Circuit Analysis and Ohm s Law

Study Unit Circuit Analysis and Ohm s Law By Robert Cecci

Circuit analysis is one of the fundamental jobs of an electrician or electronics technician With the knowledge of how voltage, current, and resistance are related in a circuit, quick and efficient troubleshooting of any circuit can easily be performed We ll begin this study unit with an examination of circuit resistance Next, Ohm s law will be used to identify the amount of current, voltage, or resistance that s in a circuit Then we ll conclude with a study of power and how to take basic meter readings in circuits When you complete this study unit, you ll be able to Find the total resistance in series, parallel, and series-parallel circuits Use Ohm s law to calculate the amount of current, voltage, or resistance in circuits Calculate the amount of power supplied and dissipated in a DC circuit List the steps for reading current, voltage, and resistance with a meter Electronics Workbench is a registered trade mark, property of Interactive Image Technologies Ltd and used with permission You ll see the symbol shown above at several locations throughout this study unit This symbol is the logo of Electronics Workbench, a computer-simulated electronics laboratory The appearance of this symbol in the text margin signals that there is an Electronics Work-bench lab experiment associated with that section of the text If your program includes Electronics Workbench as a part of your learning experience, you ll receive an experiment lab book that describes your Electronics Workbench assignments When you see the symbol in the margin of your text, follow the accompanying instructions in the lab book to complete your Electronics Workbench assignment If your program doesn t include Electronics Workbench, you may simply ignore the symbol Preview iii

CALCULATING CIRCUIT RESISTANCE 1 Introduction 1 Circuit Basics 3 Calculating Resistance in Series Circuits 4 Calculating Resistance in Parallel Circuits 5 Calculating Resistance in Series-Parallel Circuits 9 Practical Example 10 OHM S LAW 14 What Is Ohm s Law? 14 Variations of the Ohm s Law Formula 15 Practical Applications of Ohm s Law 18 Using Ohm s Law with Series Circuits 21 Practical Example 25 Using Ohm s Law with Basic Parallel Circuits 27 Using Ohm s Law with Complex Parallel Circuits 29 Using Ohm s Law with Series-Parallel Circuits 31 Practical Example 35 POWER IN DC CIRCUITS 41 The Power Formula 41 Power in Series Circuits 43 Power in Parallel Circuits 48 Power in Series-Parallel Circuits 52 Practical Example 54 MEASURING CURRENT, VOLTAGE, AND RESISTANCE 57 Analog and Digital Meters 57 Measuring Current 58 Measuring Voltage 60 Measuring Resistance 61 SELF-CHECK ANSWERS 65 Contents v

Circuit Analysis and Ohm s Law CALCULATING CIRCUIT RESISTANCE Introduction Circuit analysis is one of the most important tasks of an electrician or electronics technician In order to troubleshoot circuits and electrical devices, you must clearly understand how voltage, current, and resistance are related in circuits With this knowledge, you can perform calculations that will provide you with important information about a given circuit Then, you ll be able to quickly and easily locate the source of trouble in a faulty circuit Let s take a moment now to review some important topics Remem-ber that current is the flow of electric charge in a circuit The amount of electric current flowing through a circuit is called amperage One ampere of current is equal to the charge of 6,240,000,000,000,000,000 electrons flowing past a given point in a circuit per second Note that in this definition of ampere, it doesn t matter which direction the electrons are flowing in There are two theories of electric current flow that are used to describe the movement of electrons in a circuit You should be aware of both theories, since you ll see both of them referred to in textbooks and technical manuals In the electron theory of current flow, electrons are said to flow from the negative point of a circuit to the positive point In 1

contrast, in the conventional theory of current flow, current is said to flow from the positive point of a circuit to the negative point Why are there two theories of current flow, and which is correct? The answer is that both theories are used in the study and application of electricity The conventional theory is older and is based on the assumption that electric current flows from the positive point of a circuit to the negative point In later years, scientists determined that electrons actually flow from the negative point of a circuit to the positive Thus, the newer electron theory is the correct way to explain current flow However, the conventional theory is still used in some circles One reason for this is that many of the standard rules of electricity were developed by scientists who used the conventional theory Some reference books, manuals, and text books still refer to the conventional theory, so you should be familar with it The conventional theory is also used to explain the rules of magnetism and the electrical forces produced in motors and generators One good example of the conventional theory in practice is an everyday car battery In a car battery, the positive battery terminal is considered to be the hot terminal and is therefore painted red According to the electron theory, the negative terminal is actually the source of electron flow However, it would be too confusing to change the standards of battery production to adapt to a new theory of current flow! Therefore, the conventional theory continues to be used when referring to batteries For most practical purposes, all you need to know is that current flows out of a power source, flows through a circuit load, and then returns to the power source It doesn t really matter in which direction the current flows However, you should be aware of the two theories of current flow and what they mean For our purposes in this study unit, when we talk about circuits, we ll use the electron theory to describe current flow In addition, all our electrical diagrams are drawn using the electron or negative-to-positive standard This is because most of the actual circuit diagrams you ll see in your work are drawn according to the electron theory 2 Circuit Analysis and Ohm s Law

Circuit Basics Most DC circuits contain three basic parts: a power source, conductors, and a resistance Resistance in a circuit may come from heaters, motors, relay coils, lights, or any other electrical devices However, for instructional purposes, all the circuit examples in this text will contain resistors Now, let s briefly review the difference between a series and a parallel circuit In a series circuit, components are connected end-to-end A typical series circuit is shown in Figure 1 In this circuit, three resistors (R 1, R 2, and R 3 ) are connected end-to-end, and power is supplied by a battery Current leaving the negative battery terminal will flow through R 1, R 2, and R 3, and back to the positive battery terminal Because the components in this circuit are connected end-to-end, if any connection between the components were to break or open, the entire circuit would stop functioning In contrast, a parallel circuit is connected much differently In a parallel circuit (Figure 2), the components are connected across each other Current leaves the negative battery terminal and then splits into three different circuit paths Some of the current flows through R 1, some through R 2, and the remainder through R 3 The current then flows through the conductors back to the positive battery terminal In a parallel circuit, if one circuit path is broken or opened, current continues to flow through the other paths, and part of the circuit keeps working FIGURE 1 A typical series circuit has its components connected end-to-end All of the circuit current flows through each of the circuit components In a parallel circuit, the amount of current that flows through each resistor is related to the values of the resistors If all three resistors have exactly the same value, then the current through each resistor will be the same If one of the resistors is higher in value than the other two resistors, less current will flow through that resistor FIGURE 2 The current in this parallel circuit (I) flows from the battery and then separates onto three circuit paths (I 1, I 2 and I 3 ) Circuit Analysis and Ohm s Law 3

Calculating Resistance in Series Circuits Series circuits are very simple circuits Therefore, the process of calculating the total resistance of a series circuit is very simple To calculate the total resistance of a series circuit, simply add the values of all the resistors in the circuit Let s begin by looking at a simple tworesistor circuit (Figure 3) In this figure, a 300 (ohm) resistor (R 1 ) and a 500 resistor (R 2 ) are connected in series The total value of resistance ( ) is equal to the sum of the resistors The resistance is calculated as follows: FIGURE 3 To calculate the total resistance in this series circuit, simply add the values of the two resistors = R 1 + R 2 =300 + 500 = 800 Set up the problem Add the values of R 1 and R 2 (300 = 500 + 800) Answer: The total resistance is 800 This same addition rule applies no matter how many resistors are connected in series For example, look at the circuit in Figure 4 Here, five resistors are connected in series The total resistance in this circuit would be calculated as follows: = R 1 + R 2 + R 3 + R 4 + R 5 Set up the problem = 1,000 + 2,000 + Add the resistor values (1,000 + 100 + 100 + 60 2,000 + 100 + 100 + 60 = 3,260) = 3,260 Answer: The total resistance is 3,260 In summary, note that in a series circuit, the total circuit resistance will always be greater than the value of any one resistor 4 Circuit Analysis and Ohm s Law

FIGURE 4 In this five-resistor circuit, the total resistance can be found by adding the values of the five resistors Calculating Resistance in Parallel Circuits It s a little more difficult to calculate the total resistance in a parallel circuit There are several ways in which you can calculate resistance in parallel circuits, and you should be familiar with all of them The method you use in a particular situation will depend on the values of the resistors in the given circuit We ll learn each of these methods by looking at some specific circuit examples First, if a parallel circuit contains resistors of equal value, the total circuit resistance will be equal to a fraction of the value of one resistor So, if a parallel circuit contains two equal resistors, then the total circuit resistance is equal to one-half the value of either resistor If the circuit contains three equal resistors, the total circuit resistance is equal to one-third of the value of one resistor If the circuit contains four equal resistors, the total circuit resistance is equal to one-fourth the value of one resistor, and so on Figure 5 shows a circuit containing two 1,000 resistors connected in parallel The total resistance in this circuit is equal to one-half of 1,000 (that is, 1,000 2) Therefore, the total circuit resistance is 500 FIGURE 5 Since both parallel resistors are of the same value, the total resistance is equal to one-half the value of either resistor Circuit Analysis and Ohm s Law 5

