Physics
Resistors in Series and Parallel
When resistors are connected in series, their resistances add up, resulting in a total resistance equal to the sum of the individual resistances. In contrast, when resistors are connected in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistances. These configurations are fundamental in understanding and analyzing electrical circuits.
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11 Key excerpts on "Resistors in Series and Parallel"
eBook - PDF- Paul Peter Urone, Roger Hinrichs(Authors)
- 2012(Publication Date)
- Openstax(Publisher)
21.1 Resistors in Series and Parallel Most circuits have more than one component, called a resistor that limits the flow of charge in the circuit. A measure of this limit on charge flow is called resistance. The simplest combinations of resistors are the series and parallel connections illustrated in Figure 21.2. The total resistance of a combination of resistors depends on both their individual values and how they are connected. Figure 21.2 (a) A series connection of resistors. (b) A parallel connection of resistors. Resistors in Series When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then R 1 in Figure 21.2(a) could be the resistance of the screwdriver’s shaft, R 2 the resistance of its handle, R 3 the person’s body resistance, and R 4 the resistance of her shoes. Figure 21.3 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber- soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.) Figure 21.3 Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right). To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 21.3. 808 Chapter 21 | Circuits and DC Instruments This OpenStax book is available for free at http://cnx.org/content/col11406/1.9
eBook - PDF- J. David Irwin, R. Mark Nelms(Authors)
- 2022(Publication Date)
- Wiley(Publisher)
Resistors R 1 and R 2 are in series if they are connected end to end with one common node and carry exactly the same current. They can then be combined into a single resistor R S , where R S = R 1 + R 2 . STEP 2 Resistors in parallel. Resistors R 1 and R 2 are in parallel if they are connected to the same two nodes and have exactly the same voltage across their terminals. They can then be combined into a single resistor R p , where R p = R 1 R 2 /(R 1 + R 2 ). These two combinations are used repeatedly, as needed, to reduce the network to a single resistor at the pair of terminals. LEARNING ASSESSMENTS E2.15 Find the equivalent resistance at the terminals A-B in the circuit in Fig. E2.15. R AB A B 4 kΩ 4 kΩ 3 kΩ 12 kΩ 8 kΩ 6 kΩ FIGURE E2.15 E2.16 Find R AB in Fig. E2.16. 4 kΩ A B 3 kΩ 2 kΩ 4 kΩ 4 kΩ 2 kΩ 8 kΩ 12 kΩ 6 kΩ 2 kΩ R AB FIGURE E2.16 Answer: R AB = 3 kΩ. Answer: R AB = 12 kΩ. 2.5 Series and Parallel Resistor Combinations 53 54 Chapter 2 Resistive Circuits A standard dc current-limiting power supply shown in Fig. 2.29a provides 0–18 V at 3 A to a load. The voltage drop, V R , across a resistor, R, is used as a current-sensing device, fed back to the power supply and used to limit the current I. That is, if the load is adjusted so that the current tries to exceed 3 A, the power supply will act to limit the current to that value. The feedback voltage, V R , should typically not exceed 600 mV. If we have a box of standard 0.1-Ω, 5-W resistors, let us determine the configuration of these resistors that will provide V R = 600 mV when the current is 3 A. FIGURE 2.29 Circuits used in Example 2.21. (a) (b) (c) 0.1 Ω 0.1 Ω DC power supply R I Load 0 A 0 A All resistors 0.1 Ω V R + + − − R R Solution Using Ohm’s law, the value of R should be R = V R ___ I = 0.6 — 3 = 0.2 Ω Therefore, two 0.1-Ω resistors connected in series, as shown in Fig. 2.29b, will provide the proper feedback voltage. Suppose, however, that the power supply current is to be limited to 9 A.
