AC/DC ELECTRICAL SYSTEMS

Transcription

1 AC/DC ELECTRICAL SYSTEMS LEARNING ACTIVITY PACKET CIRCUIT ANALYSIS BB227-BC03UEN

2 LEARNING ACTIVITY PACKET 3 CIRCUIT ANALYSIS INTRODUCTION The previous LAP discussed how current, resistance, and voltage are affected by circuits with series and parallel loads. It did not, however, attempt to analyze the precise relationship among these three characteristics. This LAP will do just that through Ohm s Law and Kirchhoff s voltage and current laws. These laws are fundamental concepts used by every electrician, electronics technician, and electrical engineer. One application of these laws is to determine how much power is needed for a circuit. This LAP will explain this calculation. ITEMS NEEDED Amatrol Supplied 1 T7017 AC/DC Electrical Learning System FIRST EDITION, LAP 3, 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 POWER IN SERIES CIRCUITS OBJECTIVE 1 State the formula for calculating series resistance and give an application SKILL 1 Calculate series resistance given each load s resistance OBJECTIVE 2 State Ohm s Law, explain its importance and give an application SKILL 2 Use Ohm s Law to calculate voltage, current, and resistance in a series circuit OBJECTIVE 3 State Kirchhoff s voltage law for a series circuit and give an application Activity 1 Verifi cation of Kirchhoff s Voltage Law OBJECTIVE 4 Defi ne power and give its units of measurement OBJECTIVE 5 State a formula for calculating the total power used in an electrical circuit SKILL 3 Calculate the total power used by a series circuit SEGMENT 2 POWER IN PARALLEL CIRCUITS OBJECTIVE 6 State Kirchhoff s Current Law and give an application SKILL 4 Calculate the main line current in a parallel circuit OBJECTIVE 7 State a formula for calculating total parallel resistance SKILL 5 Calculate the total parallel resistance SKILL 6 Calculate the total power used in a parallel circuit SEGMENT 3 CIRCUIT PROTECTION DEVICES OBJECTIVE 8 Describe the function of two types of circuit protection and give an application of each OBJECTIVE 9 Describe the operation of a fuse and give its schematic symbol SKILL 7 Operate a circuit using a fuse SKILL 8 Test and replace a fuse OBJECTIVE 10 Describe the operation of two types of circuit breakers and give their schematic symbols SKILL 9 Operate a circuit using a circuit breaker SKILL 10 Test and reset a circuit breaker 3

4 SEGMENT 1 POWER IN SERIES CIRCUITS OBJECTIVE 1 STATE THE FORMULA FOR CALCULATING SERIES RESISTANCE AND GIVE AN APPLICATION The total resistance in a circuit determines how much power the circuit will draw. If the loads are connected in series, the total resistance can be calculated as follows: TOTAL RESISTANCE FOR SERIES RT = R1 R2 R3... where RT = total resistance in Ohms R1 = resistance of R1 in Ohms R2 = resistance of R2 in Ohms R3 = resistance of R3 in Ohms 24V R1 R2 R3 R4 RT Figure 1. Total Series Resistance Circuit Knowing how to calculate series resistance is useful when designing a sound system. Each speaker has a certain amount of resistance. When speakers are connected in series, their total resistance is added. Care must be taken not to overload the amplifier by connecting an excessive load. Many times multiple speakers are connected in series and parallel combinations to keep the total resistance within an acceptable range. 4

5 SKILL 1 CALCULATE SERIES RESISTANCE GIVEN EACH LOAD S RESISTANCE Procedure Overview In this procedure, you will calculate total resistance in a series circuit and verify your calculations by measuring the total resistance. 1. Calculate the total resistance of the circuit in figure 2. RT = (Ohms) 12V R1 = 10 R2 = 25 R3 = 25 RT Figure 2. Calculation of Total Resistance The answer is found as follows: RT = R1 R2 R3... RT = RT = 60 The total resistance for the circuit is 60 ohms. 5

6 2. Connect the series circuit shown in figure 3. This will allow you to prove that the resistances add together in series. NOTE You will not need to connect the power supply to take the measurements in this circuit. R1 = 10 R2 = 25 R3 = 25 POINT A POINT B Figure 3. Series Circuit 3. Measure the resistance of each load. Resistance R1 = (Ohms) Resistance R2 = (Ohms) Resistance R3 = (Ohms) R1 should be approximately 10 ohms, R2 should be approximately 25 ohms, and R3 should be approximately 25 ohms. 4. Measure the total resistance by placing one lead at point A and the other lead at point B. Total resistance = (Ohms) The total resistance should be approximately 60 ohms. There might be a slight variance. 5. Disconnect the circuit and store all components. 6