In another example, if a circuit contains four 1,000 resistors connected in parallel, the total circuit resistance is equal to onefourth of 1,000 (1,000 4) Therefore, the total circuit resistance is 250 FIGURE 6 This parallel circuit contains two resistors with different values You must use the formula shown to calculate the total resistance Now, let s look at a different type of problem If two resistors with different values are connected in parallel, you ll need to use a different method to calculate the total circuit resistance To find the total resistance in such a circuit, use the following formula: Figure 6 shows a 60 resistor (R 1 )and a 20 resistor (R 2 ) connected in parallel Let s find the total resistance in this circuit Write the formula Substitute 60 for R 1 and 20 for R 2 Multiply (60 20 = 1,200) Add (60 + 20 = 80) Divide (1,200 80 = 15) = 15 Answer: The total resistance of this circuit is 15 Note that the total resistance of a parallel circuit is always lower than the lowest value of any one resistor in the circuit Now, let s look at another example problem Suppose a circuit contains a 1,000 resistor (R 1 )and a 10 resistor (R 2 ) connected in parallel Use the formula to calculate the total resistance of the circuit Write the formula 6 Circuit Analysis and Ohm s Law

Substitute 1,000 for R 1 and 10 for R 2 Multiply (1,000 10 = 10,000) Add (1,000 + 10 = 1,010) Divide (10,000 1,010 = 99) = 99 Answer: The total circuit resistance is 99 (rounded) Note that the resistor with the lowest value in this circuit was R 2 (10 ), and the total resistance of the circuit is less than 10 In this circuit, most of the current will flow through R 2 since it has such a low resistance value Very little current will flow through R 1 since its resistance is so large compared to that of R 2 Now, let s look at a parallel circuit containing three resistors with different values (Figure 7) There are two methods of calculating the total resistance in such a circuit As shown, the values of the three resistors are 100, 200, and 300 To calculate the total resistance, first use the formula to calculate the total resistance of R 1 and R 2 Then, use that value and the value of R 3 to calculate the total circuit resistance FIGURE 7 This circuit contains these resistors connected in parallel Write the formula The abbreviation R P1 stands for the total resistance of the first two resistors Substitute 100 for R 1 and 200 for R 2 Multiply (100 200 = 20,000) Add (100 + 200 = 300) Divide (20,000 300 = 6667) Circuit Analysis and Ohm s Law 7

R P1 = 6667 Answer: The total resistance of the first two resistors is 6667 (rounded) We can now alter Figure 7 to include this value (6667 ) in place of R 1 and R 2 Figure 8 shows the updated circuit diagram The circuit now contains two resistors connected in parallel Now, use the value 6667 and the value of the third resistor to find the total circuit resistance Write the formula FIGURE 8 In this updated version of the circuit diagram in Figure 7, R P1 is used in place of R 1 and R 2 Substitute 6667 for R P1 and 300 for R 3 Multiply (6667 300 = 20,001) Add (6667 + 300 = 36667) Divide (20,001 36667 = 5455) = 5455 Answer: The total resistance of this circuit is 5455 (rounded) Note that the total resistance of this circuit (5455 ) is less than the lowest value of any one resistor (100 ) Another method you can use to find the total resistance of a parallel circuit is the reciprocal method Every number has a reciprocal The reciprocal of a given number is one divided by the number So, the reciprocal of 5 is 1 / 5 The reciprocal of 12 is 1 / 12 To find the total resistance of a parallel circuit using the reciprocal method, first find the reciprocal of each resistor value Then, add all the reciprocal values together Finally, find the reciprocal of that value to get the total resistance This isn t as difficult as it sounds! Just use the following reciprocal formula to make your calculations: 8 Circuit Analysis and Ohm s Law

Let s look at an example problem The circuit shown in Figure 9 contains three resistors with values of 20, 40, and 50 Use the reciprocal formula to find the total circuit resistance Write the formula Substitute the values of R 1, R 2, and R 3 Divide to simplify each of the fractions FIGURE 9 This circuit contains three resistors connected in parallel Add (005 + 0025 + 002 = 0095) The reciprocal of 0095 is Divide (1 0095 = 1053) = 1053 Answer: The total resistance of this circuit is 1053 (rounded) The reciprocal method can be used to calculate the total resistance of a circuit no matter how many resistors are connected in parallel Calculating Resistance in Series- Parallel Circuits When you re working with real-life circuits, you ll find that they re not always neatly laid out in series or parallel Often, circuits contain both series and parallel connections This type of circuit is called a combination or series-parallel circuit The circuit in Figure 10 contains a total of four resistors joined in both series and parallel connections Circuit Analysis and Ohm s Law 9

Now, let s calculate the total resistance of the circuit shown in Figure 10 The first step is to find the resistance of the parallel portion of the circuit This value can easily be found by using the resistance formula Write the formula Substitute the values of R 2 and R 3 FIGURE 10 This circuit contains four resistors connected in both series and parallel arrangements Multiply (20 40 = 800) Add (20 + 40 = 60) Divide (800 60 = 1333) R P = 1333 Answer: The total resistance of the parallel section of this circuit is 1333 (rounded) Now, let s substitute this value for the parallel section in our circuit diagram The revised diagram is shown in Figure 11 With this substitution, the circuit diagram in Figure 11 is now a simple series circuit Since you now have a series circuit, your calculations are easy to complete Simply add the remaining resistor values together to get the total resistance for this circuit = R 1 + R P + R 4 Set up the problem = 10 + 1333 + 30 Substitute the values for R 1, R P, and R 4 Add (10 + 1333 + 30 = 5333) FIGURE 11 In this updated version of Figure 10, the values of R 2 and R 3 have been replaced by the value R P = 5333 Answer: The total resistance of this circuit is 5333 10 Circuit Analysis and Ohm s Law

Practical Example The ability to calculate total circuit resistance in series and parallel circuits is an important skill for the industrial electrician or electronics technician This knowledge can often be used to troubleshoot a faulty machine quickly and efficiently For example, suppose you re working on a press that has two solenoid valves connected in parallel The valves are activated by a push-button When the machine is functioning properly, the operator presses the button, and both solenoid coils are energized One coil lifts a shield, and the other retracts a locator The solenoid coils each have a resistance value of 96 and are connected in parallel The circuit diagram for this machine is shown in Figure 12 Now, suppose that the machine is malfunctioning When the operator presses the button, the locator retracts, but the shield doesn t lift This problem could be the result of an open solenoid coil or a stuck solenoid valve How can you figure out what the problem is? FIGURE 12 In this circuit, two solenoid coils are connected in parallel To troubleshoot this machine, use a meter to measure the resistance of the two parallel coils Since the two solenoids have the same resistance value, the total circuit resistance should be one-half the resistance value of one solenoid coil Divide the value of one resistance by 2 = 48 Answer: The total circuit resistance is 48 The total circuit resistance is 48, so if you read 48 on your meter, the solenoid valve is stuck If you read 96, one of the two coils, the one for the shield, has opened causing the failure of the solenoid valve Now, take a few moments to review what you ve learned by completing Self-Check 1 Circuit Analysis and Ohm s Law 11

Self-Check 1 At the end of each section of Circuit Analysis and Ohm s Law, you ll be asked to pause and check your understanding of what you ve just read by completing a Self-Check exercise Answering these questions will help you review what you ve studied so far Please complete Self-Check 1 now 1 What is the total resistance of a circuit that contains three 30 V resistors connected in series? 2 A circuit contains two 90 V resistors connected in parallel What is the total circuit resistance? 3 What is the total resistance of the group of three parallel resistors shown below? (Continued) 12 Circuit Analysis and Ohm s Law

Self-Check 1 4 What is the total resistance of the series-parallel circuit shown here? Check your answers with those on page 65 Circuit Analysis and Ohm s Law 13

OHM S LAW What Is Ohm s Law? In a circuit, resistance, current, and voltage have a special relationship to one another The connection between these units was first defined by George Ohm, and his statement of this relationship is called Ohm s law Ohm s law can be defined by the following formula: In this formula, the letter I stands for current, the letter E stands for voltage, and the letter R stands for resistance The formula states that the current in a resistor is equal to the voltage applied across the resistor divided by the resistance (Current is always given in amps or in the decimal portion of an amp) Note that while Ohm s law uses the letters I and E to stand for current and voltage, some circuit diagrams will show A for current (amperage) and V for volts These values are the same no matter what letter is used to represent them Just like a screwdriver, wrench, wire cutter, or soldering iron, Ohm s law is a tool you can use to identify and repair electric or electronic circuit problems In electrical work, you ll frequently see this formula (and several variations of it) used to make calculations Using Ohm s law, a simple meter, and a little basic math, you can easily determine the voltage, current, or resistance in any circuit You can then use this knowledge to identify the source of trouble in a faulty circuit In your work as an electrician or electronics technician, you ll find many practical uses for Ohm s law Any time you know two circuit quantities (current, voltage, or resistance) the third unknown quantity can easily be found using simple multiplication or division 14 Circuit Analysis and Ohm s Law

