Sets of resistors, all having identical ohmic values, can be connected together in parallel sets of series networks, or in series sets of parallel networks. By doing this, the total power-handling capacity of the resistance can be greatly increased over that of a single resistor.
Sometimes, the total resistance of a series-parallel networkis the same as the value of any one of the resistors. This is always true if the components are identical, and are in a network called an
n-by-n matrix.That means, when nis a whole number, there are nparallel sets of n resistors in series (Fig. 4-10A), or else there are nseries sets of nresistors in parallel (Fig. 4-10B). Either arrange- ment gives the same practical result.
Engineers and technicians sometimes use series-parallel networks to obtain resistances with large power-handling capacity. A series-parallel array of nby nresistors will have n2times that of a
single resistor. Thus, a 3×3 series-parallel matrix of 2 W resistors can handle up to 32×2=9×
4-10 Series-parallel resistances. At A, sets of series resistors are connected in parallel. At B, sets of parallel resistances are connected in series. These examples show symmetricaln-by-n
2=18 W, for example. A 10 ×10 array of 1-W resistors can dissipate up to 100 W. The total power- handling capacity is multiplied by the total number of resistors in the matrix. But this is true only if all the resistors have the same ohmic values, and the same power-dissipation ratings.
It is unwise to build series-parallel arrays from resistors with different ohmic values or power ratings. If the resistors have values and/or ratings that are even a little nonuniform, one of them might be subjected to more current than it can withstand, and it will burn out. Then the current distribution in the network can change so a second component fails, and then a third. It’s hard to predict the current and power distribution in an array when its resistor values are all different.
If you need a resistance with a certain power-handling capacity, you must be sure the network can handle at least that much power. If a 50-W rating is required, and a certain combination will handle 75 W, that’s fine. But it isn’t good enough to build a circuit that will handle only 48 W. Some extra tolerance, say 10 percent over the minimum rating needed, is good, but it’s silly to make a 500-W network using far more resistors than necessary, unless that’s the only convenient combina- tion given the parts available.
Nonsymmetrical series-parallel networks, made up from identical resistors, can increase the power- handling capability over that of a single resistor. But in these cases, the total resistance is not the same as the value of the single resistors. The overall power-handling capacity is always multiplied by the total number of resistors, whether the network is symmetrical or not, provided all the ohmic values are iden- tical. In engineering work, cases sometimes arise where nonsymmetrical networks fit the need.
Quiz
Refer to the text in this chapter if necessary. A good score is at least 18 correct answers. The answers are in the back of the book.
1. Suppose you double the voltage in a simple dc circuit, and cut the resistance in half. The current will
(a) become four times as great. (b) become twice as great. (c) stay the same as it was before. (d) become half as great.
2. You can expect to find a wiring diagram
(a) on a sticker on the back of a television receiver. (b) in an advertisement for an electric oven.
(c) in the service/repair manual for a two-way radio.
(d) in the photograph of the front panel of a stereo hi-fi tuner.
For questions 3 through 11, please refer to Fig. 4-7. Remember to take significant figures into account when completing your calculations!
3. Given a dc voltage source delivering 24 V and a resistance of 3.3 kΩ, what is the current? (a) 0.73 A
(b) 138 A (c) 138 mA (d) 7.3 mA
4. Suppose the resistance is 472 Ω, and the current is 875 mA. The source voltage must therefore be
(a) 413 V. (b) 0.539 V. (c) 1.85 V.
(d) none of the above.
5. Suppose the dc voltage is 550 mV and the current is 7.2 mA. Then the resistance is (a) 0.76 Ω.
(b) 76 Ω. (c) 0.0040 Ω. (d) none of the above.
6. Given a dc voltage source of 3.5 kV and a resistance of 220 Ω, what is the current? (a) 16 mA
(b) 6.3 mA (c) 6.3 A
(d) None of the above
7. Suppose the resistance is 473,332 Ω, and the current flowing through it is 4.4 mA. The best expression for the voltage of the source is
(a) 2082 V. (b) 110 kV. (c) 2.1 kV. (d) 2.08266 kV.