eBook - PDF- J. David Irwin, R. Mark Nelms(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
Resistors R 1 and R 2 are in series if they are connected end to end with one common node and carry exactly the same current. They can then be combined into a single resistor R S , where R S = R 1 + R 2 . STEP 2 Resistors in parallel. Resistors R 1 and R 2 are in parallel if they are connected to the same two nodes and have exactly the same voltage across their terminals. They can then be combined into a single resistor R p , where R p = R 1 R 2 /(R 1 + R 2 ). These two combinations are used repeatedly, as needed, to reduce the network to a single resistor at the pair of terminals. E2.15 Find the equivalent resistance at the terminals A-B in the circuit in Fig. E2.15. R AB A B 4 kΩ 4 kΩ 3 kΩ 12 kΩ 8 kΩ 6 kΩ FIGURE E2.15 E2.16 Find R AB in Fig. E2.16. 4 kΩ A B 3 kΩ 2 kΩ 4 kΩ 4 kΩ 2 kΩ 8 kΩ 12 kΩ 6 kΩ 2 kΩ R AB FIGURE E2.16 Answer: R AB = 3 kΩ. Answer: R AB = 12 kΩ. Learning Assessments 48 CHAPTER 2 Resistive Circuits A standard dc current-limiting power supply shown in Fig. 2.29a provides 0–18 V at 3 A to a load. The voltage drop, V R , across a resistor, R, is used as a current-sensing device, fed back to the power supply and used to limit the current I. That is, if the load is adjusted so that the current tries to exceed 3 A, the power supply will act to limit the current to that value. The feedback voltage, V R , should typically not exceed 600 mV. If we have a box of standard 0.1-Ω, 5-W resistors, let us determine the configuration of these resistors that will provide V R = 600 mV when the current is 3 A. (a) (b) (c) 0.1 Ω 0.1 Ω dc power supply R I Load 0 A 0 A All resistors 0.1 Ω V R + + − − R R EX AMPLE 2.21 FIGURE 2.29 Circuits used in Example 2.21. Solution Using Ohm’s law, the value of R should be R = V R ___ I = 0.6 — 3 = 0.2 Ω Therefore, two 0.1-Ω resistors connected in series, as shown in Fig. 2.29b, will provide the proper feedback voltage. Suppose, however, that the power supply current is to be limited to 9 A.
eBook - PDF- Michael Neidle(Author)
- 2013(Publication Date)
- Butterworth-Heinemann(Publisher)
Chapter 4 D C . CIRCUITS In order to deal in a confident manner with more involved circuits, students should make a point of fully understanding simple series and parallel circuits. We commence with the 'rules of the game', but as in most parts of this work plenty of practice from earlier years' work will aid in achieving a permanent grasp of the subject. Fig. 4.1. The series circuit. 4.1. Resistors in Series The supply voltage Vis obviously equal to the sum of the potential difference across each resistor (Fig. 4.1): Voltage law V=V l +V 2 +V 3 As the same current flows through each of the resistors, V x = IR l V 2 = IR 2 V= / Ä 1 + / R 2 + IR 3 = /(R 1 +K 2 +R 3 ) But V = IR where R = total resistance IR = P 1 + R 2 + R 3 ) thus R = R l +R 2 +R 3 61 and V, = IR, 62 ELECTRICAL INSTALLATION TECHNOLOGY Example 4.1 Four similar indicator filament lamps, each rated at 5 W, 50 V, are connected in series across a 200 V supply. What is the total current taken from the supply? After a period of operation, one of the lamps fails and becomes open-circuited. Explain how a voltmeter may be used to find which lamp has failed, stating clearly what readings would be expected on the voltmeter. The only replacement lamp available is one rated at 2-5 W 50 V. What would be the voltages across this lamp and each of the other lamps if this replacement were used in the circuit, and what would be the probable result ? [T] Power in Substituting By transposition watts = voltage x current P=VI I · ' -S V 2 Thus, resistance of each of the 5 W, 50 V lamps 50x50 ~ 5 Total resistance of the 4 lamps Current in circuit 200 V 2000 Ω = 500Ω = 2000 Ω = 01 A Since an open-circuited lamp does not allow any current to pass in a series circuit Reading by voltmeter across good lamps = zero Reading across faulty lamp = practically full voltage of 200 V This is because the voltmeter would have a high resistance, being 1000 Ω/V for a good instrument, i.e.
