Resistances in series

Chapter Resistances in series

Teach Yourself Electricity and Electronics Third Edition Book
Pages 748
Views 14,417
Downloads : 46 times
PDF Size : 4.4 MiB

Summary of Contents

Teach Yourself Electricity and Electronics Third Edition Book

  • Problem 4-10Suppose that the voltmeter reads 12 V and the ammeter shows 50 mA. What is thepower dissipated by the potentiometer?Use the formula PEI. First, convert the current to amperes, getting I0.050 A.(Note that the zero counts as a significant digit.) Then PEI120.050 0.60 W.You might say that this is 600 mW, although that is to three significant figures. It’snot easy to specify the number 600 to two significant digits without using a means ofwriting numbers called scientific notation. That subject is beyond the scope of thisdiscussion, so for now, you might want to say “600 milliwatts, accurate to two significantfigures.” (You can probably get away with “600 milliwatts” and nobody will call you onthe number of significant digits.)Problem 4-11If 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 PE 2/R 33/9999/9990. 009 W 9 mW. You are justi-fied in going to only one significant figure here.Problem 4-12Suppose the resistance is 47 Kand the current is 680 mA. What is the power dissi-pated by the potentiometer?Use the formula PI 2R, after converting to ohms and amperes. Then P0.6800.68047,00022,000 W 22 kW.This is a ridiculous state of affairs. An ordinary potentiometer, such as the one youwould get at an electronics store, dissipating 22 kW, several times more than a typicalhousehold. The voltage must be phenomenal. It’s not too hard to figure out that such avoltage would burn out the potentiometer so fast that it would be ruined before the lit-tle “Pow!” could even begin to register.Problem 4-13Just from curiosity, what is the voltage that would cause so much current to be driventhrough such a large resistance?Use Ohm’s Law to find the current: EIR0.68047, 000 32,000 V 32 kV.That’s the sort of voltage you’d expect to find only in certain industrial/commercial ap-plications. The resistance capable of drawing 680 mA from such a voltage would surelynot be a potentiometer, but perhaps something like an amplifier tube in a radio broad-cast transmitter.Resistances in seriesWhen you place resistances in series, their ohmic values simply add together to get thetotal resistance. This is easy to see intuitively, and it’s quite simple to remember.Resistances in series73