Physics
Resistors in Parallel
When resistors are connected in parallel, they are arranged so that both ends of each resistor are connected to the same two points. In this configuration, the total resistance decreases compared to a single resistor, as the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistances. This allows for increased current flow in the circuit.
Written by Perlego with AI-assistance
Related key terms
1 of 5
11 Key excerpts on "Resistors in Parallel"
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)
Moreover, if the current in one device is interrupted (perhaps by an opened switch or a broken wire), the current in the other devices is not interrupted. In contrast, if household appliances were connected in series, there would be no current through any appliance if the current in the circuit were halted at any point. When two resistors R 1 and R 2 are connected as in figure 20.18, each receives current from the battery as if the other were not present. Therefore, R 1 and R 2 together draw more current from the battery than does either resistor alone. According to the definition of resistance, R = V /I, a larger current implies a smaller resistance. Thus, the two parallel resistors behave as a single equivalent resistance that is smaller than either R 1 or R 2 . Figure 20.20 returns to the water‐flow analogy to provide additional insight into this important feature of parallel wiring. In part a, two sections of pipe that have the same length are connected in parallel with a pump. In part b these two sections have been replaced with a single pipe of the same length, whose cross‐sectional area equals the combined cross‐sectional areas of section 1 and section 2. The pump (analogous to a voltage source) can push more water per second (analogous to current) through the wider pipe in part b (analogous to a wider wire) than it can through either of the narrower pipes (analogous to narrower wires) in part a. In effect, the wider pipe offers less resistance to the flow of water than either of the narrower pipes offers individually. 552 Physics FIGURE 20.20 (a) Two equally long pipe sections, with cross‐sectional areas A 1 and A 2 , are connected in parallel to a water pump. (b) The two parallel pipe sections in part a are equivalent to a single pipe of the same length whose cross‐sectional area is A 1 + A 2 .
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- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
(b) This drawing is equivalent to part a. I 1 and I 2 are the currents in R 1 and R 2 . *For clarity, two batteries are shown in Figure 20.17b, one associated with each resistance coil. In reality, both coils are connected across a single battery. 556 Chapter 20 | Electric Circuits Parallel wiring is very common. For example, when an electrical appliance is plugged into a wall socket, the appliance is connected in parallel with other appliances, as in Figure 20.19, where the entire voltage of 120 V is applied across each one of the devices: the television, the stereo, and the light bulb (when the switch is turned on). The presence of the unused socket or other devices that are turned off does not affect the operation of those devices that are turned on. Moreover, if the current in one device is interrupted (perhaps by an opened switch or a broken wire), the current in the other devices is not interrupted. In contrast, if household appliances were connected in series, there would be no current through any appliance if the current in the circuit were halted at any point. When two resistors R 1 and R 2 are connected as in Figure 20.18, each receives current from the battery as if the other were not present. Therefore, R 1 and R 2 together draw more current from the battery than does either resistor alone. According to the definition of resis- tance, R 5 V/I, a larger current implies a smaller resistance. Thus, the two parallel resistors behave as a single equivalent resistance that is smaller than either R 1 or R 2 . Figure 20.20 returns to the water-flow analogy to provide additional insight into this important feature of parallel wiring. In part a, two sections of pipe that have the same length are connected in parallel with a pump. In part b these two sections have been replaced with a single pipe of the same length, whose cross-sectional area equals the combined cross-sectional areas of section 1 and section 2.
