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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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Problem 5-12

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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In Fig. 5-5B, suppose each of the two resistors below point Z has a value of 100 , and all three resistors above Z are 10.0 . The current through each 100- resistor is 500 mA. What is the current through any of the 10.0- resistors, assuming it is equally distributed What is the voltage, then, across any of the 10. 0- resistors The total current into Z is 500 mA + 500 mA 1.00 A. This must be divided three ways equally among the 10- resistors. Therefore, the current through any one of them is 1.00/3 A 0.333 A 333 mA. The voltage across any one of the 10.0- resistors is found by Ohm s Law: E IR 0.333 10.0 3.33 V.

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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