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Voltage Divider Networks 79
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a voltage-divider circuit. Illustration for Quiz Questions 19 and 20.
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Problem 5-15 Suppose you are building an electronic circuit, and the battery supplies 9.0 V. The minus terminal is at common (chassis) ground. You need to provide a circuit point where the dc voltage is +2.5 V. Give an example of a pair of resistors that can be connected in a voltage divider configuration, such that +2.5 V appears at some point. Examine the schematic diagram of Fig. 5-8. There are infinitely many different combinations of resistances that will work here! Pick some total value, say R = R1 + R2 = 1000 . Keep in mind that the ratio R1:R will always be the same as the ratio E1:E. In this case, E1 = 2.5 V, so E1:E = 2.5/9.0 = 0.28. This means that you want the ratio R1:R to be equal to 0.28. You have chosen to
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5-8 A voltage divider
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network in which 2.5 V dc is derived from a 9.0-V dc source.
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make R equal to 1000 . This means R1 must be 280 in order to get the ratio R1:R = 0.28. The value of R2 is the difference between R and R1. That is 1000 280 = 720 . In a practical circuit, you would want to choose the smallest possible value for R. This might be less than 1000 , or it might be more, depending on the nature of the circuit and the currentdelivering capability of the battery. It s not the actual values of R1 and R2 that determine the voltage you get at the intermediate point, but their ratio.
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Problem 5-16 What is the current I, in milliamperes, drawn by the entire network of series resistances in the situation described in Problem 5-15 and its solution Use Ohm s Law to get I = E/R = 9.0/1000 = 0.0090 A = 9.0 mA. Problem 5-17 Suppose that it is all right for the voltage divider network to draw up to 100 mA of current in the situation shown by Fig. 5-8 and posed by Problem 5-15. You want to design the network to draw this amount of current, because that will offer the best voltage regulation for the circuit to be operated from the network. What values of resistances R1 and R2 should you use Calculate the total resistance first, using Ohm s Law. Remember to convert 100 mA to amperes! That means you use the figure I = 0.100 A in your calculations. Then R = E/I = 9.0/0.100 = 90 . The ratio of resistances that you need is R1:R2 = 2.5/9.0 = 0.28. You should use R1 = 0.28 90 = 25 . The value of R2 is the difference between R and R1. That is, R2 = R R1 = 90 25 = 65 .
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Quiz
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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. In a series-connected string of ornament bulbs, if one bulb gets shorted out, which of the following will occur (a) All the other bulbs will go out. (b) The current in the string will go up. (c) The current in the string will go down. (d) The current in the string will stay the same. 2. Imagine that four resistors are connected in series across a 6.0-V battery, and the ohmic values are R1 = 10 , R2 = 20 , R3 = 50 , and R4 = 100 , as shown in Fig. 5-9. What is the voltage across the resistance R2 (a) 0.18 V (b) 33 mV (c) 5.6 mV (d) 0.67 V
Quiz
5-9 Illustration for Quiz
Questions 2, 3, 8, and 9. Resistance values are in ohms.
3. In the scenario shown by Fig. 5-9, what is the voltage across the combination of R3 and R4 (a) 0.22 V (b) 0.22 mV (c) 5.0 V (d) 3.3 V 4. Suppose three resistors are connected in parallel across a battery that delivers 15 V, and the ohmic values are R1 = 470 , R2 = 2.2 k , and R3 = 3.3 k , as shown in Fig. 5-10. The voltage across the resistance R2 is (a) 4.4 V. (b) 5.0 V. (c) 15 V. (d) not determinable from the data given. 5-10 Illustration for Quiz