calculations and find that the total resistance, R, across the battery, E, is 66.67 Ω. ThenEIR66.673.0200 volts. (Some battery.)Problem 5-12In Fig. 5-5B, suppose each of the two resistors below point Z has a value of 100 Ω, andall three resistors above Z are 10.0 Ω. The current through each 100-Ω resistor is 500mA. What is the current through any of the 10.0-Ω resistors, assuming it is equally dis-tributed? What is the voltage, then, across any of the 10. 0-Ω resistors?The total current into Z is 500 mA + 500 mA 1.00 A. This must be divided threeways equally among the 10-Ω resistors. Therefore, the current through any one of themis 1.00/3 A 0.333 A 333 mA.The voltage across any one of the 10.0-Ω resistors is found by Ohm’s Law: EIR0.33310.03.33 V.Kirchhoff’s second lawThe sum of all the voltages, as you go around a circuit from some fixed point and returnthere from the opposite direction, and taking polarity into account, is always zero.This might sound strange. Surely there is voltage in your electric hair dryer, or ra-dio, or computer. Yes, there is, between different points. But no point can have an EMFwith respect to itself. This is so simple that it’s almost laughable. A point in a circuit isalways shorted out to itself.What Kirchhoff really was saying, when he wrote his second law, is a more generalversion of the second and third points previously mentioned. He reasoned that voltagecannot appear out of nowhere, nor can it vanish. All the potential differences must bal-ance out in any circuit, no matter how complicated and no matter how many branchesthere are.This is Kirchhoff’s Second Law. An alternative name might be the law of conser-vation of voltage.Consider the rule you’ve already learned about series circuits: The voltages across allthe individual resistors add up to the supply voltage. Yes, they do, but the polarities of theEMFs across the resistors are opposite to that of the battery. This is shown in Fig. 5-6. It’sa subtle thing. But it becomes clear when a series circuit is drawn with all the components,including the battery or other EMF source, in line with each other, as in Fig. 5-6.Problem 5-13Refer to the diagram of Fig. 5-6. Suppose the four resistors have values of 50, 60, 70 and80Ω, and that the current through them is 500 mA. What is the supply voltage, E?Find the voltages E1, E2, E3, and E4 across each of the resistors. This is done viaOhm’s Law. In the case of E1, say with the 50-Ω resistor, calculate E10. 500 5025 V. In the same way, you can calculate E230 V, E335 V, and E440 V. The sup-ply voltage is the sum ElE2E3E425303540 V 130 V.Kirchhoff’s Second Law tells us that the polarities of the voltages across the resis-tors are in the opposite direction from that of the supply in the above example.Kirchhoff’s second law91