Ph , FAX

3 TABLE OF CONTENTS SEGMENT 1 CHARACTERISTICS OBJECTIVE 1 Defi ne a series-parallel circuit OBJECTIVE 2 Describe a method for identifying the series and parallel sections of a circuit SKILL 1 Trace the current path in a combination circuit OBJECTIVE 3 List the seven steps for solving a combination circuit SKILL 2 Solve a combination circuit SEGMENT 2 LIGHTING CIRCUITS OBJECTIVE 4 Describe how switches are used in combination circuits and give an application SKILL 3 Connect and operate a basic lighting circuit SKILL 4 Connect and operate a ceiling fan circuit OBJECTIVE 5 Describe the function of a variable resistor and give an application Activity 1 Rheostat operation SKILL 5 Connect and operate a rheostat as a light dimmer SEGMENT 3 VOLTAGE DIVIDERS OBJECTIVE 6 Describe the function of a voltage divider and give an application OBJECTIVE 7 Describe the operation of three types of voltage dividers SKILL 6 Design a voltage divider network SKILL 7 Connect and operate a voltage divider network SEGMENT 4 TROUBLESHOOTING OBJECTIVE 8 Explain the effect of a short circuit OBJECTIVE 9 Describe the four steps for troubleshooting a short circuit SKILL 8 Locate a short circuit OBJECTIVE 10 Describe the three basic steps for troubleshooting an open circuit SKILL 9 Locate an open circuit 3

4 SEGMENT 1 CHARACTERISTICS OBJECTIVE 1 DEFINE A SERIES-PARALLEL CIRCUIT A series-parallel or combination circuit, as shown in figure 1, contains elements of both series and parallel circuits. Components that are connected in series display the characteristics of a series circuit, while components connected in parallel display the characteristics of a parallel circuit. R 1 + R 2 R 3 Figure 1. A Series-Parallel Combination Circuit Two important points to remember about the characteristics of series-parallel circuits are: Components that are connected in series have the same current flowing through them. Components that are connected in parallel have the same voltage across them. These two points will be very helpful in understanding the operation of a series-parallel or combination circuit. 4

5 OBJECTIVE 2 DESCRIBE A METHOD FOR IDENTIFYING THE SERIES AND PARALLEL SECTIONS OF A CIRCUIT The series and parallel sections of a combination circuit can be identified by tracing the current path to find nodes that split the current and nodes that recombine the current, as shown in figure 2. R 1 I 1 NODE + R 2 I 2 I 3 R 3 I 1 NODE Figure 2. Tracing the Current Path In figure 2, there are no nodes between the power supply and R 1. Therefore, R 1 is in series with the power supply and all of the current from the power supply flows through R 1. When the current reaches the node where R 2 and R 3 are located, it splits between them. This means that R 2 and R 3 are in parallel. The current in a parallel section must recombine at some point, so you should trace the two lines to find the node where they recombine. In this case, it is easy to see where the currents recombine. You can also represent that components are in parallel as R 2 R 3. The two vertical lines ( ) mean these two resistors are in parallel with each other. Being able to identify which components are in series and which are in parallel is very important when designing or troubleshooting a combination circuit. It should always be your first step in analyzing a combination circuit. 5

6 SKILL 1 TRACE THE CURRENT PATH IN A COMBINATION CIRCUIT Procedure Overview In this procedure, you will determine which components are in series and which are in parallel in a combination circuit by tracing the current path through the circuit. 1. Examine the circuit in figure 3. R 1 + R 2 R 3 R 4 Figure 3. Combination Circuit 2. Trace the current path through the circuit and note whether each component is in series or parallel. R 1 (Series/Parallel) R 2 (Series/Parallel) R 3 (Series/Parallel) R 4 (Series/Parallel) R 1 and R 4 are in series. R 2 R 3. 6

7 3. Examine the circuit in figure 4 and determine which loads are in series and which are in parallel with other loads. R 1 (Series/Parallel) R 2 (Series/Parallel) R 3 (Series/Parallel) R 4 (Series/Parallel) R 5 (Series/Parallel) R 1 and R 5 are in series. R 2 R 3 R 4. R= R= 2 R= R= 4 25 R= 5 10 Figure 4. Combination Circuit 4. Examine the circuit in figure 5 and determine which components are in series and which are in parallel. R 1 (Series/Parallel) R 2 (Series/Parallel) R 3 (Series/Parallel) R 1 R 2. R 3 is in series. + R= 1 25 R = 2 25 R = 10 3 Figure 5. Combination Circuit 7

8 OBJECTIVE 3 LIST THE SEVEN STEPS FOR SOLVING A COMBINATION CIRCUIT Solving a circuit involves determining unknown values of resistance, current, or voltage in a combination circuit. The seven general steps for solving a circuit are: Simplify the circuit to a series circuit by finding the effective equivalent resistance (R EQ ) of each parallel section of the circuit. Use R EQ to calculate the total resistance (R T ) of the circuit by adding them to the series resistances. Calculate total current (I T ) using R T (Ohms Law). Calculate the voltage drop across any series resistance or equivalent resistance you need to know using Ohms Law. Calculate the branch currents in a parallel section using the voltage drop across R EQ and Ohms Law. Use the branch currents and resistance values to calculate the voltage of the parallel resistances. Make a summary of the voltage drops and currents for each resistance to make sure they add up. You may have to alter the order of the steps sometimes, depending on the information that you are given for the circuit. But, in most cases, these seven steps should help you find the information you need to know about the circuit. 8