7 OBJECTIVE 2 STATE OHM S LAW, EXPLAIN ITS IMPORTANCE AND GIVE AN APPLICATION In the previous LAP, you saw that the current in series and parallel circuits is affected when the resistance changes. An increase in resistance causes a decrease in current and vice versa. In electrical circuits where the loads are only resistance-type loads, the voltage, current, and resistance are actually related to each other by a formula known as Ohm s Law. Ohm s Law says that one volt can push one amp of current through one ohm of resistance. It is stated mathematically as follows: OHM S LAW E = I R where E = voltage (Volts) I = current (Amps) R = resistance (Ohms) Ohm s law allows you to calculate either voltage, current or resistance if you know two of the three variables. For example, in figure 4, you can calculate the current that will run through the circuit because the voltage (24 volts) and resistance (10 ohms) are known. This is calculated as follows: E = I R 24 = I = 10 I = 24. Amps 24V I R=10 Figure 4. Ohm s Law 7

8 Ohm s law is a fundamental electrical concept. If you work with electricity, you will use it often. Four applications of Ohm s Law are: Troubleshooting Circuits - Ohm s Law can be used to diagnose problems in a circuit by comparing the actual voltage and current to the theoretical values. Calculating Power Dissipation - Ohm s Law can be used to determine how much electrical power is needed to supply a circuit. Size Component - Ohm s Law is used to size components and wires in circuits. Create Multiple Voltage Levels - Ohm s Law can be used to design a circuit to supply a specific voltage that is lower than the source voltage. This can be used for a device that needs a specific reference voltage. SKILL 2 USE OHM S LAW TO CALCULATE VOLTAGE, CURRENT, AND RESISTANCE IN A SERIES CIRCUIT Procedure Overview In this procedure, you will calculate voltage, current, and resistance in a series circuit by applying Ohm s Law. 1. Calculate the power supply s voltage for the circuit shown in figure 5 if the current is 2 amps and resistor R1 has a resistance of 20 ohms. E I = 2A A R = 20 1 Figure 5. Voltage Calculation Voltage (E) = (Volts) The solution to this problem is found by using Ohm s law as follows: E = I R E = 2A 20 Ohms E = 40 Volts 8

9 2. Calculate the voltage drop across resistor R 1 in the circuit of figure 6. The current and resistance are noted in the schematic diagram. I = 3A A E 1 V R = 10 1 R = 15 2 Figure 6. Voltage Calculation E 1 = (Volts) This is solved by using only the resistance of R 1 in the Ohm s law formula (E = 3 10). The answer is 30 V. 3. Calculate the power supply s voltage if the total circuit resistance is 25 ohms and the current measured is 5.0 amps. Voltage (E) = (Volts) The answer is E = 125 volts. In addition to using Ohm s law to solve for voltage, you can also use it to solve for current and resistance. In the applications of this unit, you will do this more often because the T7017 power supply is a constant voltage power supply. Therefore, you already know voltage. To solve for resistance, rearrange Ohm s law as follows: E R I = To solve for current, rearrange Ohm s law as follows: E I R = In the remaining steps of this skill, you will use Ohm s law to solve for current and resistance. 9

10 4. Calculate the current in the circuit shown in figure 7. The power supply is a constant voltage power supply. Current (I) = (Amps) The answer is I = 2.4 amps. 24 V I 10 Figure 7. Current Calculation Being able to calculate current in a circuit is important if you want to add a circuit protection device. You must know the total current in order to select the properly-rated protection device. If the current is much above the device s rating, the device will blow or trip when operated in the circuit, as you will see later in this LAP. 5. Calculate the current in the circuit shown in figure 8 using the voltage and resistance values. Current = (Amps) The answer is I = 2 Amps. 24V 3 9 Figure 8. Current Calculation Since the total resistance is 12 ohms (3 9), the current is (24/12). 10

11 6. Calculate the resistance of the load in figure 9 if the voltage applied to the circuit is 120V and the current is 3A. R = (Ohms) The answer is R = 40 ohms. 120V I = 3A R Figure 9. Resistance Calculation Knowing the resistance of a load is important in designing a circuit because of the effect it has on the current and voltage. Sometimes loads do not specify their resistance. However, Ohm s Law gives us a means of determining an unknown resistance value. 7. Calculate the resistance of the load shown in figure 10. The voltage and current are shown. Resistance = (Ohms) The answer is R = 10 ohms. I= 1.2A 12V R Figure 10. Resistance Calculation 11

12 8. Calculate the current in the circuit shown in figure 11. Current = (Amps) The answer is I = 0.80 Amps. 35V I Figure 11. Current Calculation 9. Calculate the unknown resistance value in the circuit shown in figure 12. HINT You will need to determine the voltage drop across both resistors. Resistance = (Ohms) The answer is R = 23 Ohms. 24V I =.8A R 7 Figure 12. Resistance Calculation 12