- Frank R. Spellman(Author)
- 2013(Publication Date)
- CRC Press(Publisher)
If the circuit is arranged so the electrons have only one possible path, the circuit is a series circuit . A series circuit, then, is defined as a circuit that contains only one path for current flow. Figure 10.29 shows a series circuit having several loads (resistors). Key Point: A series circuit is a circuit having only one path for the current to flow along. 10.14.1 S ERIES C IRCUIT R ESISTANCE To follow its electrical path, the current in a series circuit must flow through resistors inserted in the circuit (see Figure 10.30); thus, each additional resistor offers added resistance. In a series circuit, the total circuit resistance (R T ) is equal to the sum of the individual resistances , or R T = R 1 + R 2 + R 3 + … + R n (10.8) where R T = Total resistance ( Ω ). R 1 , R 2 , R 3 = Resistance in series ( Ω ). R n = Any number of additional resistors in the series. ■ EXAMPLE 10.12 Problem: Three resistors of 10 ohms, 12 ohms, and 25 ohms are connected in series across a battery whose emf is 110 volts (Figure 10.30). What is the total resistance? Solution: Given: R 1 = 10 ohms R 2 = 12 ohms R 3 = 25 ohms R T = R 1 + R 2 + R 3 R T = 10 + 12 + 25 = 47 Ω Equation 10.8 can be transposed to solve for the value of an unknown resistance; for example, transposition can be used in some circuit applications where the total resis-tance is known but the value of a circuit resistor has to be determined. ■ EXAMPLE 10.13 Problem: The total resistance of a circuit containing three resistors is 50 ohms (see Figure 10.31). Two of the circuit resistors are 12 ohms each. Calculate the value of the third resistor ( R 3 ). + – FIGURE 10.29 Series circuit. + R 1 10 ohms R 3 25 ohms R 2 12 ohms – FIGURE 10.30 Solving for total resistance in a series circuit. R 1 12 ohms R T 50 ohms R 3 ? R 2 12 ohms FIGURE 10.31 Calculating the value of one resistance in a series circuit.
eBook - PDF- William Moebs, Samuel J. Ling, Jeff Sanny(Authors)
- 2016(Publication Date)
- Openstax(Publisher)
The total resistance of this combination is intermediate between the pure series and pure parallel values ( 20.0 Ω and 0.804 Ω , respectively). b. The current through R 1 is equal to the current supplied by the battery: I 1 = I = V R eq = 12.0 V 5.10 Ω = 2.35 A. The voltage across R 1 is V 1 = I 1 R 1 = (2.35 A)(1 Ω) = 2.35 V. Chapter 10 | Direct-Current Circuits 449 10.5 The voltage applied to R 2 and R 3 is less than the voltage supplied by the battery by an amount V 1 . When wire resistance is large, it can significantly affect the operation of the devices represented by R 2 and R 3 . c. To find the current through R 2 , we must first find the voltage applied to it. The voltage across the two resistors in parallel is the same: V 2 = V 3 = V − V 1 = 12.0 V − 2.35 V = 9.65 V. Now we can find the current I 2 through resistance R 2 using Ohm’s law: I 2 = V 2 R 2 = 9.65 V 6.00 Ω = 1.61 A. The current is less than the 2.00 A that flowed through R 2 when it was connected in parallel to the battery in the previous parallel circuit example. d. The power dissipated by R 2 is given by P 2 = I 2 2 R 2 = (1.61 A) 2 (6.00 Ω) = 15.5 W. Significance The analysis of complex circuits can often be simplified by reducing the circuit to a voltage source and an equivalent resistance. Even if the entire circuit cannot be reduced to a single voltage source and a single equivalent resistance, portions of the circuit may be reduced, greatly simplifying the analysis. Check Your Understanding Consider the electrical circuits in your home. Give at least two examples of circuits that must use a combination of series and parallel circuits to operate efficiently. Practical Implications One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn (or a very long) extension cord, then this loss can be significant.
eBook - PDF- Richard C. Dorf, James A. Svoboda(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
Answer: v m 2 V 3.4 P a r a l l e l R e s i s t o r s a n d C u r r e n t D i v i s i o n Circuit elements, such as resistors, are connected in parallel when the voltage across each element is identical. The resistors in Figure 3.4-1 are connected in parallel. Notice, for example, that resistors R 1 and R 2 are each connected to both node a and node b. Consequently, v 1 v 2 , so both resistors have the same voltage. A similar argument shows that resistors R 2 and R 3 are also connected in parallel. Noticing that R 2 is connected in parallel with both R 1 and R 3 , we say that all three resistors are connected in parallel. The order of parallel resistors is not important. For example, the voltages and currents of the three resistors in Figure 3.4-1 will not change if we interchange the positions R 2 and R 3 . The defining characteristic of parallel elements is that they have the same voltage. To identify a pair of parallel elements, we look for two elements connected between the same pair of nodes. Consider the circuit with two resistors and a current source shown in Figure 3.4-2. Note that both resistors are connected to terminals a and b and that the voltage v appears across each parallel v m 8 V + – + – 75 Ω 25 Ω (a) (b) 8 V 75 Ω 25 Ω Voltmeter + – FIGURE E 3.3-1 (a) A voltage divider. (b) The voltage divider after the ideal voltmeter has been replaced by the equivalent open circuit and a label has been added to indicate the voltage measured by the voltmeter v m . R 1 v s R 2 b a + – + – – + + – v 1 R 3 v 2 v 3 FIGURE 3.4-1 A circuit with parallel resistors. R 1 R 2 i 2 i 1 i s v b a + – FIGURE 3.4-2 Parallel circuit with a current source. Parallel Resistors and Current Division 71 element. In anticipation of using Ohm’s law, the passive convention is used to assign reference directions to the resistor voltages and currents.