9 SKILL 2 SOLVE A COMBINATION CIRCUIT Procedure Overview In this procedure, you will solve combination circuits using the seven-step process. You will calculate voltage and current values for each resistance in the circuit and redraw the circuit with all the values listed. In step 1, you will be led through the process. In steps 2 and 3, you will do it yourself. 1. Perform the following substeps to simplify the combination circuit shown in figure 6. The total resistance in a combination circuit can be found by combining the equations used to find total resistance in a series circuit and a parallel circuit. For example, the total resistance of the combination circuit in figure 6 is found by first calculating the equivalent resistance of the two resistors that are in parallel, R 2 and R 3. This resistance is then added to the resistances that are in series with it, R 1 and R 4. The following substeps will lead you through this process. R 1= V R 2 = R 3 = (R R = R ) 2 3 EQ R = 25 4 Figure 6. Calculating Total Resistance 9

10 A. Calculate the equivalent resistance of R 2 and R 3 in the circuit in figure 6. REQ = (Ohms) REQ = (Ohms) This is found as follows: = + R R R EQ = + R EQ R EQ = 8.33 ohms The equivalent resistance is 8.33 ohms. B. Calculate the total resistance of the circuit. Since this equivalent resistance is in series with R1 and R4, you can now find the total resistance by applying the formula for calculating total resistance in a series circuit. R T = (Ohms) The answer is found as follows: R = R + R + R (Where R is the equivalent resistance of R and R in parallel.) T 1 EQ 4 EQ 2 3 R = R + R + R T 1 EQ 4 R = T R = T The circuit in figure 6 has the same resistance as a circuit with just one load with the resistance of ohms. This completes steps 1 and 2 of the seven step process. C. Calculate the total current in the circuit of figure 6. Since the source voltage and the total resistance are known, they can be used to calculate the total current (IT) using Ohms Law. I T = (Amps) The answer is found as follows: VT IT = R T 24 IT = I T =.41 Total current in the circuit is approximately 0.41 amps. 10

11 D. Calculate the voltage drops across the resistors in the circuit in figure 6. In figure 7, the circuit from figure 6 has been reduced to a series circuit using the equivalent resistance calculated for the two parallel resistors. Since the current is known, the voltage drop across each resistance can now be calculated using Ohms Law. R = V R = 8.33 EQ R = 25 4 Figure 7. Combination Circuit to a Series Circuit V R1 = (VDC) V REQ = (VDC) V R4 = (VDC) The answers are found as follows: VR1 = IT R1 VR1 = V = volts R1 VREQ = IT REQ VREQ = V = 3.42 volts REQ V = I R V V R4 T 4 R4 = R4 = The voltage drop across R 1 is volts. The voltage drop across R EQ is 3.42 volts. The voltage drop across R 4 is also volts. NOTE The total calculated voltage is not quite 24V because we rounded off our calculations. 11

12 Since the voltage across the equivalent resistance of R 2 R 3 is 3.42 volts, we can say that the voltage drop across each of the parallel resistors, R 2 and R 3 is 3.42 volts, as shown in figure 8. VR1= 10.25V + 24V R 2 R 3 VR2/3 = 3.42V VR4 = 10.25V Figure 8. Voltage Drops in a Combination Circuit E. Calculate the current in each branch of the circuit in figure 6. Figure 9 shows the updated circuit. V R1 = 10.25V R 1=25 I T + 24V I R2 R 2=50 IR3 R =10 V REQ=3.42V 3 I T R 4=25 V R4=10.25V Figure 9. Current Flow in a Combination Circuit 12

13 The current in each branch of the parallel section can now be calculated by dividing the voltage drop of each load by its resistance. I R2 = (Amps) I R3 = (Amps) The answers are found as follows: I R2 R2 R3 R3 V = R R IR2 = 50 I =.068A I V = R R IR3 = 10 I =.342A The current in branch R2 (I R2 ) is A or 68 ma. The current in branch R 3 (I R3 ) is A or 342 ma. The branch currents are shown in figure 10. R V I R2 = I R3 = R 2 R 3 68mA 342mA R 4 Figure 10. Branch Currents in a Combination Circuit If the branch currents are added, the sum should equal the total current that flows through the circuit. This is according to Kirchhoffs Current Law. 13

14 F. Draw the circuit from figure 9 with all values shown. Your circuit should look like figure 11. R = VR1 = 10.25V IT = 0.41mA 24V R=50 V 2 R2 = 3.42V R 3=10 VR3 = 3.42V R = 25 4 IR2 = 68mA IR3 = 342mA VR4 = 10.25V Figure 11. Circuit Solved 2. Perform the following substeps to solve the circuit in figure 12. R 1 = V R 2 = R 3 = R 5 = 100 R = 4 25 R 6 = 25 Figure 12. Combination Circuit A. Calculate the equivalent resistance of R 2 R 3 (R EQ1 ) and R 5 R 6 (R EQ2 ) for the circuit in figure 12. R EQ1 = (Ohms) R EQ2 = (Ohms) They should be R EQ1 = 25 ohms and R EQ2 = 20 ohms. 14

15 B. Calculate the total resistance (R T ) of the circuit. R T = (Ohms) R T should = 95 ohms. C. Calculate the total current (I T ) using the source voltage and R T. I T = (Amps) I T should = 1.263A. D. Calculate the voltage drops across the series resistances, including the equivalent resistances. V R1 = (Volts) V REQ1 = (Volts) V R4 = (Volts) V REQ2 = (Volts) They should be V R1 = 31.58V, V REQ1 = 31.58V, V R4 = 31.58V, and V REQ2 = 25.26V. E. Determine the current in the branches of the parallel sections. I R2 = (Amps) I R3 = (Amps) I R5 = (Amps) I R6 = (Amps) They should be I R2 and I R3 =.632A, I R5 =.253A, and I R6 = 1.01A. F. Calculate the voltage of each parallel resistance using the current and resistance values. V R2 = (Volts) V R3 = (Volts) V R5 = (Volts) V R6 = (Volts) They should be V R2 and V R3 = 31.58V, and V R5 and V R6 = 25.26V. V R2 and V R3 should be the same because they are in parallel, as should V R5 and V R6. G. Redraw the circuit and list the voltage and current for each resistor along with its resistance. 15

