Physics
Resistors in Series
When resistors are connected in series, their resistance values add up. This means that the total resistance in the circuit is the sum of the individual resistances. In a series circuit, the same current flows through each resistor, and the voltage across the resistors adds up to the total voltage of the circuit.
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10 Key excerpts on "Resistors in Series"
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
- 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- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler, Heath Jones, Matthew Collins, John Daicopoulos, Boris Blankleider(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
Note that if the current in one resistor is interrupted, the current in the other is too. This could occur, for example, if two light bulbs were connected in series and the filament of one bulb broke. Because of the series wiring, the voltage V supplied by the battery is divided between the two resistors. The drawing indicates that the portion of the voltage across R 1 is V 1 , while the portion across R 2 is V 2 , so V = V 1 + V 2 . For the individual resistances, the definition of resistance indicates that R 1 = V 1 /I and R 2 = V 2 /I, so that V 1 = IR 1 and V 2 = IR 2 . Therefore, we have V = V 1 + V 2 = IR 1 + IR 2 = I ( R 1 + R 2 ) = IR S where R S is called the equivalent resistance of the series circuit. Thus, two Resistors in Series are equivalent to a single resistor whose resistance is R S = R 1 + R 2 , in the sense that there is the same current through R S as there is through the series combination of R 1 and R 2 . This line of reasoning can be extended to any number of Resistors in Series if we note the following: Problem‐solving insight The voltage across all the Resistors in Series is the sum of the individual voltages across each resistor. The result for the equivalent resistance is Series resistors R S = R 1 + R 2 + R 3 + . . . (20.16) 548 Physics Examples 8 and 9 illustrate the concept of equivalent resistance in series circuits. EXAMPLE 8 A series circuit Suppose that the resistances in figure 20.14 are R 1 = 47 Ω and R 2 = 86 Ω, and the battery voltage is 24 V. Determine the equivalent resistance of the two resistors and the cur- rent in the circuit. Reasoning The two resistors are wired in series, since there is the same current through each one. The equivalent resistance R S of a series circuit is the sum of the individual resistances, so R S = R 1 + R 2 . The current I can be obtained from Ohm’s law as the voltage V divided by the equivalent resistance: I = V /R S .
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.
No longer available |Learn more- (Author)
- 2014(Publication Date)
- The English Press(Publisher)
Size, and position of leads (or terminals), are relevant to equipment designers; resistors must be physically large enough not to overheat when dissipating their power. ________________________ WORLD TECHNOLOGIES ________________________ Units The ohm (symbol: Ω) is the SI unit of electrical resistan ce, named after Georg Simon Ohm. Commonly used multiples and submultiples in electrical and electronic usage are the milliohm (1x10 −3 ), kilohm (1x10 3 ), and megohm (1x10 6 ). Theory of operation Ohm's law The behavior of an ideal resistor is dictated by the r elationship specified in Ohm's law: Ohm's law states that the voltage (V) across a resistor is proportional to the current (I) through it where the constant of proportionality is the resistance (R). Equivalently, Ohm's law can be stated: This formulati on of Ohm's law states that, when a voltage (V) is maintained across a resistance (R), a current (I) will flow through the resistance. This formulation is often used in practice. For example, if V is 12 volts and R is 400 ohms, a current of 12 / 400 = 0.03 amperes will flow through the resistance R. Series and parallel resistors Resistors in a parallel configuration each have the same potential difference (voltage). To find their total equivalent resistance ( R eq ): ________________________ WORLD TECHNOLOGIES ________________________ The parallel property can be represented in equations by two vertical lines || (as in geometry) to simplify equations. For two resistors, The current through Resistors in Series stays the same, but the voltage across each resistor can be different. The sum of the potential differences (voltage) is equal to the total voltage. To find their total resistance: A resistor network that is a combination of parallel and series can be broken up into smaller parts that are either one or the other. For instance, However, many resistor networks cannot be split up in this way. Consider a cube, each edge of which has been replaced by a resistor.
