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Suppose the resistance is 47 K and the current is 680 mA. What is the power dissipated by the potentiometer Use the formula P I 2R, after converting to ohms and amperes. Then P 0.680 0.680 47,000 22,000 W 22 kW. This is a ridiculous state of affairs. An ordinary potentiometer, such as the one you would get at an electronics store, dissipating 22 kW, several times more than a typical household. The voltage must be phenomenal. It s not too hard to figure out that such a voltage would burn out the potentiometer so fast that it would be ruined before the little Pow! could even begin to register.

Problem 4-11 If the resistance in the circuit of Fig. 4-7 is 999 and the voltage source delivers 3 V, what is the power dissipated by the potentiometer Use the formula P = E 2/R = 3 3/999 = 9/999 = 0.009 W = 9 mW. You are justified in going to only one significant figure here. Problem 4-12 Suppose the resistance in Fig. 4-7 is 47 k and the current is 680 mA. What is the power dissipated by the potentiometer Use the formula P = I 2R, after converting to ohms and amperes. Then P = 0.680 0.680 47,000 = 22,000 W = 22 kW. (This is an unrealistic state of affairs: an ordinary potentiometer, such as the type you would use as the volume control in a radio, dissipating 22 kW, several times more than a typical household!) Problem 4-13 How much voltage would be necessary to drive 680 mA through a resistance of 47 k , as is described in the previous problem Use Ohm s Law to find the voltage: E = IR = 0.680 47,000 = 32,000 V = 32 kV. That s the level of voltage you d expect to find on a major utility power line, or in a high-power tube-type radio broadcast transmitter.

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Although PCM is perhaps the best-known, high-quality voice digitization process, it is by no means the only one. Advances in coding schemes and improvements in the overall quality of the telephone network have made it possible for encoding schemes to be developed that use far less bandwidth that traditional PCM. In this next section, we will consider some of these techniques.

Problem 4-13

When you place resistances in series, their ohmic values add together to get the total resistance. This is easy to imagine, and it s easy to remember!

Adaptive Differential Pulse Code Modulation (ADPCM)

Just from curiosity, what is the voltage that would cause so much current to be driven through such a large resistance Use Ohm s Law to find the current: E IR 0.680 47, 000 32,000 V 32 kV. That s the sort of voltage you d expect to find only in certain industrial/commercial applications. The resistance capable of drawing 680 mA from such a voltage would surely not be a potentiometer, but perhaps something like an amplifier tube in a radio broadcast transmitter.

Problem 4-14 Suppose resistors with the following values are connected in series, as shown in Fig. 4-8: 112 , 470 , and 680 . What is the total resistance of the series combination Simply add up the values, getting a total of 112 + 470 + 680 = 1262 . You might round this off to 1260 . It depends on the tolerances of the resistors how precise their actual values are to the ones specified by the manufacturer.

Adaptive Differential Pulse Code Modulation (ADPCM) is a technique that enables toll-quality voice signals to be encoded at a half rate (32 Kbps) for transmission. ADPCM relies on the predictability that is

When you place resistances in series, their ohmic values simply add together to get the total resistance. This is easy to see intuitively, and it s quite simple to remember.

When resistances are placed in parallel, they behave differently than they do in series. One way to look at resistances in parallel is to consider them as conductances instead. In parallel, conductances add up directly, just as resistances add up in series. If you change all the ohmic values to siemens, you can add these figures up and convert the final answer back to ohms. The symbol for conductance is G. This figure, in siemens, is related to the resistance R, in ohms, by these formulas, which you learned in Chap. 2: G = 1/R R = 1/G

Problem 4-14

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