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Power Calculations
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You can calculate the power P, in watts, in a dc circuit such as that shown in Fig. 4-7, by using the formula P = EI. This formula tells us that the power in watts is the product of the voltage in volts and the current in amperes. If you are not given the voltage directly, you can calculate it if you know the current and the resistance. Recall the Ohm s Law formula for obtaining voltage: E = IR. If you know I and R but you don t know E, you can get the power P this way: P = EI = (IR)I = I 2R Suppose you re given only the voltage and the resistance. Remember the Ohm s Law formula for obtaining current: I = E/R. Therefore: P = EI = E(E/R) = E 2/R
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Problem 4-10 Suppose that the voltmeter in Fig. 4-7 reads 12 V and the ammeter shows 50 mA. What is the power dissipated by the potentiometer Use the formula P = EI. First, convert the current to amperes, getting I = 0.050 A. (Note that the last 0 counts as a significant digit.) Then multiply by 12 V, getting P = EI = 12 0.050 = 0.60 W.
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62 Direct-Current Circuit Basics
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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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Resistances in Series
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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!
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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.
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4-8 Three resistors in series.
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Illustration for Problem 4-14. Resistance values are in ohms.
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Resistances in Parallel 63
Resistances in Parallel
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 4-15 Consider five resistors in parallel. Call them R1 through R5, and call the total resistance R as shown in Fig. 4-9. Let the resistance values be as follows: R1 = 100 , R2 = 200 , R3 = 300 , R 4 = 400 , and R5 = 500 . What is the total resistance, R, of this parallel combination
4-9 Five resistors of values R1 through R5, connected in parallel,
produce a net resistance R. Illustration for Problems 4-15 and 4-16.
Converting the resistances to conductance values, you get: G1 = 1/100 = 0.01 S, G2 = 1/200 = 0.005 S, G3 = 1/300 = 0.00333 S, G4 = 1/400 = 0.0025 S, and G5 = 1/500 = 0.002 S. Adding these gives G = 0.01 + 0.005 + 0.00333 + 0.0025 + 0.002 = 0.0228 S. The total resistance is therefore R = 1/G = 1/0.0228 = 43.8 .
Problem 4-16 Suppose you have five resistors, called R1 through R5, connected in parallel as shown in Fig. 4-9. Suppose all the resistances, R1 through R5, are 4.70 k . What is the total resistance, R, of this combination When you have two or more resistors connected in parallel and their resistances are all the same, the total resistance is equal to the resistance of any one component divided by the number of components. In this example, convert the resistance of any single resistor to 4700 , and then divide this by 5. Thus, you can see that the total resistance is 4700/5 = 940 . In a situation like this, where you have a bunch of resistors connected together to operate as a single unit, the total resistance is sometimes called the net resistance. Take note, too, that R is not italicized when it means resistor, but R is italicized when it means resistance!