16 3. Now solve the circuit in figure 13. Show all your work. R 1 = 10 R 2 = V R 3 = R 4 = R 6 = 10 R = 50 5 Figure 13. Combination Circuit 16

17 SEGMENT 1 SELF REVIEW 1. A(n) circuit contains elements of both a series circuit and a parallel circuit. 2. In order to identify the series and parallel sections of a combination circuit, the current path. 3. The representation R1 R2 indicates that R1 and R2 are. 4. Identifying which components are in series and which are in parallel should be the first step in a combination circuit. 5. In a combination circuit, components connected have the same current flowing through them. 6. The last step in solving a circuit is to make a summary of the voltage drops and currents for each to make sure they add up. 7. The current will be between the branches of the parallel section of a combination circuit. 8. There are seven general steps for a combination circuit. 17

18 SEGMENT 2 LIGHTING CIRCUITS OBJECTIVE 4 DESCRIBE HOW SWITCHES ARE USED IN COMBINATION CIRCUITS AND GIVE AN APPLICATION Loads are not the only devices that can be connected in series, parallel, or combinations. Switches can also be connected in many different configurations to achieve the desired control of a circuit. This is especially true of circuits found in many residential and commercial buildings. Switches can operate several devices at one time or can operate individual components within a circuit. Three examples are: Commercial building lighting system Electrical outlets controlled by a switch Ceiling fan with lights Commercial Building Lighting System Figure 14 shows how the lighting system of a typical commercial building is connected. In this case, a switch is connected in series with the parallel lamps. This configuration allows the switch to control when the voltage is supplied to all the lamps. When the switch is closed (turned on), each of the lamps receives the source voltage. When the switch is opened (turned off), the lamps turn off. LAMPS 120 VAC LIGHT SWITCH Figure 14. A Common Commercial Building Lighting Circuit 18

19 Electrical Outlets Controlled By a Switch Another example of a combination circuit can be found in many home power outlet circuits, as shown in figure 15. In this case, the outlets are connected in parallel with each other but in series with the switch. This allows the switch to turn the outlet power on or off. Many homes have circuits like this at the front door for control of a table lamp. HOT 120 VAC GROUND TYPICAL OUTLET Figure 15. Outlets Connected to a Switch A Ceiling Fan with Lights Now, consider the ceiling fan with lights, as shown in figure 16. In this circuit, the series combination of S2 and the fan motor is in parallel with the series combination of S3 and the lamps. A switch (S1) is in series with the parallel combination of S2 and S3 to control when voltage is applied to the circuit. WALL SWITCH S1 S2 S3 120 VAC CEILING FAN MOTOR M Figure 16. A Typical Ceiling Fan Circuit 19

20 This configuration creates a variety of possibilities, as shown in figure 17. If the main switch (S1) is on and S2 and S3 are also on, the fan and the lights will be on. To run the fan with the lights off, you could turn off S3 and leave S2 on. To turn on the lights and turn off the fan, you would turn on S3 and turn off S2. Finally, you could turn the fan and the lights both off while leaving the main switch (S1) on by turning off S2 and S3. S 1(ON) S (ON) S 2 3(ON) 120 VAC FAN MOTOR M (ON) LAMPS (ON) S 1(ON) S 2(ON) S 3(OFF) 120 VAC FAN MOTOR (ON) M LAMPS (OFF) S 1(ON) S (OFF) S 2 3(ON) 120 VAC FAN MOTOR M (OFF) LAMPS (ON) S 1(ON) S 2(OFF) S 3(OFF) 120 VAC FAN MOTOR (OFF) M LAMPS (OFF) Figure 17. Operating Options for a Ceiling Fan with Lights 20

21 SKILL 3 CONNECT AND OPERATE A BASIC LIGHTING CIRCUIT Procedure Overview In this procedure, you will connect a lighting circuit such as one you might find in a commercial building. You will operate the circuit and observe the status of the lamps. You will also measure the voltage across the lamps with the switch on and off. 1. Connect the circuit shown in figure 18. Make sure that the circuit is connected to the 24 volt (outer) terminals of the power supply on the T VAC Figure 18. Lighting Circuit 2. Perform the following substeps to operate the power supply. A. Place the AC/DC switch in the AC position. B. Turn on the power supply. 3. Close (turn on) the knife switch and observe the lamps. Lamp 1 status (On/Off) Lamp 2 status (On/Off) Lamp 3 status (On/Off) All lamps should be on because the current can flow through them. 21

22 4. Measure the voltage across each lamp with the DMM. Voltage Lamp 1 = (VAC) Voltage Lamp 2 = (VAC) Voltage Lamp 3 = (VAC) Each lamp should have the source voltage across it, 24 VAC ± 2 VAC. 5. Now open (turn off) the knife switch and observe the lamps. Lamp 1 status (On/Off) Lamp 2 status (On/Off) Lamp 3 status (On/Off) All lamps should now be off because the current flow has been stopped. 6. Measure the voltage across each lamp with the DMM. Voltage Lamp 1 = (VAC) Voltage Lamp 2 = (VAC) Voltage Lamp 3 = (VAC) All the lamps should have zero volts across them. 7. Turn off the power supply, disconnect the circuit, and store all components. You have now successfully connected and operated a typical lighting circuit. 22

23 SKILL 4 CONNECT AND OPERATE A CEILING FAN CIRCUIT Procedure Overview In this procedure, you will connect a circuit that resembles a typical ceiling fan circuit. You will then control the operation of the fan motor and the lights with the three switches of the circuit. 1. Connect the circuit shown in figure 19. S1 (KNIFE SWITCH) + 12V S2 (PUSHBUTTON) S3 (SELECTOR) SWITCH M FAN LAMP 1 LAMP 2 Figure 19. A Ceiling Fan Circuit 2. Perform the following substeps to operate the T7017 power supply. A. Place the AC/DC switch in the DC position. B. Turn on the power supply. 3. Close (turn on) the knife switch (S1) and observe the lamps and the fan. Lamp 1 status (On/Off) Lamp 2 status (On/Off) Fan status (On/Off) They should all be off because S2 and S3 are off (open). 23