Let s look at an example problem Suppose the resistance in a circuit is 10 and the voltage applied across the resistor is 10 V What would the current through the resistor be? To find the value of the current in amperes, substitute the two known values into the formula, then solve Write the Ohm s law formula Substitute 10 V for E and 10 for R Divide (10 10 = 1) I = 1 A Answer: The current in this example problem is 1 A You can see that Ohm s law isn t difficult to use! Let s return to our circuit diagrams and see how Ohm s law can be used to calculate circuit values A basic one-resistor series circuit is shown in Figure 13 The circuit in Figure 13 has a 10 V power supply and a 100 resistor So, we know two of the three circuit quantities: voltage and resistance We need to find the circuit current The current can easily be found by using the Ohm s law formula Substitute the known values into the formula and solve FIGURE 13 Since we know the applied voltage and the resistance of this circuit, we can easily find the circuit current using the Ohm s law formula Write the Ohm s law formula Substitute 10 V for E and 100 for R Divide (10 100 = 01) I = 01 A Answer: The circuit current is 01 A Variations of the Ohm s Law Formula The Ohm s law formula allows you to calculate circuit current when you already know the circuit resistance and the applied voltage Now, suppose you know what the current and Circuit Analysis and Ohm s Law 15

resistance are and you need to calculate the voltage Well, the Ohm s law formula can easily be converted to allow you to find voltage Alter the Ohm s law formula as follows: This is the original formula Multiply both sides of the formula by R This alters the appearance of the formula but doesn t change its value On the right side of the formula, the two R s cancel each other out I R = E E = I R You ve now isolated the E on the right side of the equal sign Turn the formula around so that E is by itself on the left You now have a formula that can be used to find voltage By altering the Ohm s law formula in this way, you can see that voltage is equal to current times resistance Use this variation of the Ohm s law formula whenever you need to find voltage Now, suppose you know the current and voltage in a circuit, and you want to find the resistance The Ohm s law formula can be converted again to allow you to find resistance Alter the formula as follows: E = I R This is the formula for voltage Divide both sides of the formula by I This alters the appearance of the formula but doesn t change its value On the right side of the formula, the two I s cancel each other out You ve now isolated the R on the right side of the equal sign Turn the formula around so that R is by itself on the left This formula can be used to find resistance 16 Circuit Analysis and Ohm s Law

By altering the formula this way, you can now see that resistance is equal to voltage divided by current Use this variation of the Ohm s law formula whenever you need to find resistance Now that we ve shown you how to get these variations of Ohm s law, you won t have to make these conversions every time you need to use a formula Instead, use the simple chart shown in Figure 14 for reference To use this chart, place your finger over the letter that stands for the value you need to find The remaining uncovered letters will tell you what mathematical operation to perform FIGURE 14 You can use this chart to quickly find the Ohm s law formula you need So, for example, if you need to find voltage (E), place your finger over the E on the chart The remaining uncovered letters (I next to R) tell you to multiply I times R To find current (I), place your finger over the I on the chart The remaining letters (E over R) tell you to divide E by R to get current Or, if you need to find resistance (R), cover the R with your finger The remaining letters (E over I) tell you to divide E by I to get resistance Now, let s try a few example problems using variations of the Ohm s law formula Look at the circuit diagram in Figure 15 The voltage across the resistor is 12 V and the current flowing in the resistor is 002 A What is the resistance of R 1? Use the proper variation of the Ohm s law formula to calculate the resistance Substitute the known values into the formula and solve Use this formula to find resistance FIGURE 15 Since we know the applied voltage and the current for this circuit, we can easily find the resistance by using the Ohm s law formula Circuit Analysis and Ohm s Law 17

Substitute 12 V for E and 002 A for I R = 600 Divide (12 002 = 600) Answer: The circuit resistance is 600 Now, look at Figure 16 In this circuit, the current flowing in the resistor is 005 A and the resistance is 300 Use the proper variation of the Ohm s law formula to find the voltage across the resistor Substitute the known values into the formula and solve E = I R Use this formula to find voltage FIGURE 16 Using Ohm s law, you can easily find the voltage of this circuit E = 005 A 300 E = 005 300 Substitute 005 A for I and 300 for R Multiply (005 300 = 15) E = 15 V Answer: The voltage in this circuit is 15 V Practical Applications of Ohm s Law We ve already examined some circuit diagrams and used Ohm s law to calculate circuit quantities Now, let s look at a few real-life examples of how you can use Ohm s law to troubleshoot faulty circuits Suppose the power supply for an industrial system fails at regular intervals The power supply is 24 V and the load on the system is 2 The power supply is rated at 10 A but it needs to be replaced at regular intervals What could be wrong? Let s begin by looking at how much current the power supply should deliver to the system Use the Ohm s law formula Write the formula Substitute 24 V for E and 2 for R Divide (24 2 = 12) I = 12 A Answer: The current in this circuit should be 12 A 18 Circuit Analysis and Ohm s Law

The power supply is delivering 12 A of current to the load resistance when the system first energizes However, the power supply is rated at only 10 A Therefore, the power supply is being overloaded when the system turns on This is probably a result of relays, solenoids, or other loads being added to the system since it was originally built Replacing the power supply with a 15 A or 20 A supply will solve the problem Now, take a few moments to review what you ve learned by completing Self-Check 2 on the following page Circuit Analysis and Ohm s Law 19

Self-Check 2 1 What is the source voltage in a circuit where the resistance is 100 and the current is 2 A? 2 How much resistance is in a circuit with a voltage of 24 V and a current of 2 A? 3 How much current flows in a circuit with a voltage of 30 V and a resistance of 90? Check your answers with those on page 66 20 Circuit Analysis and Ohm s Law

Using Ohm s Law with Series Circuits In this section, we ll look at how Ohm s law can be used to make calculations for more complex circuits Figure 17 shows a series circuit containing a 400 resistor and a 600 resistor The circuit power supply is 10 V We can find several quantities for this circuit First, let s calculate the total circuit resistance Since this is a series circuit, simply add the values of all the resistors to get the total resistance = R 1 + R 2 Set up the problem FIGURE 17 We can use Ohm s law to find many different circuit quantities using the information in this circuit diagram = 400 + 600 = 1,000 Add the resistor values Answer: The resistance of the circuit is 1,000 Now that we know the voltage and the resistance (10 V and 1,000 ) of this circuit, we can calculate the current using the Ohm s law formula Write the formula Substitute 10 V for E and 1,000 for R Divide (10 1,000 = 001) I = 001 A Answer: The circuit current is 001 A Figure 18 shows our original circuit with the addition of the current value we just calculated Since the circuit shown in Figure 18 is a series circuit, the same amount of current (amperes) will flow through each resistor However, the voltage in a series circuit is divided up among the components The voltage that s present across each resistor is called the voltage drop In a series circuit, the sum of the voltages across the resistors equals the total supply voltage Circuit Analysis and Ohm s Law 21

FIGURE 18 The current of 001 A has been added to the circuit diagram previously shown in Figure 17 So, in Figure 18, you can see that the total voltage is 10 V The circuit contains two resistors (R 1 and R 2 ) The voltages across R 1 and across R 2 are the voltage drops in this circuit The sum of the voltage drops across R 1 and R 2 must therefore be equal to 10 V The relationship of the component voltages in a series circuit can also be explained using Kirchhoff s loop rule (developed by the nineteenth-century physicist Gustav Robert Kirchhoff) Kirchhoff s loop rule states that the sum of all the voltages in a circuit (a loop) must equal zero When using this rule, the voltage drops in a circuit are considered to be negative numbers, and the total circuit voltage is considered to be a positive number Thus, the sum of the total voltage plus the voltage drops across the components must equal zero We can use Figure 18 to illustrate Kirchhoff s loop rule In this circuit, the sum of the total voltage plus the voltage drops across the two resistors must equal zero So, we can set up the following equation to illustrate the rule: E T + E R1 + E R2 =0 10 V + E R1 + E R2 =0 Note that in this example problem, we don t know the values of E R1 and E R2 yet, so we can only substitute 10 V into the equation for E T (the total voltage) How can you find the voltage drops across each resistor? You can use Ohm s law to calculate this quantity Start by calculating the voltage drop across R 1 Substitute the value of the current for I and the value of R 1 for R E R1 = I R 1 Write the Ohm s law voltage formula E R1 stands for the voltage drop across R 1 E R1 = 001 A 400 Substitute 001 A for I and 400 for R 1 Multiply (001 400 = 4) E R1 = 4 V Answer: The voltage drop across R 1 is 4 V 22 Circuit Analysis and Ohm s Law

