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Resistance calculations
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Ohms Law can be used to find a resistance between two points in a dc circuit, when the voltage and the current are known. For the following problems, imagine that both the voltmeter and ammeter scales in Fig. 4-7 are visible, but that the potentiometer is uncalibrated.
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72 Basic dc circuits
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Problem 4-7
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If the voltmeter reads 24 V and the ammeter shows 3.0 A, what is the value of the potentiometer Use the formula R E/I and plug in the values directly, because they are expressed in volts and amperes: R 24/3.0 8. 0 . Note that you can specify this value to two significant figures, the eight and the zero, rather than saying simply 8 . This is because you are given both the voltage and the current to two significant figures. If the ammeter reading had been given as 3 A (meaning some value between 21/2 A and 31/2 A), you would only be entitled to express the answer as 8 (somewhere between 71/2 and 81/2 ). A zero can be a significant figure, just as well as the digits 1 through 9.
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What is the value of the resistance if the current is 18 mA and the voltage is 229 mV First, convert these values to amperes and volts. This gives I 0.018 A and E 0.229 V. Then plug into the equation R E/I 0.229/0.018 13 . You re justified in giving your answer to two significant figures, because the current is only given to that many digits.
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Suppose the ammeter reads 52 uA and the voltmeter indicates 2.33 kV. What is the resistance Convert to amperes and volts, getting I 0.000052 A and E 2330 V. Then plug into the formula: R 2330/0.000052 45,000,000 45 M .
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Power calculations
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You can calculate the power, in watts, in a dc circuit such as that shown in Fig. 4-7, by the formula P EI or the product of the voltage in volts and the current in amperes. You might not be given the voltage directly, but can calculate it if you know the current and the resistance. Remember the Ohm s Law formula for obtaining voltage: E IR. If you know I and R, but don t know E, you can get the power P by means of the formula P (IR)I I 2R. That is, you take the current in amperes, multiply this figure by itself, and then multiply the result by the resistance in ohms. You can also get the power if you aren t given the current directly. Suppose you re given only the voltage and the resistance. Remember the Ohm s Law formula for obtaining current: I E /R. Therefore, P E(E/R) E 2/R. Take the voltage, multiply it by itself, and divide by the resistance. Stated all together, these power formulas are: P EI I 2R E 2/R
Now you are ready to do some problems in power calculations. Refer once again to Fig. 4-7.
Resistances in series 73
Problem 4-10
Suppose that the voltmeter 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 zero counts as a significant digit.) Then P EI 12 0.050 0.60 W. You might say that this is 600 mW, although that is to three significant figures. It s not easy to specify the number 600 to two significant digits without using a means of writing numbers called scientific notation. That subject is beyond the scope of this discussion, so for now, you might want to say 600 milliwatts, accurate to two significant figures. (You can probably get away with 600 milliwatts and nobody will call you on the number of significant digits.)
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 dissipated power 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 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 4-13
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.
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