24 4. Now press (turn on) the pushbutton switch (S2) and observe the lamps and the fan. Lamp 1 status (On/Off) Lamp 2 status (On/Off) Fan status (On/Off) The fan should now be on because closing S2 allows current to flow through that branch. The lamps do not come on because S3 is not on (open). 5. Release the pushbutton and rotate (turn on) the selector switch (S3). Observe the lamps and the fan. Lamp 1 status (On/Off) Lamp 2 status (On/Off) Fan status (On/Off) The fan should be off and the lights should be on because with S3 closed, current can flow in the branch that contains the lamps. 6. Now energize (turn on) both S2 and S3 together and observe the lamps and fan. Lamp 1 status (On/Off) Lamp 2 status (On/Off) Fan status (On/Off) They should all be on because current can flow through both branches of the circuit. 7. Leave S2 and S3 on and open (turn off) the knife switch (S1) and observe the lamps and the fan. Lamp 1 status (On/Off) Lamp 2 status (On/Off) Fan status (On/Off) They should all be off because there is no current flow to the branches because S1 is open (off). 8. Turn off the power supply, disconnect the circuit, and store all components. This procedure demonstrates how a typical ceiling fan with lights operates. By using the three switches you can change the operation of the circuit. 24

25 OBJECTIVE 5 DESCRIBE THE FUNCTION OF A VARIABLE RESISTOR AND GIVE AN APPLICATION Up to this point, the resistors you have used are fixed types. Another type of resistor that is often used is a variable resistor. A variable resistor is one which can have its resistance adjusted manually. A variable resistor can have two or three terminals, usually three. The two outside terminals act like a fixed resistor, the resistance between the two does not change. The center terminal, however, is attached to a slider, which is controlled by a knob, as shown in figure 20. TERMINALS RESISTOR HOUSING KNOB Figure 20. Variable Resistor 25

26 Figure 21 shows the schematic symbols for 2-terminal and 3-terminal variable resistors. 2-TERMINAL VARIABLE RESISTOR 3-TERMINAL VARIABLE RESISTOR CENTER TERMINAL (SLIDER) Figure 21. Variable Resistor The two common types of variable resistors used in most circuits are: Rheostat - A rheostat is a variable resistor that is used primarily in highpower circuits. A rheostat often only has two terminals, but can have three. The rheostat supplied with the T7017 trainer uses three terminals. Potentiometer - A potentiometer is a variable resistor that is used primarily in low-power (e.g. electronic) circuits. It usually has three terminals and is typically smaller than a rheostat. Variable resistors are used in many devices. Your radio uses variable resistors to control the volume, tuning, balance, and so on. A television uses them to control volume, color, tint, and brightness. Sometimes they are used to dim lighting in a home or business. They can also be used to control the speed of a motor by controlling the current flow to the motor. Variable resistors are even used in some power supply units to allow the output voltage to be adjusted to a desired level. 26

27 MIN MAX BATT 1.5V BATT 200mA MAX FUSED 10A MAX FUSED NON CONTACT VOLTAGE OFF MAX 600V 600V HOLD 200 2m 200m 10 A 20m CAT 600V CAT 300V Activity 1. Rheostat Operation Procedure Overview In this procedure, you will measure the resistance of a rheostat as it is adjusted and note the effect the slider has on the resistance between it and either outside terminal. You should also note that the sum of the two resistances created by the slider should always equal the resistance across the two outside terminals. 1. Prepare the DMM to measure resistance. 2. Locate the rheostat module and place it on the construction surface of the T With the control knob on the rheostat facing you, turn the knob fully counterclockwise. 4. Measure the resistance across the two outside terminals, as shown in figure 22. Resistance = (Ohms) It should be approximately 25 ohms. 30XR DMM V m V 2 200m 20M 2M 200k 20k 2k 10 A m 1.5V 9V 20m 200 2m ma A A COM BATT 9V V 10A RHEOSTAT MODULE Figure 22. Measurement of Rheostat Resistance 5. Turn the knob fully clockwise and measure the resistance across the two outside terminals. Resistance = (Ohms) It should be the same as in step 4, approximately 25 ohms. The resistance between the two outside terminals is always the maximum amount. The slider does not affect this value. 27

28 6. Now, measure the resistance between the center terminal and one of the outside terminals. Resistance = (Ohms) 7. Measure the resistance between the center terminal and the other outside terminal. Resistance = (Ohms) 8. Add the resistances from step 6 and 7. Total resistance = (Ohms) It should equal the resistance recorded in steps 3 and 4, approximately 25 ohms. Figure 23 shows the operation of a rheostat visually. Depending on the sliders position, the resistance between it and either outside terminal varies. Therefore, it effectively creates two different resistance values. RESISTANCE BETWEEN AB=0 OHMS BC=25 OHMS AC= 25 OHMS A B C RESISTANCE BETWEEN AB=12.5 OHMS BC=12.5 OHMS AC= 25 OHMS A B C AB=25 OHMS RESISTANCE BETWEEN BC=0 OHMS AC= 25 OHMS A B C Figure 23. Operation of a Variable Resistor 9. Turn the knob counterclockwise about 1/4 turn and repeat steps 6 through Keep repeating step 8 until the knob is all the way counterclockwise. Each time the two resistances added together should equal the total across the two outside terminals. This is always true of a rheostat. 28

