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Question 15. Resistance values are in ohms.
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68 Direct-Current Circuit Basics
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16. Imagine three resistors, with values of 47 , 68 , and 82 , 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.
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Question 16. Resistance values are in ohms.
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17. Suppose you have an unlimited supply of 1-W, 100- resistors. You need to get a 100- , 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- resistors, and you need a 500- 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- resistors, and you need to get a 3000- , 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.
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CHAPTER
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Direct-Current Circuit Analysis
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IN THIS CHAPTER, YOU LL LEARN MORE ABOUT DC CIRCUITS AND HOW THEY BEHAVE UNDER VARIOUS
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conditions. These principles apply to most ac utility circuits as well.
Current through Series Resistances
Have you ever used those tiny holiday lights that come in strings If one bulb burns out, the whole set of bulbs goes dark. Then you have to find out which bulb is bad, and replace it to get the lights working again. Each bulb works with something like 10 V; there are about a dozen bulbs in the string. You plug in the whole bunch and the 120-V utility mains drive just the right amount of current through each bulb. In a series circuit, such as a string of light bulbs (Fig. 5-1), the current at any given point is the same as the current at any other point. The ammeter, A, is shown in the line between two of the bulbs. If it were moved anywhere else along the current path, it would indicate the same current.
5-1 Light bulbs in series, with an ammeter (A) in the circuit.
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70 Direct-Current Circuit Analysis
This is true in any series dc circuit, no matter what the components actually are, and regardless of whether or not they all have the same resistance. If the bulbs in Fig. 5-1 had different resistances, some of them would consume more power than others. In case one of the bulbs in Fig. 5-1 burns out, and its socket is then shorted out instead of filled with a replacement bulb, the current through the whole chain will increase, because the overall resistance of the string will go down. This will force each of the remaining bulbs to carry more current, and pretty soon another bulb would burn out because of the excessive current. If it, too, were replaced with a short circuit, the current would be increased still further. A third bulb would blow out almost right away thereafter.
Voltages across Series Resistances
The bulbs in the string of Fig. 5-1, being all the same, each get the same amount of voltage from the source. If there are a dozen bulbs in a 120-V circuit, each bulb has a potential difference of 10 V across it. This will remain true even if the bulbs are replaced with brighter or dimmer ones, as long as all the bulbs in the string are identical. Look at the schematic diagram of Fig. 5-2. Each resistor carries the same current. Each resistance Rn has a potential difference En across it equal to the product of the current and the resistance of that particular resistor. The voltages En are in series, like cells in a battery, so they add together. What if the voltages across all the resistors added up to something more or less than the supply voltage, E Then there would have to be a phantom EMF someplace, adding or taking away voltage. But that s impossible. Voltage cannot come out of nowhere! Look at this another way. The voltmeter V in Fig. 5-2 shows the voltage E of the battery, because the meter is hooked up across the battery. The voltmeter V also shows the sum of the voltages En across the set of resistances, because it s connected across the whole combination. The meter says the same thing whether you think of it as measuring the battery voltage E or as measuring the sum of the voltages En across the series combination of resistances. Therefore, E is equal to the sum of the voltages En. How do you find the voltage across any particular resistance Rn in a circuit like the one in Fig. 5-2 Remember Ohm s Law for finding voltage: E = IR. Remember, too, that you must use volts, ohms, and amperes when making calculations. In order to find the current in the circuit, I, you need to know the total resistance and the supply voltage; then I = E/R. First find the current in the whole circuit; then find the voltage across any particular resistor.
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