13 OBJECTIVE 3 STATE KIRCHHOFF S VOLTAGE LAW FOR A SERIES CIRCUIT AND GIVE AN APPLICATION Another important set of electrical relationships you often use is Kirchhoff s Laws. The first one you will learn is Kirchhoff s Voltage Law. Kirchhoff s Voltage Law states that the total voltage in a series circuit is equal to the sum of the individual voltage drops in the circuit. Stated in an equation Kirchhoff s Voltage Law would be as follows: KIRCHHOFF S VOLTAGE LAW V T = V R1 V R2 V R3 where V T = total voltage in Volts V R1 = voltage drop across R1 in Volts V R2 = voltage drop across R2 in Volts V R3 = voltage drop across R3 in Volts For example, in figure 13, the voltage drop across each resistor is shown. Therefore, the total voltage of the circuit, according to Kirchhoff s Voltage Law, is 24 V. This is equal to the power supply voltage. 24V R 1 R 2 R 3 V R1=10V V R2=5V VRT=24V V R1=9V Figure 13. Example of Kirchhoff s Voltage Law This law is very important in the design and troubleshooting of multiple load series circuits. 13

14 Activity 1. Verification of Kirchhoff s Voltage Law Procedure Overview In this procedure, you will measure the individual voltage drops across each load in a series circuit and add them to verify Kirchhoff s Voltage Law. 1. Connect the circuit in figure V R = 10 1 R = 25 2 R = 25 3 VR1 VR2 VR3 Figure 14. Example Circuit 2. Perform the following substeps to operate the circuit. A. Place the AC/DC switch in the DC position. B. Turn on the power supply. 3. Prepare the DMM to measure DC voltage. 4. Measure the voltage drop across each load. Voltage Drop R 1 = (VDC) NOTE The larger the resistance value is, the larger the voltage drop is. Voltage Drop R 2 = (VDC) Voltage Drop R 3 = (VDC) Voltage drop R 1 should be approximately 2V, voltage drop R 2 should be approximately 5V, and voltage drop R 3 should be approximately 5V. 5. Turn off the power supply, disconnect the circuit, and store all components. 14

15 6. Calculate the total voltage drop for the circuit. Total Voltage Drop = (VDC) The total voltage drop should be approximately 12V. 7. Compare the total voltage drop to the source voltage. Total Voltage Drop = (VDC) Source voltage = (VDC) The total voltage drop should equal the source voltage, which is what Kirchhoff s Voltage Law says. OBJECTIVE 4 DEFINE POWER AND GIVE ITS UNITS OF MEASUREMENT Power is the measure of the energy consumed by a circuit. It is measured in watts, abbreviated W. One watt of energy is consumed when one volt pushes one amp through a circuit. OBJECTIVE 5 STATE A FORMULA FOR CALCULATING THE TOTAL POWER USED IN AN ELECTRICAL CIRCUIT The power used by any load in a circuit can be determined using the following basic power formula: POWER FORMULA P = I E where P = power (Watts) I = current (Amps) E = voltage (Volts) This formula will work in either AC or DC circuits. The Ohm s law value for voltage, E = I R, can be substituted and the formula P = I E = I (I R) and restated as: ALTERNATE POWER FORMULA P = I 2 R where P = power (Watts) I = current (Amps) R = resistance (Ohms) 15

16 These two formulas can be used to calculate the power used by a single component or by the entire circuit. To calculate the power used by the entire circuit, E equals the power supply s voltage and I is the total current. R equals the total resistance, as shown in figure 15. E=24V I=1A R = 24 OHMS T Figure 15. Circuit Power Output Calculation The total power in a series circuit can also be calculated by adding together the power used by each load. This can be stated as: TOTAL POWER IN SERIES P T = P R1 P R2 P R3... where P T = total power in Watts P R1 = power dissipation of R 1 in Watts P R2 = power dissipation of R 2 in Watts P R3 = power dissipation of R 3 in Watts 16

17 SKILL 3 CALCULATE THE TOTAL POWER USED BY A SERIES CIRCUIT Procedure Overview In this procedure, you will calculate the total power used in a series circuit. This is a common calculation designed to determine the sizes of components needed in the circuit. 1. Calculate the total power used in the circuit shown in figure 16. P T = (Watts) The solution is found as follows: Since we know the total voltage and current, we can use the first power in series formula. P T = I T E T P T = P T = 4.56 W I =.19A 24V R 1 = 42 R 2 = 42 R = 42 3 Figure 16. Total Power Calculation 17

