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Ohm s Law problems.
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The second application of Ohm s Law is to find unknown dc voltages when the current and the resistance are known. Let s work out some problems of this kind.
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Problem 4-4 Suppose the potentiometer in Fig. 4-7 is set to 100 , and the measured current is 10 mA. What is the dc voltage Use the formula E = IR. First, convert the current to amperes: 10 mA = 0.01 A. Then multiply: E = 0.01 100 = 1.0 V. That s a little less than the voltage produced by a flashlight cell. Problem 4-5 Adjust the potentiometer in Fig. 4-7 to a value of 157 k , and suppose the current reading is 17.0 mA. What is the voltage of the source You must convert both the resistance and the current values to their proper units. A resistance of 157 k is 157,000 , and a current of 17.0 mA is 0.0170 A. Then E = IR = 0.017 157,000 = 2669 V = 2.669 kV. You should round this off to 2.67 kV. This is a dangerously high voltage. Problem 4-6 Suppose you set the potentiometer in Fig. 4-7 so that the meter reads 1.445 A, and you observe that the potentiometer scale shows 99 . 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 but to what extent This is a good time to state an important rule that should be followed in all technical calculations. The Rule of Significant Figures Competent engineers and scientists go by the rule of significant figures, also called the rule of significant digits. After completing a calculation, you should always round the answer off to the least number of digits given in the input data numbers. If you follow this rule in Problem 4-6, you must round off the answer to two significant digits, getting 140 V, because the resistance (99 ) is only specified to that level of accuracy. If the resistance were given as 99.0 , then you would round off the answer to 143 V. If the resistance were given as 99.00 , then you could state the answer as 143.1 V. However, any further precision in the resistance value would not entitle you to go to any more digits in your answer, unless the current were specified to more than four significant figures. This rule takes some getting used to if you haven t known about it or practiced it before. But after a while, it will become a habit.
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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.
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Problem 4-7 If the voltmeter in Fig. 4-7 reads 24 V and the ammeter shows 3.0 A, what is the resistance 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 8 and the 0, 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, you would only be entitled to express the answer as 8 , to one significant digit. The digit 0 can be, and often is, just as important in calculations as any of the other digits 1 through 9. Problem 4-8 What is the value of the resistance in Fig. 4-7 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 . Problem 4-9 Suppose the ammeter in Fig. 4-7 reads 52 A 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 = E/I = 2330/0.000052 = 45,000,000 = 45 M .
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