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The sum of all the voltages, as you go around a circuit from some fixed point and return there 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 radio, or computer. Yes, there is, between different points. But no point can have an EMF with respect to itself. This is so simple that it s almost laughable. A point in a circuit is always shorted out to itself. What Kirchhoff really was saying, when he wrote his second law, is a more general version of the second and third points previously mentioned. He reasoned that voltage cannot appear out of nowhere, nor can it vanish. All the potential differences must balance out in any circuit, no matter how complicated and no matter how many branches there are. This is Kirchhoff s Second Law. An alternative name might be the law of conservation of voltage. Consider the rule you ve already learned about series circuits: The voltages across all the individual resistors add up to the supply voltage. Yes, they do, but the polarities of the EMFs across the resistors are opposite to that of the battery. This is shown in Fig. 5-6. It s a 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.

Interconnecting the components of the computer is a cable known as a parallel bus. It is called parallel because the bits that make up one or more 8-bit bytes travel down the bus beside each other on individual rather than one after the next as in a serial cable on a single conductor. Both are shown schematically in Figure 4-7. The advantage of a parallel bus is speed. By pumping multiple bits into a device in the computer simultaneously, the device, such as a CPU, can process them faster. Obviously, the more leads there are in the bus,

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Refer to the text in this chapter if necessary. A good score is at least 18 correct answers. The answers are in the back of the book. 1. In a series-connected string of ornament bulbs, if one bulb gets shorted out, which of the following will occur (a) All the other bulbs will go out. (b) The current in the string will go up. (c) The current in the string will go down. (d) The current in the string will stay the same. 2. Imagine that four resistors are connected in series across a 6.0-V battery, and the ohmic values are R1 = 10 , R2 = 20 , R3 = 50 , and R4 = 100 , as shown in Fig. 5-9. What is the voltage across the resistance R2 (a) 0.18 V (b) 33 mV (c) 5.6 mV (d) 0.67 V

Problem 5-13

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3. In the scenario shown by Fig. 5-9, what is the voltage across the combination of R3 and R4 (a) 0.22 V (b) 0.22 mV (c) 5.0 V (d) 3.3 V 4. Suppose three resistors are connected in parallel across a battery that delivers 15 V, and the ohmic values are R1 = 470 , R2 = 2.2 k , and R3 = 3.3 k , as shown in Fig. 5-10. The voltage across the resistance R2 is (a) 4.4 V. (b) 5.0 V. (c) 15 V. (d) not determinable from the data given. 5-10 Illustration for Quiz

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 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 via Ohm s Law. In the case of 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 El E2 E3 E4 25 30 35 40 V 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 supply in the above example.

5. In the situation shown by Fig. 5-10, what is the current through R2 (a) 6.8 mA (b) 43 mA (c) 0.15 A (d) 6.8 A 6. In the situation shown by Fig. 5-10, what is the total current drawn from the source (a) 6.8 mA (b) 43 mA (c) 0.15 A (d) 6.8 A

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