18 2. Calculate the power used by the circuit shown in figure 17. P = (Watts) Since the current and resistance are given, we can use the I 2 R formula as follows: P = I 2 R P = (2 2 ) 10 P = 40 watts V I = 2A 5 5 Figure 17. Total Power Calculation 3. Calculate the power used by the circuit shown in figure 18. P = (Watts) The answer is 48 Watts. 24V I = 2A R Figure 18. Total Power Calculation 18

19 4. Calculate the power used by the circuit shown in figure 19. P T = (Watts) The answer is 108 Watts. V I=3A 2 10 Figure 19. Total Power Calculation 5. Calculate the power used by the circuit shown in figure 20 using the total resistance. P T = (Watts) The answer is 24 Watts. I= 2A R = 2 1 R = 4 2 Figure 20. Total Power Calculation 19

20 6. Perform the following substeps to calculate the total power used in the circuit shown in figure 20 by calculating the individual power used by each component. Figure 20 shows a circuit with the current and each resistance given. The source voltage is not known. A. Calculate the power dissipation of R1. P R1 = I 2 R 1 P R1 = P R1 = (Watts) The power used by R 1 is 8 watts. B. Repeat substep A for R 2. P R2 = (Watts) The power dissipation of R 2 is 16 watts. C. Calculate the total power dissipation by adding the individual power dissipations. P T = (Watts) The answer is found as follows: P T = P R1 P R2 P T = 8W 16W The total power used is 24 watts. This should be the same as you calculated in step Calculate the total power used by the circuit shown in figure 21. P T = (Watts) The answer is 180 Watts. I = 1.5A R = 47 1 R = 33 2 Figure 21. Total Power Calculation 20

21 SEGMENT 1 SELF REVIEW 1. Law states that one volt can push one amp through one ohm of resistance. 2. Voltage Law states that the total voltage in a series circuit is equal to the sum of the individual voltage drops in the circuit. 3. The mathematical statement of Ohm s Law is. 4. One way to calculate total power used in a series circuit is to multiply the total voltage and total. 5. The total in a circuit determines how much power the circuit will draw. 21

22 SEGMENT 2 POWER IN PARALLEL CIRCUITS OBJECTIVE 6 STATE KIRCHHOFF S CURRENT LAW AND GIVE AN APPLICATION Another of Kirchhoff s electrical laws is Kirchhoff s Current Law, which applies to parallel circuits. Kirchhoff s Current Law states that the amount of current flowing from the power supply is equal to the current flowing back to the power supply. In a parallel circuit, the total current that flows from and back to the power supply is called the main line current. As shown in figure 22, the main line current splits at node 1 with part of it flowing through each branch. These branch currents then recombine at node 2 to form the main line current again. MAIN LINE CURRENT (I T ) NODE 1 BRANCH CURRENT BRANCH CURRENT MAIN LINE CURRENT (I T ) NODE 2 Figure 22. Main Line and Branch Currents 22

23 From Kirchhoff s Current Law, we can deduce that the main line current will be equal to the sum of the currents in the branches. This is stated mathematically as: KIRCHHOFF S CURRENT LAW FORMULA I T = I R1 I R2 I R3... where I T = total current (Amps) I R1, I R2, I R3 = current of each branch (Amps) As an example, in figure 23, the main line current is 6 amps. The branch currents are 2 and 4 amps, which adds up to a total of 6 amps. This differs from series circuits where the current is the same at every point in the circuit. Kirchhoff s Current Law also applies to AC circuits. MAIN LINE = 6A V R 1 BRANCH A = 2A R 2 BRANCH B =4A Figure 23. Kirchhoff s Current Law One of the common applications of Kirchhoff s Current Law is in house wiring systems. As appliances and devices are added to the circuit, the main line current is increased because all of the items are in parallel. A special circuit protection device called a circuit breaker is installed in the main line to prevent the main line current from exceeding a certain level. Knowing Kirchhoff s Current Law will help to determine how many devices can be added to a circuit without exceeding the limit. 23

24 SKILL 4 CALCULATE THE MAIN LINE CURRENT IN A PARALLEL CIRCUIT Procedure Overview In this procedure, you will calculate the total current in a parallel circuit using Kirchhoff s current law. You will then connect a circuit and measure the current in each branch to verify your calculations. 1. Use Ohm s Law to calculate the current in each branch of the circuit in figure 24. Current I 1 = (Amps) Current I 2 = (Amps) Current I 3 = (Amps) They should be I 1 = 1.2A, I 2 =.48A, and I 3 =.48A. I T I 1 I 2 I 3 12V R = 10 R 2 = 25 1 R = 25 3 Figure 24. Main Line Current Calculation 2. Calculate the main line current using the branch currents you calculated in step 1 and Kirchhoff s Current Law. I T = I 1 I 2 I 3 Main line current = (Amps) It should be 2.16A. 24

