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The first way to use Ohm s Law is to find current values in dc circuits. In order to find the current, you must know the voltage and the resistance, or be able to deduce them. Refer to the schematic diagram of Fig. 4-7. It consists of a variable dc generator, a voltmeter, some wire, an ammeter, and a calibrated, wide-range potentiometer. Component values have been left out of this diagram, so it s not a wiring diagram. But
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70 Basic dc circuits values can be assigned for the purpose of creating sample Ohm s Law problems. While calculating the current in the following problems, it is necessary to mentally cover up the meter.
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Circuit for working Ohm s Law problems.
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Suppose that the dc generator (Fig. 4-7) produces 10 V, and that the potentiometer is set to a value of 10 . Then what is the current This is easily solved by the formula I E/R. Just plug in the values for E and R; they are both 10, because the units were given in volts and ohms. Then I 10/10 1 A.
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Problem 4-2
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The dc generator (Fig. 4-7) produces 100 V and the potentiometer is set to 10 K . What is the current First, convert the resistance to ohms: 10 K 10,000 . Then plug the values in: I 100/10,000 0.01 A. This might better be expressed as 10 mA. Engineers and technicians prefer to keep the numbers within reason when specifying quantities. Although it s perfectly all right to say that a current is 0.01 A, it s best if the numbers can be kept at 1 or more, but less than 1,000. It is a little silly to talk about a current of 0.003 A, or a resistance of 107,000 , when you can say 3 mA or 107 K .
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Problem 4-3
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The dc generator (Fig. 4-7) is set to provide 88.5 V, and the potentiometer is set to 477 M . What is the current This problem involves numbers that aren t exactly round, and one of them is huge. But you can use a calculator. The resistance is first changed to ohms, giving E/R 88.5/ 477,000,000 . Then you plug into the Ohm s Law formula: I 477,000,000 0.000000186 A 0.186 uA. This value is less than 1, but there isn t much
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Problem 4-1
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Resistance calculations 71 you can do about it unless you are willing to use units of nanoamperes (nA), or billionths of an ampere. Then you can say that the current is 186 nA.
Voltage calculations
The second use of Ohm s Law is to find unknown voltages when the current and the resistance are known. For the following problems, uncover the ammeter and cover the voltmeter scale instead in your mind.
Problem 4-4
Suppose the potentiometer (Fig. 4-7) is set to 100 ohms, and the measured current is 10 mA. What is the dc voltage Use the formula E I R. First, convert the current to amperes: 10 mA = 0. 01 A. Then multiply: E 0. 01 100 1 V. That s a low, safe voltage, a little less than what is produced by a flashlight cell.
Problem 4-5
Adjust the potentiometer (Fig. 4-7) to a value of 157 K , and let the current reading be 17 mA. What is the voltage of the source Now you have to convert both the resistance and the current values to their proper units. A resistance of 157 K is 157,000 ; a current of 17 mA is 0. 017 A. Then E IR 0.017 157,000 2669 V 2.669 kV. You might want to round this off to 2.67 kV. This is a dangerous voltage. If you touch the terminals you ll get clobbered.
Problem 4-6
You set the potentiometer (Fig. 4-7) so that the meter reads 1.445 A, and you observe that the potentiometer scale shows 99 ohms. What is the voltage These units are both in their proper form. Therefore, you can plug them right in and use your calculator: E IR 1. 445 99 143.055 V. This can, and should, be rounded off to 143 V. A purist would go further and round it to the nearest 10 volts, to 140 V. It s never a good idea to specify your answer to a problem with more significant figures than you re given. The best engineers and scientists go by the rule of significant figures: keep to the least number of digits given in the data. If you follow this rule in Problem 4-6, you must round off the answer to two significant figures, getting 140 V, because the resistance specified (99 ) is only accurate to two digits.