eBook - PDF- James A. Svoboda, Richard C. Dorf(Authors)
- 2013(Publication Date)
- Wiley(Publisher)
CHAPTER 3 Resistive Circuits I N T H I S C H A P T E R 3.1 Introduction 3.2 Kirchhoff’s Laws 3.3 Series Resistors and Voltage Division 3.4 Parallel Resistors and Current Division 3.5 Series Voltage Sources and Parallel Current Sources 3.6 Circuit Analysis 3.7 Analyzing Resistive Circuits Using MATLAB 3.8 How Can We Check . . . ? 3.9 DESIGN EXAMPLE— Adjustable Voltage Source 3.10 Summary Problems Design Problems 3.1 I n t r o d u c t i o n In this chapter, we will do the following: Write equations using Kirchhoff’s laws. Not surprisingly, the behavior of an electric circuit is determined both by the types of elements that comprise the circuit and by the way those elements are connected together. The constitutive equations describe the elements themselves, and Kirchhoff’s laws describe the way the elements are connected to each other to form the circuit. Analyze simple electric circuits, using only Kirchhoff’s laws and the constitutive equations of the circuit elements. Analyze two very common circuit configurations: series resistors and parallel resistors. We will see that series resistors act like a “voltage divider, ” and parallel resistors act like a “current divider.” Also, series resistors and parallel resistors provide our first examples of an “equivalent circuit.” Figure 3.1-1 illustrates this important concept. Here, a circuit has been partitioned into two parts, A and B. Replacing B by an equivalent circuit, B eq , does not change the current or voltage of any circuit element in part A. It is in this sense that B eq is equivalent to B. We will see how to obtain an equivalent circuit when part B consists either of series resistors or of parallel resistors. Determine equivalent circuits for series voltage sources and parallel current sources. Determine the equivalent resistance of a resistive circuit.
eBook - PDF- Richard C. Dorf, James A. Svoboda(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
CHAPTER 3 Resistive Circuits I N T H I S C H A P T E R 3.1 Introduction 3.2 Kirchhoff’s Laws 3.3 Series Resistors and Voltage Division 3.4 Parallel Resistors and Current Division 3.5 Series Voltage Sources and Parallel Current Sources 3.6 Circuit Analysis 3.7 Analyzing Resistive Circuits Using MATLAB 3.8 How Can We Check . . . ? 3.9 DESIGN EXAMPLE— Adjustable Voltage Source 3.10 Summary Problems Design Problems 3.1 I n t r o d u c t i o n In this chapter, we will do the following: Write equations using Kirchhoff’s laws. Not surprisingly, the behavior of an electric circuit is determined both by the types of elements that comprise the circuit and by the way those elements are connected together. The constitutive equations describe the elements themselves, and Kirchhoff’s laws describe the way the elements are connected to each other to form the circuit. Analyze simple electric circuits, using only Kirchhoff’s laws and the constitutive equations of the circuit elements. Analyze two very common circuit configurations: series resistors and parallel resistors. We will see that series resistors act like a “voltage divider, ” and parallel resistors act like a “current divider.” Also, series resistors and parallel resistors provide our first examples of an “equivalent circuit.” Figure 3.1-1 illustrates this important concept. Here, a circuit has been partitioned into two parts, A and B. Replacing B by an equivalent circuit, B eq , does not change the current or voltage of any circuit element in part A. It is in this sense that B eq is equivalent to B. We will see how to obtain an equivalent circuit when part B consists either of series resistors or of parallel resistors. Determine equivalent circuits for series voltage sources and parallel current sources. Determine the equivalent resistance of a resistive circuit.
