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Problem 5-13 Refer to the diagram of Fig. 5-6. Suppose the four resistors have values of 50 , 60 , 70 , and 80 , and that the current through each of them is 500 mA. What is the battery voltage, E Find the voltages E1, E2, E3, and E4 across each of the resistors. This can be done using Ohm s Law. For E1, say with the 50- resistor, calculate E1 = 0.500 50 = 25 V. In the same way, you can calculate E2 = 30 V, E3 = 35 V, and E4 = 40 V. The supply voltage is the sum E1 + E2 + E3 + E4 = 25 + 30 + 35 + 40 = 130 V. Kirchhoff s Second Law tells us that the polarities of the voltages across the resistors are in the opposite direction from that of the battery. Problem 5-14 In the situation shown by Fig. 5-6, suppose the battery provides 20 V. Suppose the resistors labeled with voltages E1, E2, E3, and E4 have ohmic values in the ratio 1:2:3:4 respectively. What is the voltage E3 This problem does not provide any information about current in the circuit, nor does it give you the exact resistances. But you don t need to know these things to solve for E3. Regardless of what the actual ohmic values are, the ratio E1:E2:E3:E4 will be the same as long as the resistances are in the ratio 1:2:3:4. We can plug in any ohmic values we want for the values of the resistors, as long as they are in that ratio. Let Rn be the resistance across which the voltage is En, where n can range from 1 to 4. Now that we have given the resistances specific names, suppose R1 = 1.0 , R2 = 2.0 , R3 = 3.0 , and R4 = 4.0 . These are in the proper ratio. The total resistance is R = R1 + R2 + R3 + R4 = 1.0 + 2.0 + 3.0 + 4.0 = 10 . You can calculate the current as I = E/R = 20/10 = 2.0 A. Then the voltage E3, across the resistance R3, is given by Ohm s Law as E3 = IR3 = 2.0 3.0 = 6.0 V.
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Voltage Divider Networks
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Resistances in series produce ratios of voltages, and these ratios can be tailored to meet certain needs by means of voltage divider networks. When a voltage divider network is designed and assembled, the resistance values should be as small as possible without causing too much current drain on the battery or power supply. (In practice, the optimum values depend on the nature of the circuit being designed. This is a matter for engineers, and specific details are beyond the scope of this course.) The reason for choosing the smallest possible resistances is that, when the divider is used with a circuit, you do not want that circuit to upset the operation of the divider. The voltage divider fixes the intermediate voltages best when the resistance values are as small as the current-delivering capability of the power supply will allow. Figure 5-7 illustrates the principle of voltage division. The individual resistances are R1, R2, R3, . . . , Rn. The total resistance is R = R1 + R2 + R3 + . . . + Rn. The supply voltage is E, and the current in the circuit is therefore I = E/R. At the various points P1, P2, P3, . . . , Pn, the potential differences relative to the negative battery terminal are E1, E2, E3, . . . , En, respectively. The last voltage, En, is the same as the battery voltage, E. All the other voltages are less than E, and ascend in succession, so that E1 < E2 < E3 < . . . < En. (The mathematical symbol < means is less than. ) The voltages at the various points increase according to the sum total of the resistances up to each point, in proportion to the total resistance, multiplied by the supply voltage. Thus, the voltage E1 is equal to ER1/R. The voltage E2 is equal to E(R1 + R2)/R. The voltage E3 is equal to E(R1 + R2 + R3)/R. This process goes on for each of the voltages at points all the way up to En = E(R1 + R2 + R3 + . . . + Rn)/R = ER/R = E.
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