29 SKILL 5 CONNECT AND OPERATE A RHEOSTAT AS A LIGHT DIMMER Procedure Overview In this procedure, you will connect and operate a rheostat in a circuit and use it to dim a lamp by controlling the voltage across the lamp. 1. Connect the circuit shown in figure 24. Make sure the knob on the rheostat is facing you before you connect it. + AC SOURCE SELECT DC LEFT TERMINAL 24V RHEOSTAT 24V 12V 12V LAMP CENTER TERMINAL RHEOSTAT MODULE LAMP MODULE Figure 24. Light Dimmer Circuit 2. Place the AC/DC switch in the DC position. 3. Make sure the rheostat knob is turned fully clockwise before turning on the power supply. 4. Turn on the power supply and observe the lamps status. Lamp status (On/Off) Lamp status (Bright/Dim) The lamp should be on and bright. 5. Prepare the DMM to measure voltage. 6. Measure the voltage across the lamp. Voltage = (Volts) The voltage should be approximately 23.5 V. 29

30 7. Turn the knob about 1/4 turn counterclockwise and observe the lamp. Lamp status (Brighter/Dimmer) The lamp should start to dim somewhat. 8. Measure the voltage across the lamp. Voltage = (Volts) It should be lower than it was in step Turn the knob to about the half-way point and observe the lamp. Lamp status (Brighter/Dimmer) The lamp should be even dimmer. 10. Measure the voltage across the lamp. Voltage = (Volts) It should be lower than in step Turn the knob about 3/4 counterclockwise and observe the lamp. Lamp status (Brighter/Dimmer) The lamp should be even dimmer than before. 12. Measure the voltage across the lamp. Voltage = (Volts) It should be lower than in step Now turn the knob fully counterclockwise and observe the lamp. Lamp status (Brighter/Dimmer) The lamp should be very dim. 14. Measure the voltage across the lamp. Voltage = (Volts) It should be lower than in step Adjust the knob back and forth several more times and observe the lamp to familiarize yourself with the operation. 16. Turn off the power supply, disconnect the circuit, and store all components. 30

31 SEGMENT 2 SELF REVIEW 1. can be connected in series, parallel, or combinations just like loads. 2. A(n) often has a switch to apply power to the circuit, a switch to control the lights, and a switch to control the fan motor. 3. A(n) resistor can be manually adjusted. 4. A(n) is a variable resistor used primarily in high-power circuits. 5. A variable resistor most commonly used in low-power circuits is a(n). 6. Depending on the position of the on a variable resistor, the resistance between it and either outside terminal will vary. 7. A variable resistor can be used to lighting in a home or business. 31

32 SEGMENT 3 VOLTAGE DIVIDERS OBJECTIVE 6 DESCRIBE THE FUNCTION OF A VOLTAGE DIVIDER AND GIVE AN APPLICATION A common application of Ohm s Law and series circuits is a voltage divider network. The function of a voltage divider network is to create a voltage that is lower than the source voltage. As shown in figure 25, the voltage divider consists of two or more resistors in series with the source voltage. The voltage at point B with reference to ground is some value between the source voltage and zero volts. +12V SOURCE VOLTAGE POINT B R 1 R 2 V R1 V R2 Figure 25. A Voltage Divider Network The main application of voltage divider networks is with electronic circuits, which need different voltage levels such as +12 volts, +5 volts, and +3 volts to operate. The voltage divider network provides these voltages from one power supply instead of using three separate power supplies. This saves space and money. Another application of a voltage divider is to provide a reduced voltage to drive a load. 32

33 OBJECTIVE 7 DESCRIBE THE OPERATION OF THREE TYPES OF VOLTAGE DIVIDERS There are three points you must consider when designing or using a voltage divider network: How constant the load voltage must be How much power the circuit will consume Whether load resistance is high or low For these reasons, there are three types of voltage divider networks: Firm Voltage Divider Stiff Voltage Divider Loaded Voltage Divider All three of these networks look like the network in figure 26. The difference is in how the resistances for R 1 and R 2 are selected. Each of the three types of networks is designed to deal with a certain type of application. The reason for this is the load resistance affects the voltage level from the voltage divider. Before you learn more about each type of network it is important to understand why the voltage level supplied by any voltage divider changes if the load resistance changes. To understand why, first examine the divider network shown in figure 26. As shown, a voltage divider network is actually a combination circuit because the load is connected in parallel with one part of the circuit. In this case, it is R 2. This resistor is called the bleeder resistor. The voltage, V RL, supplied to the load R L depends on the amount of the equivalent resistance of R 2 and R L. If R L or R 2 change the voltage V RL will also change. V T = 24 VOLTS + 24V R 1 = 100 VOLTAGE DIVIDER NETWORK V RL= 6 VOLTS R 2 = 10 RL = LOAD RESISTANCE Figure 26. A Combination of Circuit Serving as a Voltage Divider 33

34 Firm and Stiff Voltage Dividers To deal with the effects of changes in load, firm and stiff voltage dividers are designed so that the bleeder resistor, in parallel with the load, is much less than the load resistance. In this case, the voltage level is determined mostly by the bleeder resistor, R 2, not the load resistor, R L. A firm voltage divider uses a bleeder resistor, R 2, which is 1/10 the resistance of the load. A stiff voltage divider uses a bleeder resistor which is 1/100 the resistance of the load, as shown in figure 27. The firm voltage divider is used if the load voltage does not need to be very constant. Its advantage is that it does not consume as much power as the stiff voltage divider. The stiff voltage divider is used when the load voltage must be more constant. FIRM VOLTAGE DIVIDER STIFF VOLTAGE DIVIDER R 1 R R L R= 2 R= 10 R L R L R L Figure 27. A Firm and a Stiff Voltage Divider The relationship between the voltages and resistances for either a stiff or firm voltage divider is shown in the following formula. This is actually an approximation because it assumes R L is infinite. This formula can be used to calculate the load voltage if you know the resistance. It can also be used to size resistor R 1 if you know what the load voltage needs to be. STIFF AND FIRM VOLTAGE DIVIDER FORMULA V RL R2 = VT R + R 1 2 where V RL = load voltage in volts V T R 1 R 2 = total voltage drop in volts = resistance of R 1 in Ohms = resistance of R 2 in Ohms The firm and stiff voltages dividers are commonly used to supply voltages to devices which have relatively high resistances. An electronic amplifier is an example. 34