eBook - PDF- Leslie Basford(Author)
- 2013(Publication Date)
- Made Simple(Publisher)
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. 14, 1 = 1 + 1 + 1 R 2 + 3 + 6 _ 3 + 2 + l 6 A ~ 3 + 2 + 1 = i n So the total resistance is less than any one of the resistors. SERIES 2 ohms 3 ohms 6 ohms total resistance 11 ohms PARALLEL 2 ohms 3 ohms 6 ohms total resistance 1 ohm Fig. 14. Resistors joined in series and in parallel How the formula is obtained for three Resistors in Parallel is shown in Fig. 15; the same reasoning can be applied to any number of resistors. The total current flowing is I amps. It divides (at A) into the three currents, /i, h, and h passing through the resistors JRI, 2 *2, and Rz respectively. Hence / = / l + / 2 + /3 Since all three resistors are connected between points A and B, the p.d. across each is the p.d. between A and B, namely V volts. Applying Ohm's law to each resistor in turn Resistance 25 h Applying Ohm's law to the whole network of resistors, Fig. 15. Circuit diagram of three Resistors in Parallel where R is the combined resistance of Ri, R 2 , and Rs 9 i.e. the value of the single resistor which could exactly replace the three. Substituting these relationships, the equation becomes dividing by V / = / i + / 2 + / 3 V V_ V v_ R Rx^ R 2 ^~ Rs R Rx^ R 2 ^ Rs EXAMPLE: What is the combined resistance of a 20-Cl resistor in parallel with an 80-iQ resistor? V Ri V R* V Rz 26 Electricity Made Simple 2on R2 = 80 ft 1 k + 1 R k + *a 1 20 + 1 R 20 + 80 = R = = 16 ft Answer (Again, the combined resistance is less than either of the individual resistors.) SERIES-PARALLEL CIRCUITS A circuit in which some components are joined in series while others are joined in parallel, as shown by the network of resistors in Fig.
eBook - PDF- Michael Neidle(Author)
- 2013(Publication Date)
- Butterworth-Heinemann(Publisher)
4.3. Parallel Circuits In the basic parallel circuit (Fig. 4.5), the main current I is the sum of the individual branch currents. Current law / = / 1 + / 2 + / 3 Since the voltage V is common and I = — V _ V V V where R is the equivalent ^ ^ ^ ^ resistance of the group 66 ELECTRICAL INSTALLATION TECHNOLOGY Fig. 4.5. Currents through resistors connected in parallel. I* It Ri I 2 R 2 I3 R3 -»»—1 h Dividing through by V, K /^! K 2 #3 Occasionally, this reciprocal form can be usefully replaced by J J / 2 ' ί | Λ 3 T /V2/V3 The expression above is obtained through using the product R l R 2 R 3 as an L.C.M. More often, for two Resistors in Parallel, the expression R = KjR 2 Ä!+R 2 makes for easier working. For series-parallel circuits, Resistors in Parallel should be seen as a single group in series with other resistors. Example 4.3 Four resistors, AB, BC, AD and DC, are connected together to form a closed square ABCD. The known resistance values are: AD 12 Ω, AB 35 Ω, and DC 12 Ω. A d.c. supply of 120 V is connected to A and C so that current enters the combination at A and leaves at C. A high-resistance voltmeter is connected between B and D and, whilst carrying negligible current, registers a voltage drop of 10 V from B to D. (a) Calculate the value of the resistance BC, and the total current taken from the supply. (b) Calculate also the value of BC, such that the potential difference between B and D is in the reverse direction i.e. from D to B. [C] D.C. CIRCUITS 67 -120V Fig. 4.6. Circuit conditions of balanced bridge. (a) The circuit conditions are given in Fig. 4.6. The supply current / enters the junction at A and splits up into the currents l x and I 2 as shown; also the network consists of two parallel branches each comprising two resistors in series, since negligible current passes through the voltmeter. As the resistors AD and DC are equal and form one parallel branch, a p.d. of 60 V must exist between the ends of these resistors.
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.
- Frank R. Spellman, Nancy E. Whiting(Authors)
- 2013(Publication Date)
- CRC Press(Publisher)
11.7.7.7 Product Over the Sum Method When any two unequal resistors are in parallel, it is often easier to calculate the total resistance by multiplying the two resistances and then dividing the product by the sum of the resistances: R R R R R T = × + 1 2 1 2 (11.38) where R T is the total resistance in parallel, and R 1 and R 2 are the two Resistors in Parallel. R 1 2 ohms R 2 4 ohms R 3 8 ohms R T = ? FIGURE 11.50 Illustration for Example 11.41. 284 Handbook of Mathematics and Statistics for the Environment ■ EXAMPLE 11.42 Problem: What is the equivalent resistance of a 20-ohm and a 30-ohm resistor connected in parallel? Solution: Given: R 1 = 20 ohms R 2 = 30 ohms R R R R R T = × + = × + = 1 2 1 2 20 30 20 30 12 ohms 11.7.7.8 Power in Parallel Circuits As in the series circuit, the total power consumed in a parallel circuit is equal to the sum of the power consumed in the individual resistors. Note: Because power dissipation in resistors consists of a heat loss, power dissipations are additive regardless of how the resistors are connected in the circuit. P T = P 1 + P 2 + P 3 + … + P n (11.39) where P T is the total power, and P 1 , P 2 , P 3 , … P n are the branch powers. Total power can also be calculated by the following equation: P T = E × I T (11.40) where P T is the total power, E is the voltage source across all parallel branches, and I T is the total current. The power dissipated in each branch is equal to E × I and equal to V 2 / R . Note: In both parallel and series arrangements, the sum of the individual values of power dissi-pated in the circuit equals the total power generated by the source. The circuit arrangements cannot change the fact that all power in the circuit comes from the source. 11.7.8 S ERIES –P ARALLEL C IRCUITS So far, we have discussed series and parallel DC circuits; however, operators will seldom encounter a circuit that consists solely of either type of circuit. Most circuits consist of both series and parallel elements.