eBook - PDF- Leslie Basford(Author)
- 2013(Publication Date)
- Made Simple(Publisher)
Resistance 23 the output voltage is tapped off between the sliding contact C and the fixed end B. When the sliding contact is moved to A it taps off the whole of the input voltage E. When it is moved to B there is no output voltage at all. At some intermediate point such as C, the output voltage is K, as shown. Applying Ohm's law to the input circuit: E=IR Applying Ohm's law to the output circuit: V=IR' Dividing the second equation by the first: E R Hence the ratio of the output to input voltages is the ratio of the tapped-off resistance to the total resistance of the potentiometer. The same formula would apply, of course, if the variable resistor were replaced by two fixed resistors between AC and BC. R E S I S T O R S I N S E R I E S A N D I N P A R A L L E L When a number of components are wired up end to end in a continu-ous chain they are said to be joined in series. Fairy lights on a Christmas tree are usually wired in series. If one bulb is removed all the rest go out because the same current flows through every bulb in the chain. The current is the same everywhere throughout a series circuit. When components are wired up in a kind of ladder arrangement, with all the left-hand ends connected together, and all the right-hand ends connected together, they are said to be joined in parallel. Ordinary room lights are wired in parallel across the mains supply. If one bulb is removed the others will not be affected. Fig. 14 shows three resistors joined in series and in parallel. In the first case, the current has to pass through all three resistors; the total resistance met by the current is obviously the sum of the individual resistances (in this case 11 ft). Using symbols, the total or combined resistance is R = R x + R 2 + Ra + any number of additional resistances In the case of resistors joined in parallel, the total or combined resistance R is given by the formula 24 Electricity Made Simple Applying this formula to the resistors in Fig.
eBook - PDF- Michael Brumbach(Author)
- 2016(Publication Date)
- Cengage Learning EMEA(Publisher)
59 C H A P T E R 5 Basic Resistive Electrical Circuits O B J E C T I V E S After studying this chapter, the student will be able to: ■ Identify a series resistive circuit. ■ Identify a parallel resistive circuit. ■ Identify a combination resistive circuit. ■ Perform all necessary calculations to analyze a resistive electrical circuit. Electrical circuits are the building blocks that comprise all electrical devices, no matter how simple or complex. A technician must gain a solid understanding of the basics if he or she ever hopes to understand more complicated devices. Regardless of the complexity, all electrical circuits consist of series, parallel, or combination circuits. This chapter will introduce you to these fundamental circuits upon which more complex circuits are built. Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 60 CHAPTER 5 circuit consists of a power source (indicated by the AC sine wave), conductors, loads (the three light bulbs), and a control device (the switch). When you look at this circuit, imagine yourself as the current flowing from the power source. As you leave the power source, you flow through the conductor to the switch. In Figure 5–2, the switch is open. This means that the current flow cannot continue to the loads. There is not a complete path because of the open switch. Therefore, the light bulbs will not light. Now look at Figure 5–3.
eBook - ePub- John E. Ayers(Author)
- 2024(Publication Date)
- CRC Press(Publisher)
CDR may be used to find how a total current splits among parallel resistances. It states that the fraction of total current flowing in one parallel resistor is equal to the equivalent resistance for the other resistors divided by the sum of the equivalent resistance for the other resistors and the resistance for the branch in question.Some configurations or resistors may not be simplified by the use of parallel and series combinations. These configurations generally contain wye-connected or delta-connected combinations of resistors, and an example is the Wheatstone bridge. Circuits of this type may be simplified by using delta–wye or wye–delta transformations.The principle of superposition is a useful tool for solving some circuit problems. It states that when a linear system is driven by more than one independent source, the total response is the sum of the individual responses associated with each of the independent sources. Here, the total response is a voltage or a current, and we can find it by summing the individual responses, each determined by leaving one independent source active while deactivating all others.Problems
- Problem 1.1. Consider the electrical circuit of Figure P1.1 containing a battery and two resistors.
Long Description for Figure P1.1Circuit involving the series connection of a battery and two resistors in a single mesh. The positive terminal of a 9 V battery is connected to one terminal of a resistor having the color code red, yellow, brown, gold. The other terminal of this resistor is connected to a second resistor with the color code brown, blue, brown, gold. The other terminal of this resistor is connected to the negative terminal of the 9 V battery.
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