- Dennis L. Eggleston(Author)
- 2011(Publication Date)
- Cambridge University Press(Publisher)
number of resistors in series may be replaced by a single equivalent resistor given by: R eq = i R i (1.9) where the sum is over all the resistors in series. To see this, consider the circuit shown in Fig. 1.7 . We would like to replace the circuit on the left by the equivalent circuit on the right. The circuit on the right will be equivalent if the current supplied by the battery is the same. By KCL, the current in each resistor is the same. Applying KVL around the circuit loop and Ohm’s Law for the drop across the resistors, we obtain V = IR 1 + IR 2 + IR 3 = I ( R 1 + R 2 + R 3 ) = IR eq (1.10) where R eq = R 1 + R 2 + R 3 . (1.11) This derivation can be extended to any number of resistors in series, hence Eq. (1.9) . 1.2.1.2 Resistors in parallel Components connected in parallel are connected in a head-to-head and tail-to-tail fashion. The components are often drawn in parallel lines, hence the name. When forming equivalent circuits, any number of resistors in parallel may be replaced by a single equivalent resistor given by: 1 R eq = i 1 R i (1.12) where the sum is over all the resistors in parallel. To see this, consider the circuit shown in Fig. 1.8 . Again, we would like to replace the circuit on the left by the equivalent circuit on the right. 8 Basic concepts and resistor circuits V R 1 R 2 R 3 I I 1 I 2 I 3 V R eq I Figure 1.8 Equivalent circuit for resistors in parallel. First, note that KCL requires I = I 1 + I 2 + I 3 . (1.13) Since the resistors are connected in parallel, the voltage across each one is the same, and, by KVL is equal to the battery voltage: V = I 1 R 1 , V = I 2 R 2 , V = I 3 R 3 . Solving these for the three currents and substituting in Eq. (1.13) gives I = V R 1 + V R 2 + V R 3 = V 1 R 1 + 1 R 2 + 1 R 3 = V R eq (1.14) where 1 R eq = 1 R 1 + 1 R 2 + 1 R 3 . (1.15) Again, this derivation can be extended to any number of resistors in parallel, hence Eq. (1.12) . A frequent task is to analyze two resistors in parallel.
eBook - PDF- J. David Irwin, R. Mark Nelms, Amalendu Patnaik(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
52 CHAPTER 2 RESISTIVE CIRCUITS E2.14 Find the equivalent resistance at the terminals A-B in the network in Fig. E2.14. A B 6 kΩ 3 kΩ 18 kΩ 10 kΩ 6 kΩ R AB Figure E2.14 ANSWER: R AB = 22 kΩ. LEARNING ASSESSMENT E2.15 Find the equivalent resistance at the terminals A-B in the circuit in Fig. E2.15. R AB A B 4 kΩ 4 kΩ 3 kΩ 12 kΩ 8 kΩ 6 kΩ Figure E2.15 ANSWER: R AB = 3 kΩ. LEARNING ASSESSMENTS PROBLEM-SOLVING STRATEGY SIMPLIFYING RESISTOR COMBINATIONS When trying to determine the equivalent resistance at a pair of terminals of a network com- posed of an interconnection of numerous resistors, it is recommended that the analysis begin at the end of the network opposite the terminals. Two or more resistors are combined to form a single resistor, thus simplifying the network by reducing the number of components as the analysis continues in a steady progression toward the terminals. The simplification involves the following: STEP 1. Resistors in series. Resistors R 1 and R 2 are in series if they are connected end to end with one common node and carry exactly the same current. They can then be combined into a single resistor R S , where R S = R 1 + R 2 . STEP 2. Resistors in parallel. Resistors R 1 and R 2 are in parallel if they are connected to the same two nodes and have exactly the same voltage across their terminals. They can then be combined into a single resistor R p , where R p = R 1 R 2 /(R 1 + R 2 ). These two combinations are used repeatedly, as needed, to reduce the network to a single resistor at the pair of terminals. SECTION 2.5 SERIES AND PARALLEL RESISTOR COMBINATIONS 53 A standard dc current-limiting power supply shown in Fig. 2.29a provides 0–18 V at 3 A to a load. The voltage drop, V R , across a resistor, R, is used as a current-sensing device, fed back to the power supply and used to limit the current I. That is, if the load is adjusted so that the current tries to exceed 3 A, the power supply will act to limit the current to that value.
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