Copyright Goodheart-Willcox Co., Inc. 62 Electricity Copyright Goodheart-Willcox Co., Inc. Another circuit problem is presented in Figure 7-6. In this problem, we know the applied voltage (100 V) and the voltage drops across the resistors, but we do not know the value of the resistors. The total current is given as 100 mA (0.1 A). According to Ohm’s law: R = E __ I The voltage drop across R1 is 50 V, so: R1 = 50 V _____ 0.1 A = 500 Ω The voltage drop across R2 is 25 V, so: R2 = 25 V _____ 0.1 A = 250 Ω The voltage drop across R3 is 25 V, so: R3 = 25 V _____ 0.1 A = 250 Ω The problem can also be proved by working it backward. The total resistance equals R1 + R2 + R3 (500 Ω + 250 Ω + 250 Ω) equals 1000 Ω or 1 kΩ. Now, IT = ET ___ RT or I = 100 V _______ 1000 Ω = 0.1 A which proves that we did our calculations correctly. There are thousands of applications of series circuits and voltage drops in electrical and electronic circuits. Time used to practice these problems and formulas will be time well spent. Note that voltmeters are always connected in parallel or across the voltage drop to be measured. Applications Project 1—Experimenter, can be found in Chapter 22. Conduct the experiments labeled Problem 2 and Problem 3. Problem 2 involves the use of only one lamp. Problem 3 uses three lamps wired in series. The lights in Problem 3 will glow much more dimly than the single lamp used in Problem 2. Why is this so? The answer can be found in Ohm’s law. Each light is a resistance unit. When three lights are in series, consider the sum of the resistances. Increased resistance will decrease the current. R3 = ? I = 100 mA or 0.1 A E = 100 V R2 = ? R1 = ? V 25 V 25 V V V 50 V Voltmeter Goodheart-Willcox Publisher Figure 7-6. The voltage drops and current are known. What are the values of the resistors?