25 BATT 1.5V BATT 200mA MAX FUSED 10A MAX FUSED NON CONTACT VOLTAGE MAX 600V 600V 3. Connect the circuit shown in figure Prepare the DMM to measure DC current. Be sure you have the test leads connected to the correct terminals of the DMM to measure current. 5. Place the AC/DC selector switch on the power supply in the DC position. 6. Perform the following substeps to measure the current in each branch of the circuit, as shown in figure 25. A. Connect the circuit shown in figure 25 with the meter connected to measure current in branch 1. Remember, the meter must be in series with the load. SCHEMATIC SOURCE SELECT I T I 1 I 2 I 3 AC 12V 24V 12V DC 12V A A A SELECTOR SWITCH MODULE MIN MAX V 600 OFF m 30XR HOLD V m 200 NOTE: THE METER IS SHOWN CONNECTED IN BRANCH 1 ONLY. 20M 2m 2M 20m 200k 200m 20k 10 A 2k 10 A 200 A 200m 1.5V 9V 20m 200 2m ma CAT 600V A CAT 300V COM BATT 9V V 10A 25 RESISTOR OHM MODULE 25 RESISTOR OHM MODULE 10 RESISTOR OHM MODULE Figure 25. Parallel Circuit Current Measurement B. Turn on the power and measure the current in branch 1. Current I 1 = (Amps) It should be approximately the same as the value you calculated in step 1. C. Turn off the power supply. D. Move the meter to branch 2, as shown in the schematic of figure

26 E. Turn on the power supply and measure current in branch 2. Current I 2 = (Amps) F. Turn off the power supply. G. Move the meter to branch 3 as shown in the schematic of figure 25. H. Turn on the power supply and measure the current in branch 3. Current I 3 = (Amps) It should be approximately the same as the values you calculated in step 1. I. Turn off the power supply. 7. Place the DMM in the circuit as shown in figure 26 to measure the main line current. 8. Turn on the power supply and record the current. Main line current = (Amps) The main line current should approximately match the calculations you made in step 2. NOTE There may be some variance in the measured values and the calculated values depending on how close the actual resistance of each resistor is to the theoretical value. The variance can be as much as or - 10%. A IT I 1 I 2 I 3 12V R =10 1 R2 R3 =25 =25 I T Figure 26. Main Line Current Measurement 9. Turn off the power supply. 10. Disconnect the circuit and store the components. 26

27 OBJECTIVE 7 STATE A FORMULA FOR CALCULATING TOTAL PARALLEL RESISTANCE The calculation of total resistance in a parallel circuit is different than in a series circuit where the total resistance is equal to the sum of the resistances. The formula used to calculate resistance in a parallel circuit is: TOTAL PARALLEL RESISTANCE FORMULA 1 RT = R R R where R T = total resistance in ohms R 1, R 2, R 3 = individual resistances in ohms The result of this formula is that the total resistance of the parallel circuit actually decreases as you add more resistors in parallel. You will see this demonstrated in the skill that follows. This is just the opposite of a series circuit. R T R 1 R 2 R 3 Figure 27. Resistance in a Parallel Circuit 27

28 SKILL 5 CALCULATE THE TOTAL PARALLEL RESISTANCE Procedure Overview In this procedure, you will calculate the total resistance of the circuit using the parallel resistance formula. In the first step you will be given an example. 1. Calculate the total resistance of the circuit in figure 28. R T = (Ohms) The solution is found as follows: R R R R R T T T T T 1 = R R R 1 = = = 002. = 50Ω The total resistance is 50 ohms. I = 0.24A T 12V R= 1 R= 2 R= Figure 28. Total Resistance Calculation 28

29 NOTE Each time another load is added to a parallel circuit, the effective resistance is lowered. This is exactly the opposite effect of adding a load to a series circuit, which would increase the effective resistance of that circuit. This is the reason the main line current increases as more branches are added. Since the total resistance is reduced, Ohm s Law says that the current must increase since the voltage is constant. 2. Calculate the total resistance of the parallel circuit shown in figure 29. R T = (Ohms) The total resistance should be 5.56 ohms. Notice that this is lower than any one of the three individual resistances. 12V R 1 = 10 R 2 = 25 R 3 = 25 Figure 29. Total Resistance Calculation 29