eBook - PDF- Richard C. Dorf, James A. Svoboda(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
The power absorbed by the three resistances is now p 24 2 50,000 70,000 5 mW Finally, the power supply current is i a 24 50,000 70,000 0 2 mA which is well below the 100 mA that the voltage sources are able to supply. The design is complete. 94 CHAPTER 3 Resistive Circuits 3.10 S U M M A R Y Kirchhoff’s current law (KCL) states that the algebraic sum of the currents entering a node is zero. Kirchhoff’s voltage law (KVL) states that the algebraic sum of the voltages around a closed path (loop) is zero. Simple electric circuits can be analyzed using only Kirchhoff’s laws and the constitutive equations of the circuit elements. Series resistors act like a “voltage divider,” and parallel resistors act like a “current divider.” The first two rows of Table 3.10-1 summarize the relevant equations. Series resistors are equivalent to a single “equivalent resis- tor.” Similarly, parallel resistors are equivalent to a single “equivalent resistor.” The first two rows of Table 3.10-1 summarize the relevant equations. Series voltage sources are equivalent to a single “equivalent voltage source. ” Similarly, parallel current sources are equivalent to a single “equivalent current. ” The last two rows of Table 3.10-1 summarize the relevant equations. Often circuits consisting entirely of resistors can be reduced to a single equivalent resistor by repeatedly replacing series and/or parallel resistors by equivalent resistors.
- 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)
STEP 2. Using Ohm’s law, define a voltage across each resistor in terms of the defined current. STEP 3. Apply KVL to the single-loop circuit. STEP 4. Solve the single KVL equation for the current i(t). If i(t) is positive, the current is flowing in the direction assumed; if not, then the current is actually flowing in the opposite direction. SINGLE-LOOP CIRCUITS SECTION 2.4 SINGLE-NODE-PAIR CIRCUITS 45 R 2 R 1 υ(t) i(t) i 1 (t) i 2 (t) + − Figure 2.22 Simple parallel circuit. CURRENT DIVISION An important circuit is the single-node-pair circuit. If we apply KVL to every loop in a single-node-pair circuit, we discover that all of the elements have the same volt- age across them and, therefore, are said to be connected in parallel. We will, however, apply Kirchhoff’s current law and Ohm’s law to determine various unknown quantities in the circuit. Following our approach with the single-loop circuit, we will begin with the simplest case and then generalize our analysis. Consider the circuit shown in Fig. 2.22. Here we have an independent current source in parallel with two resistors. Since all of the circuit elements are in parallel, the voltage υ (t) appears across each of them. Furthermore, an examination of the circuit indicates that the current i(t) is into the upper node of the circuit and the currents i 1 (t) and i 2 (t) are out of the node. Since KCL essen- tially states that what goes in must come out, the question we must answer is how i 1 (t) and i 2 (t) divide the input current i(t). Applying Kirchhoff’s current law to the upper node, we obtain i(t) = i 1 (t) + i 2 (t) and, employing Ohm’s law, we have i(t) = υ (t) — R 1 + υ (t) — R 2 = ( 1 — R 1 + 1 — R 2 ) υ (t) = υ (t) — R p where 1 — R p = 1 — R 1 + 1 — R 2 2.16 R p = R 1 R 2 _______ R 1 + R 2 2.17 Therefore, the equivalent resistance of two resistors connected in parallel is equal to the product of their resistances divided by their sum.
Index pages curate the most relevant extracts from our library of academic textbooks. They’ve been created using an in-house natural language model (NLM), each adding context and meaning to key research topics.