8. A source delivers 12 V and the current is 777 mA. The best expression for the resistance is (a) 15 Ω.
(b) 15.4 Ω. (c) 9.3 Ω. (d) 9.32 Ω.
9. Suppose the voltage is 250 V and the current is 8.0 mA. The power dissipated by the potentiometer is
(a) 31 mW. (b) 31 W. (c) 2.0 W. (d) 2.0 mW.
10. Suppose the voltage from the source is 12 V and the potentiometer is set for 470 Ω. The power dissipated in the resistance is approximately
(a) 310 mW. (b) 25.5 mW. (c) 39.2 W. (d) 3.26 W.
11. If the current through the potentiometer is 17 mA and its resistance is set to 1.22 kΩ, what is the power dissipated by it?
(a) 0.24 µW (b) 20.7 W (c) 20.7 mW (d) 350 mW
12. Suppose six resistors are hooked up in series, and each of them has a value of 540 Ω. What is the resistance across the entire combination?
(a) 90 Ω (b) 3.24 kΩ (c) 540 Ω
(c) None of the above
13. If four resistors are connected in series, each with a value of 4.0 kΩ, the total resistance is
(a) 1 kΩ. (b) 4 kΩ. (c) 8 kΩ. (d) 16 kΩ.
14. Suppose you have three resistors in parallel, each with a value of 0.069 MΩ. Then the total resistance is
(a) 23 Ω. (b) 23 kΩ. (c) 204 Ω. (d) 0.2 MΩ.
15. Imagine three resistors in parallel, with values of 22 Ω, 27 Ω, and 33 Ω. If a 12-V battery is connected across this combination, as shown in Fig. 4-11, what is the current drawn from the battery? (a) 1.4 A (b) 15 mA (c) 150 mA (d) 1.5 A Quiz 67
4-11 Illustration for Quiz Question 15. Resistance values are in ohms.
16. Imagine three resistors, with values of 47 Ω, 68 Ω, and 82 Ω, connected in series with a 50-V dc generator, as shown in Fig. 4-12. The total power consumed by this network of resistors is
(a) 250 mW. (b) 13 mW. (c) 13 W.
(d) impossible to determine from the data given.
4-12 Illustration for Quiz Question 16. Resistance values are in ohms.
17. Suppose you have an unlimited supply of 1-W, 100-Ωresistors. You need to get a 100-Ω, 10-W resistor. This can be done most cheaply by means of a series-parallel matrix of
(a) 3 ×3 resistors. (b) 4 ×3 resistors. (c) 4 ×4 resistors. (d) 2 ×5 resistors.
18. Suppose you have an unlimited supply of 1-W, 1000-Ωresistors, and you need a 500-Ω resistance rated at 7 W or more. This can be done by assembling
(a) four sets of two resistors in series, and connecting these four sets in parallel. (b) four sets of two resistors in parallel, and connecting these four sets in series. (c) a 3 ×3 series-parallel matrix of resistors.
(d) a series-parallel matrix, but something different than those described above.
19. Suppose you have an unlimited supply of 1-W, 1000-Ωresistors, and you need to get a 3000-Ω, 5-W resistance. The best way is to
(a) make a 2 ×2 series-parallel matrix. (b) connect three of the resistors in parallel. (c) make a 3 ×3 series-parallel matrix. (d) do something other than any of the above.
20. Good engineering practice usually requires that a series-parallel resistive network be assembled (a) from resistors that are all different.
(b) from resistors that are all identical.
(c) from a series combination of resistors in parallel but not from a parallel combination of resistors in series.
(d) from a parallel combination of resistors in series, but not from a series combination of resistors in parallel.
IN THIS CHAPTER,YOU’LL LEARN MORE ABOUT DC CIRCUITS AND HOW THEY BEHAVE UNDER VARIOUS
conditions. These principles apply to most ac utility circuits as well.