Now, let s calculate the voltage drop across R 2 Use the same formula as the previous example, but substitute R 2 for R this time E R2 = I R 2 Write the Ohm s law voltage formula E R2 stands for the voltage drop across R 2 E R2 = 001 A 600 Substitute 001 A for I and 600 for R 2 Multiply (001 600 = 6) E R2 = 6 V Answer: The voltage drop across R 2 is 6 V Note that if you add the two voltage drops (E R1 + E R2 ) the sum equals the total applied voltage in the circuit (4 V + 6 + 10 V) You can use this addition method to check your answers in circuit problems like these You can also use Kirchoff s loop rule to check your answers Remember, according to Kirchhoff s rule, the sum of the total circuit voltage and the voltage drops must equal zero Using Kirchoff s rule, you must also remember that the voltage drops are considered to be negative numbers 10 V +E R1 + E R2 = 0 Set up the problem The sum of the total voltage plus the voltage drops must equal zero 10 V + ( 4 V) + ( 6 V) = 0 Substitute 4 V for E R1 and 6 V for E R2 10 4 6 = 0 Simplify the equation 6 6 = 0 The sum of all the voltages equals zero, so 0 = 0 your voltage drops were correct You can use either the addition method or Kirchhoff s loop rule to determine the voltage drops in a series circuit Most people find the addition method easier to understand and use However, Kirchhoff s loop rule will have other useful applications when you eventually study more advanced circuit analysis, so you should be aware of the rule Now, let s examine another circuit Figure 19 shows a special type of series circuit called a voltage divider A voltage divider is often used to create multiple voltages from one power supply It can also be used to apply the correct voltage to transistors or other solid-state circuit components FIGURE 19 This diagram shows a typical three-resistor voltage divider circuit Circuit Analysis and Ohm s Law 23

Let s begin calculating values for the circuit shown in Figure 19 Start by finding the total resistance in the circuit Add the values of all the resistors = R 1 + R 2 + R 3 Set up the problem = 200 + 400 + 600 Substitute the resistor values for R 1, R 2, and R 3 Add (200 + 400 + 600 = 1,200) = 1,200 Answer: The total resistance of this circuit is 1,200 Now, to find the current in this circuit, use the Ohm s law formula Write the formula Substitute 12 V for E and 1,200 for Divide (12 1,200 = 001) I = 001 A Answer: The circuit current is 001 A Now that you know the circuit current and the total resistance, you can calculate the voltage drop across each resistor in the voltage divider circuit Use Ohm s law to calculate the voltage drop across R 1 E R1 = I R 1 Write the Ohm s law voltage formula E R1 = 001 A 200 Substitute 001 A for I and 200 for R 1 Multiply (001 200 = 2) E R1 = 2 V Answer: The voltage drop across R 1 is 2 V Let s add the current value and the voltage of R 1 to our circuit diagram Figure 20 shows the updated diagram Next, let s look at the voltage drop across R 2 Use Ohm s law to calculate this value E R2 = I R 2 Write the Ohm s law voltage formula E R2 = 001 A 400 Substitute 001 A for I and 400 for R 2 Multiply (001 400 = 4) FIGURE 20 This circuit diagram is an updated version of the diagram E R2 = 4 V Answer: The voltage drop across R 2 is 4 V 24 Circuit Analysis and Ohm s Law

The total voltage in this circuit is 12 V The voltage drop across R 1 is 2 V and the voltage drop across R 2 is 4 V Therefore, the voltage drop across R 3 should be 6 V, the remaining voltage in the circuit Let s calculate it to make sure E R3 = I R 3 Write the Ohm s law voltage formula E R3 = 001 A 600 Substitute 001 A for I and 600 for R 3 Multiply (001 600 = 6) E R3 = 6 V Answer: The voltage drop across R 3 is 6 V Figure 21 shows our voltage divider with all the values filled in Note how the voltage of the circuit changes as it moves through the various resistors in the circuit Practical Example Let s look at a practical example problem that illustrates the use of both Ohm s law and Kirchoff s loop rule Suppose you re assigned the task of installing a lightemitting diode (LED) indicator on a circuit board The circuit board has a 24 V power supply, and the LED must be limited to 0005 A of current At this current, the voltage shown in Figure 19 This diagram shows the voltage drop across R 1 At point A, the circuit voltage is total (12 V) The voltage at point B is equal to the total supply voltage minus the voltage dropped across R 1 (12 V 2 V = 10 V) across the LED is 2 V (Figure 22) What size resistor should you use to keep the current at 0005 A? First, before Ohm s law can be applied to the resistor, the voltage across the resistor must be calculated According to Kirchhoff s loop rule, the sum of all the voltages in the circuit must equal zero We know the total circuit voltage (24 V) and the voltage across the diode (2 V) Use Kirchhoff s rule to find the voltage across R 1 FIGURE 21 This diagram of the voltage divider circuit shows all the voltage values At point D, all the circuit voltage has been dropped by the resistors, and the circuit voltage reaches 0 V Note how three different voltages are created by one supply voltage Circuit Analysis and Ohm s Law 25

FIGURE 22 The resistor in this circuit keeps the current through the LED at 0005 A E T + E D + E R1 = 0 Set up the problem The total circuit voltage plus the voltage across the diode (E D ) plus the voltage across the resistor (E R1 ) equals 0 24 V + ( 2 V) + E R1 = 0 Remember that the voltage drops are considered to be negative numbers Substitute 24 V for E T and 2 V for E D 24 2 + E R1 = 0 Simplify the equation Add (24 2 = 22) 22 + E R1 = 0 You can now see from this equation that the value of E R1 must be 22, because 22 22 = 0 E R1 = 22 V Answer: The voltage drop across the resistor is 22V Now that we know the value of the voltage drop across R 1, we can use Ohm s law to calculate the resistance across R 1 Write the standard Ohm s law formula We can rewrite the formula to reflect the values in this problem Substitute 22 V for R 1 and 0005 A for I R1 Divide (22 0005 = 4,400) R 1 = 4,400 Answer: The resistance across R 1 should be 4,400 Therefore, as this calculation shows, a 4,400 resistor (44 k ) should be placed in series with the LED 26 Circuit Analysis and Ohm s Law

Using Ohm s Law with Basic Parallel Circuits Parallel circuits are slightly more complicated than series circuits because parallel circuits contain more than one path for current to flow through You ll need to use a slightly different method to calculate circuit quantities in parallel circuits Let s examine this method now A simple two-resistor parallel circuit is shown in Figure 23 This circuit contains a 100 resistor and a 600 resistor connected in parallel First, let s calculate the total resistance of the circuit Write the formula Substitute the values of R 1 and R 2 into the formula Multiply (100 600 = 60,000) Add (100 + 600 = 700) FIGURE 23 We can use Ohm s law to find the total current of this circuit Divide (60,000 700 = 857) = 857 Answer: The total resistance of the circuit is 857 Now that we know the total resistance, we can easily calculate the total current by using Ohm s law Write the formula Substitute 12 V for E and 857 for R Divide (12 857 = 014) I = 014 A Answer: The total current in the circuit is 014 A Circuit Analysis and Ohm s Law 27

Now, let s see how much current is flowing through each of the resistors Since the same voltage is applied across each of the resistors, we can use Ohm s law to calculate the current in each resistor Note that R 1 has the lower value of the two resistors, so more current should flow through R 1 Write the formula I R1 stands for the current of R 1 Substitute 12 V for E and 100 for R 1 Divide (12 100 = 012) I R1 = 012 A Answer: The current flowing through R 1 is 012 A Now, use Ohm s law again to calculate the current flowing through R 2 Write the formula Substitute 12 V for E and 600 for R 2 Divide (12 600 = 002) I R2 = 002 A Answer: The current flowing through R 2 is 002 A To check that our calculations are correct, let s add the current values of both resistors together The result should be the total current in the circuit I T = I R1 + I R2 I T = 012 A + 002 A Write the formula I T stands for the total circuit current I R1 and I R2 stand for the currents in each of the resistors Substitute the values for I R1 and I R2 into the formula Add (012 + 002 = 014) I T = 014 A Answer: The total current is 014 A The total current is 014 A, which is the same value we found earlier Therefore, our calculations are correct 28 Circuit Analysis and Ohm s Law

Using Ohm s Law with Complex Parallel Circuits Complex parallel circuits contain more than two resistors The circuit shown in Figure 24 contains four resistors connected in parallel The total voltage in the circuit is 24 V With this value and the resistance of each resistor, we can calculate the current flowing through each resistor FIGURE 24 This circuit is a complex circuit containing four resistors connected in parallel Let s start with R 1 Use Ohm s law to find the current in R 1 Write the formula Substitute 24 V for E and 80 for R Divide (24 80 = 03) I R1 = 03 A Answer: The current flowing through R 1 is 03 A Now, find the current in R 2 Write the formula Circuit Analysis and Ohm s Law 29