35 If the load resistance is relatively low, these two types of voltage dividers waste a lot of power. For example, if you had a load which required 50 ma of current, your divider would waste 500 ma of current using a firm divider (using the 10:1 ratio of the load resistor to the bleeder resistor) or 5 Amps using a stiff divider (100:1 ratio). For these applications, a loaded voltage divider network should be used. Loaded Voltage Divider A loaded voltage divider network looks exactly like a firm or stiff network except the bleeder resistor is 10 times the resistance of the load resistor instead of 1/10 or 1/100, as shown in figure 28. With this approach, most of the power is used by the load. The power wasted by the divider resistor is only 10%. V T R 1 + V RL R = 10 x R 2 L R L Figure 28. Loaded Voltage Divider The relationship between the resistances and the voltages is shown in the following formula. This is the same formula that is used for a stiff or firm divider except that R EQ (parallel resistance of R L and R 2 ) is inserted for R 2. This formula is the exact calculation for the load voltage. STIFF AND FIRM VOLTAGE DIVIDER FORMULA V RL REQ = VT R + R 1 EQ where V RL = load voltage in volts V T R 1 R EQ = total voltage drop in volts = resistance of R 1 in Ohms = resistance of R 2 and R L in Ohms 35

36 SKILL 6 DESIGN A VOLTAGE DIVIDER NETWORK Procedure Overview In this procedure, you will design circuits with all three types of voltage dividers. You will be led through the first design of each type and will then work through the others yourself. 1. Perform the following substeps to design a firm voltage divider network for the circuit shown in figure 29. R 1 12V V RL = 8V R 2 R L = 1M Figure 29. Voltage Divider Circuit A. Calculate the resistance of R 2. R2 = (Ohms) The answer is as follows: R R RL = 10 1,000,000 = 10 R = 100,000 The resistance of R2 is 100,000 ohms. 36

37 B. Calculate the resistance of R 1. R 1 =(Ohms) The basic formula is: V RL R R + R 2 = 1 2 V T The answer is found by rearranging the formula as follows: R T V RL 1 V R = R ,000 VRL = 100,000 8 R = 150, ,000 The resistance of R 1 is 50,000 ohms. 2. Design a firm voltage divider network for the circuit shown in figure 29 for an R L of 75 K. R 2 = (Ohms) R 1 = (Ohms) R 2 is 7,500 ohms and R 1 is 3,750 ohms. 3. Perform the following substeps to design a stiff voltage divider for the circuit shown in figure 30. R 1 24V V RL = 16V R 2 R L = 50 k Figure 30. Voltage Divider Circuit 37

38 A. Calculate the resistance of R 2. R 2 = (Ohms) R R 2 2 RL = ,000 = 100 The resistance of R 2 is 500 ohms. B. Calculate the resistance of R 1. R 1 = (Ohms) V R R = R T V RL R1 = R = The resistance of R 1 is 250 ohms. 4. Design a stiff voltage divider network for the circuit shown in figure 30 for an R L of 85K. R 2 = (Ohms) R 1 = (Ohms) R 2 is 850 ohms and R 1 is 425 ohms. 5. Perform the following substeps to design a loaded voltage divider for the circuit shown in figure 31. R V V RL = 3.6V R 2 R L = 5 Figure 31. Loaded Voltage Divider 38

39 A. Calculate the resistance of R 2. R 2 = (Ohms) Resistor R 2 is chosen to be 10 times larger than R L as follows: R = R 10 2 L R = The resistance of R 2 is 50 ohms. B. Calculate R EQ (the resistance of R 2 R L ). NOTE The design of the loaded voltage dividers considers the effect of the load in parallel with the bleeder resistor. R EQ = (Ohms) R R R EQ EQ EQ 1 = R R 2 L 1 = =.22 The solution is as follows: The resistance of R EQ is 4.55 ohms. C. Calculate the value of R 1. R 1 = (ohms) This calculation uses the same voltage divider formula used to design stiff and firm dividers except the combined resistance of R 2 and R L (R EQ ) is used instead of just R 2. This can be rearranged as before: V R RL 1 EQ T EQ 1 = REQ V RL 1 R EQ = V R + R V R R 1 = R = T The resistance of R 1 is 10.1 ohms. 39

40 6. Design a loaded voltage divider for the circuit shown in figure 31. Use a V RL of 10V and an R L of 10 ohms. R 2 = (Ohms) R 1 = (Ohms) R 2 is 100 ohms and R 1 is 1.45 ohms. 7. Design a loaded voltage divider for the circuit shown in figure 31. Use a V RL of 4V and an R L of 2 ohms. R2 = (Ohms) R1 = (Ohms) R 2 is 20 ohms and R 1 is 3.46 ohms. SKILL 7 CONNECT AND OPERATE A VOLTAGE DIVIDER NETWORK Procedure Overview In this procedure, you will connect and operate a loaded voltage divider network. You will then take measurements to verify its operation. 1. Perform the following substeps to set the rheostat (R L ) to the required load resistance value. NOTE In the next step, the rheostat will be used in a voltage divider circuit to simulate the load (R L ). The two (2) 25-ohm resistors will be used in series to simulate R 2. A. Set the DMM to measure resistance. 40

