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In Fig. 5-6, suppose the battery provides 20 V. Let the resistors, having voltage drops E1, E2, E3, and E4, have their ohmic values in the ratio 1:2:3:4 respectively. What is E3 This problem does not tell you the current in the circuit, nor the exact resistance values. But you don t need to know these things. Regardless of what the actual ohmic values are, the ratio E1:E2:E3:E4 will be the same. This is a sort of corollary to Kirchhoffs Second Law. You can just invent certain ohmic values with the necessary ratio. Let s have them be R1 1 , R2 2 , R3 3 , and R4 4 . Then the total resistance is R R1 R2 R3 R4 1 2 3 4 10 . You can calculate the current as I E/R 20/10 2 A. Then the voltage E3, across R3, is given by Ohm s Law as E3 I(R3) 2 3 6 V. You are encouraged to calculate the other voltages and observe that they add up to 20 V. In this problem, there is freedom to literally pick numbers out of the air so that calculations are easy. You could have chosen ohmic values like 47, 94, 141, and 188 (these too are in the ratio 1:2:3:4), and you d still get E3 6 V. (Go ahead and try it.) But that would have made needless work for yourself. Series combinations of resistors are often used by electronic engineers to obtain various voltage ratios, to make circuits work just right. These resistance circuits are called voltage divider networks.

(c) 036 A (d) Not determinable from the information given 14 Look at Fig 5-5B Let each resistor have a value of 820 Suppose the top three resistors all lead to identical light bulbs If I1 = 50 mA and I2 = 70 mA, what is the power dissipated in the resistor carrying current I4 (a) 33 W (b) 40 mW (c) 13 W (d) It can t be found using the information given 15 Refer to Fig 5-6 Suppose the resistances R1, R2, R3, and R4 are in exactly the ratio 1:2:4:8 from left to right, and the battery supplies 30 V What is the voltage E2 (a) 40 V (b) 80 V (c) 16 V (d) It is not determinable from the data given 16 Refer to Fig 5-6 Suppose the resistances are each 33 k , and the battery supplies 12 V.

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Mainframes were fine for the computing requirements of large, homogeneous user communities, but as the technology became cheaper and more.

Earlier, you were assured that this course would not drag you through ridiculously complicated circuits. You can imagine, by now nightmarish series-parallel matrixes of resistors drawn all over whole sheets of paper, captioned with wicked queries: What is the current through R135 But that stuff is best left to professional engineers, and even they aren t likely to come across it very often. Their job is to make things as neat and efficient as possible. If an engineer actually is faced with such a scenario, the reaction will probably be, How can this circuit be simplified

If the plus terminal of a dc voltmeter is placed between resistances R1 and R2 (with voltages E1 and E2 across them, respectively), and the minus terminal of the voltmeter is placed between resistances R3 and R4 (with voltages E3 and E4 across them, respectively), what will the meter register (a) 00 V (b) 30 V (c) 60 V (d) 12 V 17 In a voltage divider network, the total resistance (a) should be large to minimize current drain (b) should be as small as the power supply will allow (c) is not important (d) should be such that the current is kept to 100 mA 18 The maximum voltage output from a voltage divider (a) is a fraction of the power supply voltage (b) depends on the total resistance (c) is equal to the supply voltage (d) depends on the ratio of resistances 19.

Voltage divider networks 93 Resistances in series produce ratios of voltages, and these ratios can be tailored to meet certain needs. When designing voltage divider networks, the resistance values should be as small as possible, without causing too much current drain on the 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, voltages will be E1, E2, E3, ..., En. The last voltage, En, is the same as the supply voltage, E. All the other voltages are less than E, so E1 E2 E3 ... En E. (The symbol means is less than. )

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