Substitute 24 V for E and 20 for R 2 Divide (24 20 = 12) I R2 = 12 A Answer: The current flowing through R 2 is 12 A Continue your calculations by finding the current in R 3 Write the formula Substitute 24 V for E and 100 for R 3 Divide (24 100 = 024) I R3 = 024 A Answer: The current flowing through R 3 is 024 A Finally, find the current in R 4 Write the formula Substitute 24 V for E and 60 for R 4 Divide (24 60 = 04) I R4 = 04 A Answer: The current flowing through R 4 is 04 A You know that the higher the value of a resistor, the lower the amount of current that will flow through it These calculations demonstrate this fact, since the lowest value resistor (the 20 resistor) has the greatest amount of current flowing through it (12 A) The highest value resistor (the 100 resistor) will have the least amount of current flowing through it (024 A) Now, let s check our calculations by finding the total resistance of the circuit Use the reciprocal method to find the total resistance of this circuit Write the formula Substitute the values of R 1, R 2, R 3, and R 4 Simplify by dividing each term (1 80 = 00125; 1 20 = 005; 1 100 = 001; 1 60 = 00167) 30 Circuit Analysis and Ohm s Law

Add the values (00125 + 005 + 001 + 00167 = 00892) Find the reciprocal of 00892 Divide (1 00892 = 1121) = 1121 Answer: The total resistance of this circuit is 1121 (rounded) Now, let s calculate the total circuit current using Ohm s law Write the formula I T stands for total circuit current Substitute 24 V for E and 1121 for Divide (24 1121 = 214) I T = 214 A Answer: The total current of this circuit is 214 A (rounded) Finally, you can check your work by adding together the values of the currents in the individual resistors The sum of the currents should equal the total current in the circuit I T = I R1 + I R2 + I R3 + I R4 Write the formula I T = 03 A + 12 A + 024 A + 04 A Substitute the current values for each resistor Add (03 + 12 + 024 + 04 = 214) I T = 214 A Answer: The total current is 214 A The total current value of 214 A is the same value we obtained by using Ohm s law Therefore, all your calculations were correct Using Ohm s Law with Series-Parallel Circuits A combination series-parallel circuit contains components that are connected in both series and parallel arrangements While Ohm s law can still be used to calculate circuit quantities, you ll need to make some extra calculations when working with these circuits Circuit Analysis and Ohm s Law 31

The circuit in Figure 25 contains a parallel resistor group that includes an 80 resistor and a 180 resistor The parallel resistor group is connected in series with a 100 resistor and a 30 resistor To calculate the total resistance in this circuit, you ll first need to find the total value of the two parallel resistors FIGURE 25 Ohm s law can be used to calculate all the circuit quantities you need to know in this circuit Write the formula R P stands for the total resistance of the parallel group Substitute the values of R 2 and R 3 into the formula Multiply (80 180 = 14,400) Add (80 + 180 = 260) R P = 554 Divide (14,400 260 = 5538) Answer: The total resistance of the parallel group is 554 (rounded) Now, let s update our circuit diagram to include the total resistance value for the parallel group Figure 26 shows the updated circuit diagram By updating the circuit diagram in this way, we ve turned the diagram into a simple series circuit Now, calculate the total resistance of the circuit by adding together the values of the three resistors = R 1 + R P + R 4 Set up the problem = 100 + 554 + 30 Substitute the values of FIGURE 26 In this updated version of Figure 25, we ve replaced the parallel group of R 2 and R 3 with R P the three resistors, R 1, R P, and R 4 = 1854 Add (100 + 554 + 30 = 1854) Answer: The total resistance of the circuit is 1854 32 Circuit Analysis and Ohm s Law

Next, we ll use Ohm s law to calculate the total circuit current Write the formula Substitute 20 V for E and 1854 for Divide (20 1854 = 011) I = 011 A Answer: The total current of the circuit is 011 A (rounded) To continue our calculations, we can now use Ohm s law to find the voltage drops across each of the resistors (R 1, R P, and R 4 ) First, find the voltage drop across R 1 E R1 = I R 1 Write the Ohm s law voltage formula E R1 = 011 A 100 Substitute 011 A for I and 100 for R 1 Multiply (011 100 = 11) E R1 = 11 V Answer: The voltage drop across R 1 is 11 V Now, let s use Ohm s law to find the voltage drop across R P E RP = I R P Write the Ohm s law voltage formula E RP stands for the voltage drop across R P E RP = 011 A 554 Substitute 011 A for I and 554 for R P Multiply (011 554 = 6) E RP = 6 V Answer: The voltage drop across R P is 6 V (rounded) Now, use Ohm s law to find the voltage drop across R 4 E R4 = I R 4 Write the Ohm s law voltage formula E R4 = 011 A 30 Substitute 011 A for I and 30 for R 4 Multiply (011 30 = 33) E R4 = 33 V Answer: The voltage drop across R 4 is 33 V In Figure 27, the circuit diagram has again been updated to include the voltage drop across each resistor Remember that the total voltage in the circuit should equal the sum of the resistors voltages Add the resistors voltage drops to see if this is true Circuit Analysis and Ohm s Law 33

FIGURE 27 This updated version of Figure 25 shows the voltage drops across each resistor E T = E R1 + E RP + E R4 Set up the problem E T = 11 V + 6 + 33 V Substitute the values for E R1, E RP, and E R4 Add (11 + 6 + 33 = 203) E T = 203 V Answer: The total voltage is 203 V The sum of the resistors voltage drop is the same as the total voltage for the circuit Therefore, our calculations were correct Note that a difference of 03 V exists between the original voltage and our calculations due to the rounding off of decimal places The final two quantities to calculate in this circuit are the currents through the parallel resistors R 2 and R 3 Let s calculate the current through R 2 first Write the formula I R2 stands for the current through R 2 Substitute 6 V for E RP and 80 for R 2 Divide (6 80 = 008) I R2 = 008 A Answer: The current through R 2 is 008 A (rounded) 34 Circuit Analysis and Ohm s Law

Now, calculate the current through R 3 Write the formula I R3 stands for the current through R 3 Substitute 6 V for E RP and 180 for R 3 Divide (6 180 = 003) I R3 = 003 A Answer: The current through R 3 is 003 A (rounded) Remember that the current through each resistor in a series circuit is the same In our combination circuit, the parallel resistor group is connected in series to the other resistors in the circuit Therefore, the current through the parallel group should be the same as the current through either series resistor To check to see if this is true, add the currents of the two parallel resistors together I P = I R2 + I R3 I P = 008 A + 003 A I P = 011 A Set up the problem I P stands for the current in the parallel group Substitute the values for I R2 and I R3 Add (008 + 003 = 011) Answer: The current in the parallel group is 011 A The current in the parallel group is 011 A, which is the same as the current in the series circuit Therefore, our calculations are correct Practical Example Let s look at a real-life troubleshooting situation in which you ll analyze a combination series-parallel circuit Figure 28 shows a DC motor circuit that contains three sets of windings The windings are connected in a combination series-parallel arrangement The first winding, R A, is an armature winding on the rotor of the motor This winding is powered by brushes and spins with the rotor Two additional field windings are attached to the circular case of the motor (F 1 and F 2 ) These two windings remain stationary FIGURE 28 This circuit diagram shows a DC motor containing three sets of windings connected in a series-parallel arrangement Circuit Analysis and Ohm s Law 35

If the motor isn t running properly, you may be asked to test the resistance of the motor When the motor is disconnected from the power supply, how much resistance should be present in the motor s circuit? To calculate the circuit s total resistance, first calculate the resistance of the parallel circuit group containing F 2 and the armature Write the formula R P stands for the total resistance of the parallel circuit R F2 stands for the resistance of the field winding, and R A stands for the resistance of the armature Substitute the values into the formula Multiply (10 3 = 30) Add (10 + 3 =13) Divide (30 13 = 231) R P = 231 Answer: The resistance of the parallel circuit is 231 (rounded) Now, add the resistance of F 1 to the resistance of the parallel group to find the total resistance value you would read at the motor s terminals = R F1 + R P Set up the problem = 20 + 231 Substitute the values of R F1 and R P Add (20 + 231 = 2231) = 2231 Answer: The total resistance of the circuit is 2231 The total resistance of the motor should be 2231 when it s disconnected from a power supply Suppose you measure the resistance of the motor with a meter and determine that the resistance is correct Your next troubleshooting step would be to connect the motor to a power supply and measure the current How much current should be present in the motor? Use Ohm s law to find this value 36 Circuit Analysis and Ohm s Law