41 MIN MAX BATT 1.5V BATT 200mA MAX FUSED 10A MAX FUSED NON CONTACT VOLTAGE OFF MAX 600V 600V HOLD 200 2m 200m 10 A 20m CAT 600V CAT 300V B. With the control knob of the rheostat facing you, measure the resistance between the left terminal and the center terminal as shown in figure XR DMM V m 20M 2M 200k 20k 2k 10 A m 1.5V 9V 20m 200 2m ma A V 2 200m A LEFT TERMINAL COM BATT 9V V 10A RHEOSTAT MODULE CENTER TERMINAL Figure 32. Resistance Across teh Rheostat C. Turn the knob until the resistance between the left and center terminals is 5.0 ohms. 2. Connect the circuit shown in figure 33 using the rheostat you just set in step 2 for R L and the two (2) 25 ohm resistors for R 2. The circuit is equivalent to the one that you calculated the values for in the previous skill (step 5). R= V 25 R= 2 50 TWO 25 ( RESISTORS IN SERIES) 25 R L =5 RHEOSTAT Figure 33. Circuit to Simulate Figure 31 41

42 3. Perform the following substeps to operate the circuit and make measurements. A. Place the AC/DC switch in the DC position. B. Turn on the power supply on the T7017. C. Set the DMM to measure DC Volts. D. Now measure the voltage across R L. V RL = (Volts) It should be approximately 3.6V. NOTE Actual values will vary depending on the exact values of the resistors and the source voltage. E. Turn off the power supply. In the next two steps you will measure the unloaded voltage produced by the divider network. 4. Disconnect R L from the circuit and turn on the power supply. 5. Now measure the voltage across R 2 (the series combination of the two 25 ohm resistors) to determine if unloaded voltage is close to the 3.6 V value produced when the load was attached? Unloaded voltage close to loaded voltage? (Yes/No) You should find that the unloaded voltage is much greater than 3.6 V. This is because the loaded divider network requires that the load be constant for proper operation. 6. Turn off the power supply. 7. Disconnect the circuit and store all components. 42

43 SEGMENT 3 SELF REVIEW 1. A(n) network creates a voltage that is lower than the source voltage. 2. A loaded voltage divider network looks exactly like a firm or stiff network, except the bleeder resistor is times the resistance of the load resistor. 3. To deal with the effects of changes in load, firm and stiff voltage dividers are designed so that the resistor is much less than the load resistance. 4. The main application of voltage divider networks is with electronic circuits that need voltage levels to operate. 5. The firm and stiff voltage dividers are commonly used to supply voltage to devices that have relatively high. 43

44 SEGMENT 4 TROUBLESHOOTING OBJECTIVE 8 EXPLAIN THE EFFECT OF A SHORT CIRCUIT A short circuit across any branch of a parallel circuit causes the entire circuit to be short circuited, as shown in figure 34. All current flows through the short circuited branch and bypasses the other branches. This causes an excessive amount of current to flow since there is practically no resistance. I T + 20V R 1 R 2 R 3 I T SHORT Figure 34. A Short in a Parallel Circuit 44

45 If a short circuit occurs in a series circuit, excessive current flow may or may not occur. For example, if a single component is shorted, as shown in figure 35, the current continues to flow through the rest of the circuit. Only the shorted component is bypassed. This means the circuit flow will be higher, but it may not be excessive. SERIES CIRCUIT I T + SHORT Figure 35. A Shorted Componet in Series Circuit When excessive current occurs, something in the circuit is usually damaged. It might damage the power supply or it may cause a wire to burn up. In most cases, a short eventually creates an open, such as a wire burning up. The open can be in the main line, in the branch containing the short, or anywhere along the path to the short. Adding a fuse or circuit breaker to the main line of a parallel circuit helps to protect the components should a short develop in any branch. If this happens, the fuse or circuit breaker opens, as shown in figure I T FUSE BLOWS SHORT Figure 36. A Fuse in a Parallel Circuit 45

46 OBJECTIVE 9 DESCRIBE THE FOUR STEPS FOR TROUBLESHOOTING A SHORT CIRCUIT The four steps for troubleshooting a circuit to find a short are: Step 1. Look for visible signs of damage - There may be visible signs, such as smoke or soot that will indicate where the short occurred. If there are no visible signs, you will have to continue on to the next step. Step 2. Disconnect the power supply - Do not replace or reset any circuit protection devices at this time. Wait until the problem has been resolved. Step 3. Connect an ohmmeter across the main line - This is shown in figure 37. OHMMETER ACROSS MAIN LINE DISCONNECTED POWER SUPPLY R=? T R 1 R 2 R 3 DISCONNECT BRANCHES ONE AT A TIME Figure 37. Connecting an Ohmeter to Locate a Short Step 4. Disconnect each branch one at a time - If the resistance stays close to zero when the branch is disconnected, the short is in another branch. Reconnect that branch and move on to the next one. If the resistance suddenly rises when a branch is disconnected, that is the branch where the short occurred. If disconnecting all branches still results in a zero ohm reading, the short has developed directly across the main line and not in any branch. Once you have located the short, you can correct the problem and then return the circuit to operation. You should make sure that all the branches are reconnected, the circuit protection device is replaced or reset, and the power supply is reconnected. This 4-step procedure is one of the most common techniques used by electricians and electronic technicians to troubleshoot for shorts. 46

47 SKILL 8 LOCATE A SHORT CIRCUIT Procedure Overview In this procedure, you will troubleshoot a circuit to locate a short circuit using the four-step process. You will then repair the short and return the circuit to operation. NOTE Consult your instructor as to whether you should continue with this procedure or not. You will need an extra 3-amp fuse. 1. Connect the circuit shown in figure 38 and turn on the AC power supply. Observe the fuse. Fuse status (Blown/Not blown) The fuse should blow immediately indicating a short. 3A FUSE SHORTING LEAD 12V R = R = 1 R = Figure 38. Parallel Circuit with a Short 47