30 3. Now connect the circuit shown in figure 29. Do not connect the circuit to the power supply. Any time you make resistance measurements in a circuit, you should disconnect the circuit from the power supply. 4. Measure the resistance of each branch of the circuit with the DMM. Remember to disconnect one side of the load in that branch from the circuit before measuring. Reconnect each load after you make your measurement. R 1 = (Ohms) R 2 = (Ohms) R 3 = (Ohms) The resistances should be R 1 = 10 ohms, R 2 = 25 ohms, and R 3 = 25 ohms. 5. Measure the total resistance of the circuit, as shown in figure 30. Total resistance = (Ohms) It should be approximately 5.56 ohms, the same value you calculated earlier. This shows that the formula for calculating parallel resistance works. DISCONNECTED SUPPLY R T R=10 1 R=25 2 R=25 3 Figure 30. Total Resistance Measurement 6. Disconnect the circuit and store all components. 30

31 SKILL 6 CALCULATE THE TOTAL POWER USED IN A PARALLEL CIRCUIT Procedure Overview In this procedure, you will calculate the total power used in a circuit. This calculation uses the same principles used to calculate the power in a series circuit except that the calculation of total resistance is different. In steps 1 and 2 you will be given some examples. Then you will do it yourself. 1. Calculate the total power used in the circuit shown in figure 31. The power in a parallel circuit can be calculated using any of the three methods used for series circuits: Method A - Calculate total resistance of the parallel circuit and use the I 2 R formula. Method B - Determine the total voltage and current and use the I E formula Method C - Calculate the power consumed by each component and add them together, as follows: P T = P 1 P 2 P 3... I = 3 AMPS E = 24V R 1 = R 2 = R = 3 Figure 31. Total Power Used by a Parallel Circuit P T = (Watts) The total power used by this circuit is 72 watts. Since the circuit in figure 33 shows the total voltage and current, method B (I E) can be used as follows: P T = I T E T P T = 3 24 P T = 72 watts 31

32 2. Perform the following substeps to calculate the total power used in the circuit shown in figure 32. I = 2A R 1 = 10 R 2 = 25 Figure 32. Total Power Calculation A. Calculate the total resistance, R T. 1 R = T 1 1 R R 1 2 R T = Ohms The total resistance is approximately 7.14 ohms. B. Calculate the total power. P T = I 2 R T P T = (Watts) Your answer should be 28.6 watts. 32

33 3. Perform the following substeps to calculate the total power used in the circuit shown in figure 33. Use method C to solve it. MAXIMUM POWER =80W 20V R 1 = R 2 = R 3 = Figure 33. Total Power Calculation A. Calculate the current in each branch of the circuit. I R1 = (Amps) I R2 = (Amps) I R3 = (Amps) The current I R1 = 0.4A. The current I R2 = 1A. The current I R3 = 2A. B. Now calculate the power dissipation of each load using the basic power formula (P = I E). P 1 = (Watts) P 2 = (Watts) P 3 = (Watts) They should be P R1 = 8W, P R2 = 20W, and P R3 = 40W. Since this is a parallel circuit, the voltage across each branch is equal to the source voltage. C. Calculate the total power used by the circuit. P T = P 1 P 2 P 3 P T = (Watts) The total power used should be 68W. 33

34 4. Calculate the power used by the circuit shown in figure 34. V = 24V 10W 25W Figure 34. Total Power Calculation P T = (Watts) Your answer should be 35 watts. 5. Calculate the power used by the circuit shown in figure I = 2A Figure 35. Total Power Calculation P T = (Watts) Your answer should be 20 Watts. 6. Solve the following design problem. Determine how many 100W light bulbs can be connected to a common commercial building s circuit if the source voltage is 120 VAC and the maximum current that can be provided is 20A. HINT: What is the maximum output power that the source can provide with the 20A limit? 34

35 SEGMENT 2 SELF REVIEW 1. The current is the total current in a parallel circuit. 2. Kirchhoff s Current Law says that the amount of current flowing from the source is always to the current flowing back to the source. 3. The total resistance of a parallel circuit (increases/ decreases) as more resistors are added in parallel. 4. Each time another branch is added to a circuit, it will (increase/decrease) the load on the power supply. 5. Kirchhoff s Current Law says that the main line current will be equal to the of the currents in the branches. 35

36 SEGMENT 3 CIRCUIT PROTECTION DEVICES OBJECTIVE 8 DESCRIBE THE FUNCTION OF TWO TYPES OF CIRCUIT PROTECTION AND GIVE AN APPLICATION OF EACH High current can damage electrical components. This condition can occur when too many loads are connected to the circuit or if a short circuit (short) occurs. A short circuit occurs when there is a direct path with little or no resistance created between the positive and negative terminals of the power supply, and therefore potentially high, damaging current. There are two types of devices commonly used to protect electrical components from high current: Fuse Circuit Breaker Both of these devices interrupt the flow of current. The fuse is a low cost device that must be replaced each time an overload or short circuit condition occurs. The circuit breaker can be reset. Figure 36. Fuse (Left) and Circuit Breaker (Right) Fuses are used in applications where a problem rarely occurs. A car s light system is an example. Circuit breakers are used where overloads commonly occur. A power supply and your house wiring are two examples. 36