Write the formula Substitute 50 V for E and 2231 for Divide (50 2231 = 224) I = 224 A Answer: The total current in the circuit should be 224 A (rounded) When you measure the actual current in the motor with a meter, if you measure a value higher than 224 A, there could be a mechanical drag problem or the motor could be overloaded Or, brush problems could cause the motor failure Now, pause to check your learning by completing Self- Check 3 Circuit Analysis and Ohm s Law 37

Self-Check 3 1 How much current flows through the circuit shown in the figure below? 2 What is the circuit current in the figure shown below? (Continued) 38 Circuit Analysis and Ohm s Law

Self-Check 3 3 What voltage will deliver a current of 0072 A through the parallel resistors shown below? 4 In the figure shown below, the total voltage is 10 V and the total current is 01 A The values of two of the three resistors are labeled What is the resistance of the third resistor? (Continued) Circuit Analysis and Ohm s Law 39

Self-Check 3 5 What is the voltage drop across R 2 in the circuit shown below? Check your answers with those on page 67 40 Circuit Analysis and Ohm s Law

POWER IN DC CIRCUITS The Power Formula Power is the work performed in an electrical circuit The amount of work that can be performed by a circuit is measured in units called watts (W) The amount of power a circuit produces indicates the amount of energy the circuit uses The more power a circuit uses, the more energy it uses So, for example, a 40 W lightbulb burning for two hours uses twice the amount of energy as a 20 W lightbulb burning for two hours Also, a 20 W bulb burning for two hours uses twice as much energy as a 20 W bulb burning for one hour In daily practice, energy (the amount of power delivered over a period of time) is measured in units called watthours or kilowatthours The electric meters found outside most homes and businesses measures the amount of energy used inside the building in these units You already know that voltage from a power source creates current in a circuit Power is the amount of work produced by a circuit So, you can see that voltage, current, and power have a relationship to one another In fact, in a circuit, power is equal to voltage times current This relationship is expressed with the following formula: P = E I, or P = EI In this formula, P stands for power in watts, E stands for voltage in volts, and I stands for current in amperes Let s look at an example circuit (Figure 29) and perform some calculations with it This circuit contains a 10 V power supply, and the current is 05 A To calculate the power in this circuit, simply multiply the voltage times the current P = E I Write the power formula P = 10 V 05 A Substitute 10 V for E and 05 A for I Multiply (10 05 = 5) P = 5W Answer: The power in this circuit is 5 W Circuit Analysis and Ohm s Law 41

When current flows through a resistor, heat is produced The heat is a result of the work being done to move electrons through the resistor Since power is dissipated (lost) by a resistor, we can use a variation of the power formula to calculate power when a resistance is present in a circuit The following calculations show how we obtain this variation of the power formula FIGURE 29 This circuit contains a 10 V power supply and a 20 resistor P = E I Write the original power formula P = (I R) I Now, since E is equal to I R, substitute I R for E in the formula P = I I R Multiply the two I s together to get I 2 P = I 2 R The result is a variation of the power formula Look at the circuit in Figure 29 again Note that the circuit contains a 20 resistor Using this variation of the power formula, let s calculate the power in this circuit again P = I 2 R Write the formula P = (05 A) 2 20 Substitute 05 A for I and 20 for R P = 05 05 20 The expression (05 A) 2 means to multiply P = 025 20 05 by itself (05 05 = 025) P = 025 20 Multiply (025 20 = 5) P = 5W Answer: The power in this circuit is 5 W Note that this is the same answer you got before (5 W) Therefore, you can find the power in a circuit whenever you know the resistance and the current Now, suppose you know only the resistance and voltage in a circuit Well, you can rewrite the power formula again so that you can calculate power in this situation P = E I Write the original power formula P = E Since current is equal to, substitute for I in the formula 42 Circuit Analysis and Ohm s Law

P = E Multiply the two E s together The result is another variation of the power formula Let s use the circuit in Figure 29 one more time Use this variation of the power formula to calculate the power in the circuit Write the formula Substitute 10 V for E and 20 for R Multiply 10 by itself (10 10 = 100) Divide (100 20 = 5) P = 5W Answer: The power in the circuit is 5W We got the same answer again by using this formula So, as you can see, if you know any two of the three circuit quantities (voltage, current, or resistance), you can easily find the power in a DC circuit Power in Series Circuits In any resistive circuit, the total power supplied by the power source is equal to the power dissipated through the circuit resistors That is, the amount of power that goes into the circuit is equal to the amount of power that comes out of the circuit You can remember this important rule with the phrase power in equals power out This rule is true for series, parallel, and combination series-parallel circuits This rule allows us to perform useful calculations with any DC circuit Let s start by examining a series circuit and performing power calculations for it Figure 30 shows a circuit that contains three resistors connected in series The power supply is a battery In this circuit, the sum of the voltage drops across the three Circuit Analysis and Ohm s Law 43

FIGURE 30 This series circuit contains three resistors connected in series resistors is equal to the battery voltage Also, the sum of the power across each resistor will equal the total power supplied by the battery Figure 30 shows us the total voltage of the circuit and the values of the three resistors We can use this information to calculate the total power in the circuit, and also the amount of power that s dissipated by each of the three resistors Let s start by calculating the total power in this circuit Add the three resistances together to find the total circuit resistance Then, use a variation of the power formula to calculate the total power = R 1 + R 2 + R 3 Set up the problem = 2,000 + 8,000 + 4,000 Substitute the values for R 1, R 2, and R 3 Add (2,000 + 8,000 + 4,000 = 14,000) = 14,000 Answer: The total resistance in the circuit is 14,000 44 Circuit Analysis and Ohm s Law

You now know the total resistance, so you can use a variation of the power formula to find the total power in the circuit Write the formula Substitute 28 V for E and 14,000 for R Multiply 28 by itself (28 28 = 784) Divide (784 14,000 = 0056) P = 0056 W Answer: The total power in the circuit is 0056 W Now, suppose you need to find the power dissipated by one resistor Let s calculate the power through the first resistor, R 1 To make this calculation, you ll need to know the voltage drop across R 1 So, your first step will be to find the total circuit current using Ohm s law Write the Ohm s law formula Substitute 28 V for E and 14,000 for Divide (28 14,000 = 0002) I = 0002 A Answer: The total current in the circuit is 0002 A Now, to find the voltage drop across R 1, use the Ohm s law voltage formula E R1 = I T R 1 Write the Ohm s law voltage formula E R1 = 0002 A 2,000 Substitute 0002 A for I and 2,000 for R 1 E R1 = 0002 2,000 Multiply (0002 2,000 = 4) E R1 = 4 V Answer: The voltage drop across R 1 is 4 V Circuit Analysis and Ohm s Law 45

Now, find the power dissipated across R 1 Use a variation of the power formula Write the power formula In our problem, P will be P R1, E will be E R1, and R will be R 1 Substitute 4 V for E R1 and 2,000 for R 1 Multiply 4 by itself (4 4 = 16) Divide (16 2,000 = 0008) P R1 = 0008 W Answer: The total power dissipated across R 1 is 0008 W Let s continue our calculations by finding the voltage drops across R 2 and R 3 First, use the Ohm s law voltage formula to find the voltage drop across R 2 E R2 = I R 2 Write the Ohm s law voltage formula E R2 = 0002 A 8,000 Substitute 0002 A for I and 8,000 for R 2 E R2 = 0002 8,000 Multiply (0002 8,000 = 16) E R2 = 16 V Answer: The voltage drop across R 2 is 16 V Use the Ohm s law formula again to find the voltage drop across R 3 E R3 = I R 3 Write the Ohm s law voltage formula E R3 = 0002 A 4,000 Substitute 0002 A for I and 4,000 for R 3 E R3 = 0002 4,000 Multiply (0002 4,000 = 8) E R3 = 8 V Answer: The voltage drop across R 3 is 8 V 46 Circuit Analysis and Ohm s Law

Let s pause a moment to check our calculations thus far You know that the total voltage in a circuit is equal to the voltage drops across the resistors in the circuit Therefore, add the three voltage drops together to see if they add up to the correct total voltage for this circuit E T = E R1 + E R2 + E R3 Set up the problem 28 V = 4 + 16 V + 8 V Substitute 28 V for E T, the total voltage in the circuit Then, substitute the voltage drops for E R1, E R2, and E R3 28 = 4 +16 +8 Add (4 + 16 + 8 = 28) 28 V = 28 V Answer: The three voltage drops add up to the correct total voltage, so our calculations so far are correct Since our calculations check out, let s find the power dissipated by R 2 and R 3 Use a variation of the power formula Write the power formula In our problem, P will be P R2, E will be E R2, and R will be R 2 Substitute 16 V for E R2 and 8,000 for R 2 Multiply 16 by itself (16 16 = 256) Divide (256 8,000 = 0032) P R2 = 0032 W Answer: The total power dissipated by R 2 is 0032 W Now, find the power dissipated by R 3 Write the power formula In our problem, P will P R3, E will be E R3, and R will be R 3 Circuit Analysis and Ohm s Law 47