48 2. Perform the following substeps to troubleshoot the circuit. A. Look at the circuit to see if there are any visible signs of a short (smoke or soot). B. Turn off and disconnect the power supply from the circuit. C. Set the DMM to measure resistance and connect it across the circuit, as shown in figure V DMM R=10 1 R=25 2 R=25 3 DISCONNECT HERE ONE BRANCH AT A TIME Figure 39. Troubleshooting with a DMM D. Disconnect Branch 1 (R 1 ) of the circuit and observe the resistance reading. Resistance - Branch 1 disconnected = (Ohms) E. Reconnect branch 1. F. Repeat substeps D and E for the other two branches. Resistance - Branch 2 (R 2 ) disconnected = (Ohms) Resistance - Branch 3 (R 3 ) disconnected = (Ohms) The resistance readings when branches 1 and 2 are disconnected should be zero. When branch 3 is disconnected the resistance reading should rise. That means this is the branch where the short occurred. 3. Correct the short circuit by removing the wire that is shorting the resistor. NOTE This wire is used to simulate what happens when a component, such as a resistor, actually shorts out. In an actual situation, you would need to replace the shorted component. In this case, you will simply remove the wire. 48

49 4. Make sure all the branches are reconnected and observe the resistance reading. Resistance reading = (Ohms) The reading should be 5.56 ohms, which is the effective resistance of the circuit. Therefore, the short circuit has been eliminated. If the reading is still zero, another short is present and must be found. 5. Replace the blown fuse with a 3-amp fuse. 6. Reconnect the power supply to the circuit. 7. Operate the circuit and observe the fuse. Fuse status (Blown/Not blown) The fuse should not blow since the short has been repaired. 8. Turn off the power supply. 9. Disconnect the circuit and store all components. OBJECTIVE 10 DESCRIBE THE THREE BASIC STEPS FOR TROUBLESHOOTING AN OPEN CIRCUIT In many cases, an open is caused by a short. This happens when a short continues for a while, causing a wire or a component to burn and open. It also causes an open in a fuse or circuit breaker. If one branch of a parallel circuit has opened, the rest of the circuit will still operate but with a reduced total current. This occurs because the effective resistance of the circuit has been increased. If an open occurs in a series circuit, current will not flow. The method for troubleshooting a circuit to locate an open is much the same as for locating a short. The three basic steps are: Step 1. Disconnect the power supply from the circuit Step 2. Connect the ohmmeter across the main line - This allows you measure the total resistance of the circuit. Step 3. Disconnect each branch, one at a time - The total resistance should increase as each branch is disconnected. If disconnecting a branch has no effect on the total resistance, the branch you are testing has an open somewhere. More than one branch may have an open. Therefore, you should continue to check the rest of the branches to be sure that another branch doesnt have an open. NOTE If the open occurs in the main line, the circuit will not operate at all. 49

50 Once a branch with an open has been found, it must be determined if a short caused the open. If this is the case, you will have to repair or replace the component to correct the short and open. You should then check the whole circuit for any other shorts. SKILL 9 LOCATE AN OPEN CIRCUIT Procedure Overview In this procedure, you will troubleshoot a parallel circuit to determine where an open has occurred. You will then repair the open and return the circuit to operation. 1. Connect the circuit shown in figure 40 and then turn on the power supply. Note the status of the lamp in branch 2. Lamp status (On/Off) The lamp should be off because there is no path for current to flow in branch 2. DISCONNECTED WIRE + 12V R=10 1 R=3.1 2 R=25 3 Figure 40. A Parallel Circuit with an Open 2. Perform the following three substeps to troubleshoot the circuit for an open. A. Turn off and disconnect the power supply from the circuit. B. Set the DMM to measure resistance, connect it across the main line of the circuit and observe the reading. Total resistance = (Ohms) The total resistance should be approximately 7.2 ohms. 50

51 C. Disconnect branch 1 and record the resistance reading. Resistance - Branch 1 disconnected = (Ohms) D. Reconnect branch 1. E. Repeat substeps C and D for the other two branches. Resistance - Branch 2 disconnected = (Ohms) Resistance - Branch 3 disconnected = (Ohms) When branches 1 and 3 are disconnected, the total resistance reading should increase. However, when branch 2 is disconnected, the total resistance reading does not change. This means branch 2 contains the open. Also, since branch 2 contains a lamp, we could have determined that branch had an open since the lamp did not come on when the circuit was operated. 3. Repair the open by connecting the loose wire to the component in branch 2, as shown in figure 41. RECONNECT WIRE + 12V R=10 1 R=3.1 2 R=25 3 Figure 41. Connecting a Loose Wire 4. Perform the four steps to troubleshoot the circuit to determine if there is a short present in the circuit. Short (Yes/No) You should not find a short present. 5. Measure the total resistance of the circuit. Total resistance = (Ohms) The total resistance should now be lower than it was in step 2B of Skill 7. This is because the resistance of the lamp is now affecting the circuit. 51

52 6. Reconnect the power supply to the circuit and turn on the circuit. Observe the lamp status. Lamp status (On/Off) The lamp should now be on since the open has been repaired. 7. Turn off the power supply. 8. Disconnect the circuit and store all components. NOTE The same results you saw in this skill would have occurred if the lamp had been burned out, which happens frequently with lamps. When a lamp burns out, the filaments inside the lamp open. 52

53 SEGMENT 4 SELF REVIEW 1. A short circuit across any branch of a(n) circuit causes the entire circuit to be short circuited. 2. Adding a fuse or circuit breaker to the of a parallel circuit will help protect the components from damage. 3. Eventually, a short in a parallel circuit will cause a(n). 4. There are basic steps in troubleshooting a parallel circuit for opens. 5. An open in a branch of a parallel circuit will cause the effective resistance to. 6. In a series circuit, if a single component is, the current continues to flow through the rest of the circuit. 7. In most cases, repairing an open in a circuit requires a component that has opened. 8. If an open occurs in the main line, the will not operate at all. 53