37 OBJECTIVE 9 DESCRIBE THE OPERATION OF A FUSE AND GIVE ITS SCHEMATIC SYMBOL A fuse is a low cost protection device that is placed in series with an electrical circuit to protect the power supply and the components from damage due to excess current flow. As shown in figure 37, a fuse consists of a conductive wire or metal foil strip encased in a glass tube. When the current flow exceeds the rated value of the fuse, the wire or foil strip melts and opens the circuit (the fuse is blown). If a fuse is good, it has continuity just like an unbroken wire. However, if a fuse is blown, there is no longer continuity. Figure 37 also shows the schematic symbol for a fuse. SCHEMATIC SYMBOL GLASS TUBE ELEMENT FUSE WIRE METAL CAP Figure 37. A Fuse and Its Schematic Symbol Many electrical devices have fuses. They are one of the first things a technician checks when troubleshooting a circuit. Fuses are easy to replace. Fuses are rated for a maximum current value. If for some reason that maximum value is exceeded, the fuse blows to protect the components of the circuit. Special care needs to be taken to make sure that the properly rated fuse is installed to provide the needed protection. 37

38 SKILL 7 OPERATE A CIRCUIT USING A FUSE Procedure Overview In this procedure, you will operate a circuit using a fuse for circuit protection. 1. Connect the circuit as shown in figure 38. SCHEMATIC SOURCE SELECT AC 24V 12V 12V DC 12 VAC FUSE PUSHBUTTON SWITCH LAMP FUSE MODULE PUSH BUTTON SWITCH MODULE LAMP MODULE Figure 38. A Fuse in a Circuit 2. Perform the following substeps to operate the circuit. A. Place the AC-DC selector switch in the AC position. B. Turn on the power supply. C. Press and hold the pushbutton switch. Lamp status (On/Off) The lamp should come on. 38

39 D. Release the pushbutton switch. Lamp status (On/Off) The lamp should go off. Since the lamp did come on when the pushbutton switch was pressed, we know that the fuse did not blow. If the fuse was blown, the lamp would not have come on when the pushbutton switch was pressed. E. Turn off the power supply. Leave the circuit connected and continue to Skill 8. SKILL 8 TEST AND REPLACE A FUSE NOTE This procedure is optional. Consult your instructor to determine whether you should perform this procedure or not. Procedure Overview In this procedure, you will test a fuse to determine if it has blown and replace it with the proper rated fuse if necessary. 1. Make sure the power supply is off. 2. Use a wire to short out the lamp, as shown in figure 39. LAMP MODULE 12 VAC FUSE SHORTING WITH A WIRE Figure 39. Shorting Out the Lamp 39

40 3. Turn on the power supply. 4. Push the pushbutton switch. Lamp status (On/Off) The light should not come on because the fuse is blown. The fuse is blown because of the excessive current caused by the short circuit. 5. Release the pushbutton switch. 6. Turn off the power supply. 7. Check the fuse for continuity using the continuity tester of the DMM. Fuse status (Continuity/No Continuity) You should not hear a beep because the fuse does not have continuity. It is blown. 8. Perform the following substeps to replace the blown fuse, as shown in figure 40. A. Pry up on the end of the fuse using the tip of one of the test leads until the end is completely free of the clip. This is shown in step A of figure 40. B. Pull the other end out of the clip as shown in step B of figure 40. C. Place a new fuse into the clip as shown in step C of figure 40 and press down until the fuse is secure. You should hear a snapping sound when the fuse is clipped. STEP A STEP B STEP C FUSE TEST LEAD PULL UP PRESS DOWN FUSE CLIPS Figure 40. Removing and Replacing a Fuse 9. Remove the wire shorting out the lamp. 10. Turn on the power supply. 11. Operate the pushbutton switch again. Lamp status (On/Off) The light should come on and stay on. You have now successfully replaced the fuse. 40

