Problem 5-14In Fig. 5-6, suppose the battery provides 20 V. Let the resistors, having voltagedrops 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 resistancevalues. But you don’t need to know these things. Regardless of what the actual ohmicvalues are, the ratio E1:E2:E3:E4 will be the same. This is a sort of corollary to Kirch-hoffs Second Law. You can just invent certain ohmic values with the necessary ratio.Let’s have them be R11Ω, R22Ω, R33Ω, and R44Ω. Then the total resis-tance is RR1R2R3R4123410Ω. You can calculate the cur-rent as IE/R20/102 A. Then the voltage E3, across R3, is given by Ohm’s Lawas E3I(R3)236 V.You are encouraged to calculate the other voltages and observe that they add up to20 V.In this problem, there is freedom to literally pick numbers out of the air so that cal-culations are easy. You could have chosen ohmic values like 47, 94, 141, and 188 Ω(these too are in the ratio 1:2:3:4), and you’d still get E36 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 obtainvarious voltage ratios, to make circuits work just right. These resistance circuits arecalled voltage divider networks.Voltage divider networksEarlier, you were assured that this course would not drag you through ridiculously com-plicated circuits. You can imagine, by now nightmarish series-parallel matrixes of resis-tors drawn all over whole sheets of paper, captioned with wicked queries: “What is thecurrent through R135?” But that stuff is best left to professional engineers, and eventhey aren’t likely to come across it very often. Their job is to make things as neat and ef-ficient as possible. If an engineer actually is faced with such a scenario, the reaction willprobably be, “How can this circuit be simplified?”92 Direct-current circuit analysis5-6Kirchhoff’s Second Law. The sum of the voltages across the resistors is equal to, buthas opposite polarity from, the supply voltage E. Thus EE1E2E3E40.Also see quiz questions 15 and 16.