Substitute 8 V for E R3 and 4,000 for R 3 Multiply 8 by itself (8 8 = 64) Divide (64 4,000 = 0016) P R3 = 0016 W Answer: The total power dissipated by R 3 is 0016 W Now, let s conclude this section by checking our power calculations The total power in the circuit should be equal to the sum of the power values dissipated by the three resistors P T = P R1 + P R2 + P R3 Set up the problem P T = 0008 + 0032 + 0016 W P T = 0008 + 0032 + 0016 Substitute the three power values for P R1, P R2, and P R3 Add (0008 + 0032 + 0016 = 0056) P T = 0056 W Answer: The total power in the circuit is 0056 W The value 0056 W is the same value that we originally found with the power formula, so our calculations were all correct Power in Parallel Circuits Power calculations for parallel circuits (and series-parallel circuits) are performed in the same way as those for series circuits Simply use the proper formulas for calculating current, voltage, and resistance The power in equals power out rule also applies to these circuits So, the sum of the power dissipated by each resistor is equal to the total power generated by the power source Figure 31 shows a circuit with three resistors connected in parallel The power source is 50 V, and the three resistors have values of 10, 50, and 25 48 Circuit Analysis and Ohm s Law

The first step in analyzing this circuit is to calculate the current through each resistor This is easily done by using the Ohm s law formula Start by finding the current through R 1 Write the Ohm s law formula Substitute 50 V for E and 10 for R 1 Divide (50 10 = 5) I R1 = 5 A Answer: The current through R 1 is 5 A FIGURE 31 In this circuit, three resistors are connected in parallel Now, find the current through R 2 Write the Ohm s law formula Substitute 50 V for E and 50 for R 2 Divide (50 50 = 1) I R2 = 1 A Answer: The current through R 2 is 1 A Find the current through R 3 Write the Ohm s law formula Substitute 50 V for E and 25 for R 3 Divide (50 25 = 2) I R3 = 2 A Answer: The current through R 3 is 2 A Circuit Analysis and Ohm s Law 49

Now that we know the current through each resistor, we can find the power dissipated by each resistor Start by finding the power dissipated by R 1 P R1 = (I R1 ) 2 R 1 Write the power formula P R1 = (5 A) 2 10 Substitute 5 A for I R1 and 10 for R 1 P R1 = 5 5 10 Multiply 5 by itself (5 5 = 25) P R1 = 25 10 Multiply (25 10 = 250) P R1 = 250 W Answer: The power dissipated by R 1 is 250 W Now, calculate the power dissipated by R 2 P R2 = (I R2 ) 2 R 2 Write the variation of the power formula P R2 = (1 A) 2 50 Substitute 1 A for I R2 and 50 for R 2 P R2 = 1 1 50 Multiply 1 by itself (1 1 = 1) P R2 = 1 50 Multiply (1 50 = 50) P R2 = 50 W Answer: The power dissipated by R 2 is 50 W Calculate the power dissipated by R 3 P R3 = (I R3 ) 2 R 3 Write the variation of the power formula P R3 = (2 A) 2 25 Substitute 2 A for I R3 and 25 for R 3 P R3 = 2 2 25 Multiply 2 by itself (2 2 = 4) P R3 = 4 25 Multiply (4 25 = 100) P R3 = 100 W Answer: The power dissipated by R 3 is 100 W Now, you can easily find the total power in the circuit by adding the power values for each resistor P T = P R1 + P R2 + P R3 P T = 250 + 50 + 100 W Set up the problem Substitute the power values for each resistor P T = 250 + 50 + 100 Add (250 + 50 + 100 = 400) P T = 400 W Answer: The total power for the circuit is 400 W 50 Circuit Analysis and Ohm s Law

Of course, we can also find the total circuit power in several other ways For example, we can add the currents in the resistors to find the total circuit current, then use the power formula to find the total circuit power I T = I R1 + I R2 + I R3 I T = 5 + 1 + 2 A Set up the problem Substitute the current values for each resistor I T = 5 + 1 + 2 Add (5 + 1 + 2 = 8) I T = 8 A Answer: The total current through all three branches of this circuit is 8 A Now, use the power formula to find the total power P = E I Write the power formula P = 50 V 8 A Substitute 50 V for E and 8 A for I P = 50 8 Multiply (50 8 = 400) P = 400 W Answer: The total power in the circuit is 400 W Note that this is the same answer we obtained previously (400 W) We can use still another method to calculate the total power in this circuit We can calculate the total resistance of the circuit and use a variation of the power formula to find the total power Let s use the reciprocal formula to find the total resistance of the circuit Write the reciprocal formula Substitute the resistance values for R 1, R 2, and R 3 Divide to simplify each of the fractions Add (01 + 002 + 004 = 016) The reciprocal of 016 is Circuit Analysis and Ohm s Law 51

Divide (1 016 = 625) = 625 Answer: The total resistance of the circuit is 625 Now, use the power formula to find the total power in the circuit Write the variation of the formula Substitute 50 V for E and 625 for Multiply 50 by itself (50 50 = 2,500) Divide (2,500 625 = 400) P = 400 W Answer: The total power in this circuit is 400 W Again, we ve gotten the same answer, 400 W The method you use to calculate power is dependent upon the quantities you know Always use the easiest method available to calculate the quantity you need to find FIGURE 32 This series-parallel circuit contains a group of two parallel resistors that s connected in series to two other resistors Power in Series-Parallel Circuits To conclude our study of power in DC circuits, let s look at a typical series-parallel circuit (Figure 32) This circuit has a 24 V power supply and four resistors Two of the resistors are connected in parallel The parallel group is then connected to two other resistors in series Let s assume that you need to calculate the total power in this circuit To do this, you ll need to find the total resistance of this circuit 52 Circuit Analysis and Ohm s Law

First, find the resistance of the parallel branch of the circuit Since both resistors in the parallel group have the same value, simply divide the value of one resistor by two Divide 40 by 2 (40 2 = 20) R P = 20 Answer: The total resistance of the parallel group is 20 Now, add the values of R 1, R P, and R 4 to get the total resistance in the circuit = R 1 + R P + R 4 Set up the problem = 20 + 20 + 80 Substitute the values for R 1, R P, and R 4 = 20 + 20 + 80 Add (20 + 20 + 80 = 120) = 120 Answer: The total resistance of the circuit is 120 Now, use the power formula to determine the total power in the circuit Write the variation of the power formula Substitute 24 V for E and 120 for Multiply 24 by itself (24 24 = 576) Divide (576 120 = 48) P = 48 W Answer: The total power in the circuit is 48 W Since you know the voltage and the total resistance in this circuit, you can use Ohm s law to calculate the circuit current Write the Ohm s law formula Circuit Analysis and Ohm s Law 53

Substitute 24 V for E and 120 for Divide (24 120 = 02) I = 02 A Answer: The circuit current is 02 A Practical Example FIGURE 33 This diagram shows an automobile headlamp circuit Let s take a look at a practical example in which you can use the DC power formulas Figure 33 shows an automobile headlamp circuit The circuit contains two 60 W headlamps connected in parallel The headlamp group is then connected in series with a fuse and a 12 V battery The problem in this circuit is that the fuse fails every time the headlamps are turned on The correct fuse rating is unknown Calculate the current flowing in the fuse so that you can install the proper fuse and fix the circuit The total power that must be delivered by the battery is the sum of the amounts of power used by the headlamps So, add the power amounts of the two headlamps Then, substitute this value into the power formula, and solve to find the correct current P T = 60 W + 60 W P T = 120 W P = EI Add the two headlamp power amounts together Write the power formula Divide each side of the equation by E This will isolate the I (the current) and produce a variation of the power formula Substitute 120 W for P and 12 V for E 54 Circuit Analysis and Ohm s Law

Divide (120 12 = 10) 10 A = I Answer: The current is 10 A According to our calculations, there is 10 A flowing through the fuse in Figure 33 Therefore, the fuse size must be slightly greater than 10 A Replace the present fuse with the next larger size fuse (a 12 A fuse) in order to repair the circuit Now, pause to check your learning by completing Self- Check 4 Circuit Analysis and Ohm s Law 55

Self-Check 4 1 The voltage in a circuit is 24 V and the current is 005 A What is the power in this circuit? 2 A circuit has a resistance of 100 and a current of 2 A What is the power of this circuit? 3 What is the power dissipated by both resistors in the circuit shown below? 4 What is the power dissipated by all of the resistors in the circuit shown below? Check your answers with those on page 69 56 Circuit Analysis and Ohm s Law

