Voltage calculations

Chapter Voltage calculations

Teach Yourself Electricity and Electronics Third Edition Book
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Teach Yourself Electricity and Electronics Third Edition Book

  • you can do about it unless you are willing to use units of nanoamperes (nA), or bil-lionths of an ampere. Then you can say that the current is 186 nA.Voltage calculationsThe second use of Ohm’s Law is to find unknown voltages when the current and the re-sistance are known. For the following problems, uncover the ammeter and cover thevoltmeter scale instead in your mind.Problem 4-4Suppose the potentiometer (Fig. 4-7) is set to 100 ohms, and the measured current is10 mA. What is the dc voltage?Use the formula EI R. First, convert the current to amperes: 10 mA = 0. 01 A.Then multiply: E0. 01 1001 V. That’s a low, safe voltage, a little less than whatis produced by a flashlight cell.Problem 4-5Adjust the potentiometer (Fig. 4-7) to a value of 157 K , and let the current reading be17 mA. What is the voltage of the source?Now you have to convert both the resistance and the current values to their properunits. A resistance of 157 Kis 157,000 ; a current of 17 mA is 0. 017 A. Then EIR0.017157,0002669 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 4-6You set the potentiometer (Fig. 4-7) so that the meter reads 1.445 A, and you observethat the potentiometer scale shows 99 ohms. What is the voltage?These units are both in their proper form. Therefore, you can plug them right inand use your calculator: EIR1. 44599143.055 V. This can, and should, berounded off to 143 V. A purist would go further and round it to the nearest 10 volts, to140 V.It’s never a good idea to specify your answer to a problem with more significant fig-ures than you’re given. The best engineers and scientists go by the rule of significantfigures: keep to the least number of digits given in the data. If you follow this rule inProblem 4-6, you must round off the answer to two significant figures, getting 140 V, be-cause the resistance specified (99 ) is only accurate to two digits.Resistance calculationsOhms’ Law can be used to find a resistance between two points in a dc circuit, when thevoltage and the current are known. For the following problems, imagine that both thevoltmeter and ammeter scales in Fig. 4-7 are visible, but that the potentiometer is un-calibrated.Resistance calculations71