41 12. Release the pushbutton switch. 13. Perform the following substeps to turn OFF and secure the power supply. A. Turn the power supply off. B. Disconnect any wires or components connected to the output terminals of the power supply and store them. OBJECTIVE 10 DESCRIBE THE OPERATION OF TWO TYPES OF CIRCUIT BREAKERS AND GIVE THEIR SCHEMATIC SYMBOLS A circuit breaker performs the same protective service as a fuse. However, it can be reset and used again. Just as a fuse blows, a circuit breaker opens or trips when an excessive current is present. Most circuit breakers are either thermally triggered (tripped due to heat caused by excessive current) or magnetically triggered (tripped due to the strength of the magnetic field created by excessive current). In some cases a circuit breaker is a combination of both. The thermally triggered type has a lag time (delay) before it trips because the temperature must increase enough to trip the breaker. The magnetically triggered type trips immediately when a surge of excessive current is present. However, a slow increase in current may not cause the magnetically triggered breaker to trip. This is why it is better to have a circuit breaker that uses both types. Circuit breakers can have different types of reset switches, as shown in figure 41. Some have lever type resets, while some have pushbutton resets. You will be using a circuit breaker with a lever reset. Most newer homes and businesses use circuit breakers in their electrical control panels. Figure 41 also shows a typical circuit breaker panel you would see in a home or business. ON TRIPPED OFF RESET PUSHBUTTON RESET LEVER TYPE RESET CIRCUIT BREAKER PANEL BOX Figure 41. Different Styles of Circuit Breaker Resets and a Typical Circuit Breaker Panel 41

42 Figure 42 shows the schematic symbols for these circuit breakers. MAGNETIC THERMAL THERMAL-MAGNETIC Figure 42. Schematic Symbols for Circuit Breakers SKILL 9 OPERATE A CIRCUIT USING A CIRCUIT BREAKER Procedure Overview In this procedure, you will connect and operate a circuit with a circuit breaker as a protection device. 1. Connect the circuit shown in figure 43. SCHEMATIC SOURCE SELECT AC DC 24V 12V 24V 12V CIRCUIT BREAKER 25 CIRCUIT BREAKER MODULE SWITCH MODULE 25 RESISTOR OHM MODULE Figure 43. A Circuit with a Circuit Breaker 42

43 2. Perform the following substeps to operate the circuit. A. Place the AC-DC selector switch in the DC position. B. Turn on the power supply. C. Energize (close) the knife switch and leave it closed. Wait for about 30 seconds to see if the circuit breaker trips off. Circuit breaker status (Tripped/Not Tripped) The circuit breaker should not trip because the 25 ohm resistor only draws a current of 0.96A (I = E/R). This is below the 1 amp rating of the circuit breaker. D. Open the knife switch. 3. Turn off the power supply. 4. Replace the 25 ohm resistor with a 10 ohm resistor. 5. Repeat step 2 and observe the circuit breaker s operation. Circuit breaker status (Tripped/Not Tripped) The circuit breaker should now trip immediately. Now the current is 2.4 amps. Over twice the rating of the circuit breaker. 6. Turn off the power supply and open the knife switch. Leave the circuit breaker tripped and continue to Skill

44 SKILL 10 TEST AND RESET A CIRCUIT BREAKER Procedure Overview In this procedure, you will check a circuit breaker to see if it has tripped by testing its continuity. You will also correct the problem that made the circuit breaker trip and then reset the circuit breaker. This procedure is one that you might perform at home if one of the circuit breakers in your circuit breaker panel tripped. 1. Make sure the T7017 power supply is off. 2. Prepare the DMM to measure continuity using the continuity function. 3. Test the circuit breaker for continuity by measuring across it s terminals. Circuit breaker status (Tripped/Not Tripped) You should not hear a beep because there is no continuity. The breaker has been tripped. 4. Replace the 10 ohm resistor with a 25 ohm resistor. 5. Reset the circuit breaker by pushing the lever up into the ON position. 6. Retest the circuit breaker for continuity. Circuit breaker status (Tripped/Not Tripped) You should now find that there is continuity, which means that the circuit breaker is not tripped. 7. Turn on the power supply. 8. Energize (close) the knife switch and leave it closed. Wait to see if the circuit breaker trips off. Circuit breaker status (Tripped/Not Tripped) The circuit breaker should not trip. Once again the current is below the rating of the circuit breaker. 9. Open the knife switch. 10. Perform the following substeps to turn off and secure the power supply. A. Turn off the power supply. B. Unplug the power cord from the wall outlet. C. Disconnect any wires that may be connected to the power supply output terminals. 11. Turn off the DMM, remove the test leads and store them. 44

45 SEGMENT 3 SELF REVIEW 1. A(n) consists of a conductive wire or metal foil strip encased in a glass tube. 2. When the current flow exceeds the rated value of the fuse, the fuse will. 3. If a fuse is blown, there is no longer. 4. The difference between a fuse and a circuit breaker is that a circuit breaker can be and used again, while a fuse cannot. 5. A thermally-triggered circuit breaker will have a delay time before it trips because the must build up. 6. A magnetically-triggered circuit breaker will trip immediately when a(n) of excessive current is present. 7. Most newer homes and businesses use in their electrical control panels. 8. A(n) occurs when there is a direct path with little or no resistance created between the positive and negative terminals of a power supply. 45