MEASURING CURRENT, VOLTAGE, AND RESISTANCE Analog and Digital Meters In this study unit, you learned how to calculate the current, voltage, resistance, and power in a circuit by using formulas In order to calculate any of these quantites, you must already know two circuit quantities In all the example problems we examined so far, you were given two circuit values However, in real life, you ll need to find these values for yourself by using a meter Meters are very useful tools that you can use to measure circuit quantities There are two basic types of meters used today, the analog meter and the digital meter Let s start by looking at the analog meter An analog meter contains a moving pointer and a calibrated scale (Figure 34) The pointer is attached to a coil of wire suspended on jeweled bearings The pointer assembly is then placed between the poles of a permanent magnet When current flows through the pointer s coil, the coil creates a magnetic field that interacts with the field of the permanent magnet As a result, the pointer rotates a certain number of degrees The number of degrees that the pointer moves depends on the amount of current flowing in the coil The current that causes the pointer to move all the way to the right end of the scale is called full-scale current Typically, full-scale current is between 10 microamperes and 100 milliamperes If you need to measure more current than that, you must adjust the meter s range switch The range switch places shunts (resistors) across the coil that divert most of the current around the meter movement Because the proportion between the coil current and the shunt current is known, the meter scale can therefore be interpreted for larger currents FIGURE 34 This illustration shows the pointer assembly of an analog meter Circuit Analysis and Ohm s Law 57

Two types of analog meters are used in measuring current: the ammeter and the voltohm-milliammeter or VOM An ammeter can only be used to measure current Depending on the model used, an ammeter can measure amperage in a range from microamperes to several hundred amperes FIGURE 35 A typical VOM has input jacks, a range switch, a zero-ohms adjust, and other features (Photo used with permission of Simpson Electric) A typical VOM is shown in Figure 35 A VOM has multiple scales for measuring current, and can generally measure current in a range from microamperes up to about 10 A A VOM can also be used to measure voltage and resistance, making it a very versatile and compact testing instrument Digital versions of the ammeter and VOM are also available Digital meters are called digital multimeters or DMMs A digital multimeter gives readouts of current values in a four-digit display (Figure 36) Because this type of meter provides the user with a digital readout, its measurements can be more exact than those provided by an analog meter Digital readouts remove the possibility of the user making improper pointer-to-scale readings Measuring Current When a troubleshooter checks a circuit, he or she generally measures voltage and resistance first However, current can also be measured with an ammeter to see if the circuit is drawing unusually high or low levels of current FIGURE 36 This illustration shows a digital multimeter This model measures voltage, resistance, current, and frequency (Reproduced with permission of Fluke Corporation) When measuring current with an analog meter, the first step is to remove all power from the circuit This must be done since you ll need to break open the circuit and insert the meter leads into it (Figure 37) 58 Circuit Analysis and Ohm s Law

That is, you must connect the meter in series with the circuit Once the circuit is broken, connect the alligator clips on the meter s test leads to the conductors at the point where the circuit was broken You must be careful to insert the meter s leads properly into the circuit The negative lead must be placed on the negative side of the circuit, and the positive lead must be placed on the positive side of the circuit Reversing the placement of the leads can damage an analog meter by moving the pointer against the stop very abruptly Note: Never use an ammeter to measure the voltage across a power source; you could destroy the meter FIGURE 37 When measuring current, you must break the circuit and install the ammeter in series with the circuit Note that the circled letter A is the symbol for the ammeter in the circuit Next, set the meter s range switch to a current value that s higher than what you expect to measure in the circuit By doing this, you can prevent damage to the meter Now, reapply power to the circuit and measure the current directly on the meter If the pointer of an analog meter is too far to the right side of the scale, remove the power, change the range selector switch, reapply power, and then retake the measurement The most accurate position of the pointer on an analog meter s scale is toward the center After you complete your measurements, remove the power once again and reconnect the broken ends of the circuit conductors to each other With some modern meters and stand-alone, clamp-type meters, you don t need to break the circuit to take measurements Instead, these meters have split jaws that are clamped on a circuit conductor The meter then measures current as a function of the magnetic field that surrounds the conductor Circuit Analysis and Ohm s Law 59

Measuring Voltage Measuring voltage in a circuit or component is much easier than measuring current A voltmeter is used to measure voltage When measuring voltage, the voltmeter s leads are placed across the component or the power source Figure 38 shows a voltmeter set up to measure the voltage across a resistor (R 2 ) This is called a parallel connection Follow these steps to measure voltage FIGURE 38 Voltage can easily be measured across any circuit component Step 1: To protect the meter, set the voltage range to a higher voltage than you expect to measure in the circuit Step 2: Install the meter s negative lead on the circuit point or component lead that will be used as the negative reference Note that the circled letter V stands for the voltmeter in this diagram Step 3: Step 4: Install the meter s positive lead on the circuit point or component lead that will be used as the positive reference If you re using an analog meter, read the meter scale and adjust the voltage range switch until the pointer is near the center of the scale If you re using a digital meter, adjust the voltage range switch to the lowest range possible without losing the first digit The range selector switch on the voltmeter must be set to a higher range than the voltage level you expect to find in the circuit Meter damage can result if the meter isn t set to the proper range This is especially true with analog meters Digital meters aren t as easily damaged, and they ll usually display the letters OL (overload) if the wrong range is chosen 60 Circuit Analysis and Ohm s Law

CAUTION! You must use extreme caution when measuring voltages in circuits that contain more than 48 V The possibility of electrical shock exists High-voltage circuits (over 1,000 V) should never be measured without the use of an accessory probe called a high-voltage probe A high-voltage probe is heavily insulated and converts high voltages to lower voltages that a standard meter can accept Many digital meters have features such as autoranging (the range is automatically set inside the meter); autopolarity (the meter displays the polarity of the voltage); and overvoltage/ overcurrent protection One type of digital meter in use today even includes a memory feature This feature allows a technician to hold the meter probes on the circuit until a beep sounds The memory then displays the measured value, even after the probes are removed Measuring Resistance Most VOMs provide for multiple ranges in resistance readings Such readings are direct (R 1); 10 times the reading (R 10); 100 times the reading (R 100); and so forth So, for example, when you take a reading at the R 100 range, the actual resistance is 100 times the value you read from the meter s scale An ohmmeter contains an internal battery, a milliammeter, and a scaling resistor The ohmmeter s leads are attached to each side of the component to be measured The ohmmeter then generates current in its battery The current is measured as it flows through the milliammeter, through the scaling resistor, through the component being measured, and back to the battery A diagram of such a circuit is shown in Figure 39 In Figure 39, note that R S is a variable resistor This resistor adjusts the meter pointer to the zero mark on the ohms scale when the resistance between the leads is zero ohms that is, when the leads are shorted together In this situation, it s actually the amount of current flowing through the milli - ammeter that s adjusted When this zero-ohms adjustment is completed, the meter pointer is at zero on the resistance scale, and full-scale current is flowing through the milliammeter Circuit Analysis and Ohm s Law 61

FIGURE 39 This illustration shows a resistor being tested with an ohmmeter The circled letters ma stand for the milliammeter in the circuit When the resistance between the leads is infinite (that is, when there s nothing connected between the leads or when an open circuit is being measured) no current flows through the meter The pointer stays at the left end of the scale which is marked with an infinity sign ( ) The remainder of the scale is calibrated according to the proportion of full-scale current that will flow with various values of resistance connected across the leads The procedure for measuring the resistance of a component is as follows: Step 1: Step 2: Step 3: Step 4: Remove power from the system Remove one lead of the component to be measured from the circuit board or terminal block Touch the two meter leads together Zero the meter using the ohms adjust, or zero adjust, control (the scaling resistor) Measure the resistance of the component directly on the meter Polarity and range selection aren t important, since neither will destroy the meter Always observe two cautions when you re taking resistance measurements First, be sure to remove all circuit power Second, remove any parallel circuit paths to the component Parallel circuit paths will greatly lower ohmmeter readings Loose components (those not yet installed in circuits) may be measured directly with an ohmmeter Ohmmeters can also be used to check the integrity of a conductor or to locate conductor shorts This type of test is known as a continuity test 62 Circuit Analysis and Ohm s Law

Digital meters also measure resistance by measuring the current flow through a component However, the internal components and circuits in the digital meter convert a current, voltage, or resistance value to a frequency that s counted and then displayed on the front panel of the meter (the liquid crystal display, or LCD) Now, take a few moments to complete Self-Check 5 Circuit Analysis and Ohm s Law 63

Self-Check 5 Fill in the blanks in each of the following statements 1 When you re measuring, the circuit must be broken and the meter leads must be inserted into the circuit 2 The circuit you re measuring shouldn t be energized (turned on) when you re measuring 3 The type of meter is the most accurate for making circuit measurements 4 When you re using an analog meter to measure resistance, the first step is to the circuit Check your answers with those on page 71 64 Circuit Analysis and Ohm s Law