|Teach Yourself Electricity and Electronics, 5th edition|
|Explanations for Quiz Answers in Chapter 11|
|1. We have a capacitor whose value increases as the temperature rises. Therefore, by definition, the capacitor exhibits a positive temperature coefficient. The extent of the increase per degree Celsius doesn't matter in this definition. The correct choice is (c).|
|2. If we place a substance with a high dielectric constant between the plates of an air-variable capacitor, we increase the capacitance of the device. The same effect results from rotating the shaft to increase the mutual surface area between the sets of plates (assuming that the capacitor's plates aren't fully meshed already). If we place a trimmer capacitor in parallel with the air-variable capacitor, we increase the overall capacitance. The best choice is (d), "Any of the above."|
|3. As the value of a capacitor increases, it takes more time to completely charge up after we connect a 12-V battery if all other factors remain constant. The correct choice is (a).|
|4. We can look back at the chapter text and see straightaway that of the four choices given here, only 0.05 µF represents a value that we'd likely find for a disk-ceramic capacitor. The correct answer is (d).|
|5. Air variables generally have maximum capacitance values of 1000 pF or less, and we won't find that value very often. The best answer here is (c), 8200 pF.|
|6. Technically, a millifarad represents 0.001 F or 1000 µF, which in turn represents 1,000,000,000 pF. If we have a component with a value of 0.005 millifarads, we multiply 1,000,000,000 (picofarads per millifarad) by 0.005 (millifarads) to get 5,000,000 pF. The correct answer is (a).|
|7. Although we can solve this problem by "brute force" using the formula for capacitances in series, let's recall the solution to Problem 11-3. It tells us that if we connect n capacitors of identical value in series, the net value equals 1/n of the capacitance of one component by itself. Therefore, when we connect two 100-pF capacitors in series, we get 100 pF x 1/2, or 50 pF, as the capacitance of the combination. The correct answer is (b).|
8. We have two capacitors in parallel, so we can simply add their values. One of the components has a value of 220 pF, and the other has a value of 680 pF. When we add them, we get 900 pF. The correct choice is (d).
|9. When we have several capacitors with different values C1, C2,
C3, ..., Cn connected in series, we can find the net
capacitance C using the formula
C = 1 / (1/C1 + 1/C2 + 1/C3 + ... + 1/Cn)
If we let C1 = 220 pF and C2 = 680 pF, then the formula tells us that
C = 1 / (1 / 220 + 1 / 680)
The correct answer is (a).
|10. When we check the chapter text, we find information saying that of the four choices here, only 10 µF represents a value that we'd likely find for an aluminum electrolytic capacitor. The best choice is (d).|
|11. Air has a low dielectric constant -- the same as a vacuum for most practical purposes. Because of this property, air does not allow us to get very much capacitance per unit volume. The correct answer is (a). You might suppose, for a moment, that (b) would work, but air actually exhibits high efficiency in the role of a dielectric, so (b) is wrong. Choices (c) and (d) are both false.|
|12. Remember that 1 pF equals a millionth (0.000001) of a microfarad; conversely, 1
µF equals a million (1,000,000) picofarads. To convert picofarads to microfarads, we must
multiply by 0.000001 or divide by 1,000,000. Therefore,
33,000 pF =
33,000 / 1,000,000
The correct answer is (a).
|13. Tantalum capacitors exhibit relatively high capacitance values per unit volume. Clearly, the characteristic indicated by choice (c) does not apply to these components. The correct answer is therefore (c). All of the other three characteristics apply to tantalum capacitors, so we can rule them out.|
|14. If we connect four 100-pF capacitors in parallel, their values simply add, so we get a net capacitance of 400 pF. The correct answer is (d).|
|15. Note that one of the two capacitors (68 µF) has a value many times larger than the other one (10 pF). When we parallel-connect two capacitors in a situation of this sort, we can consider the net capacitance equal to the larger of the two values "for all intents and purposes." In this case that's 68 µF, so the correct choice is (a).|
|16. We have the same two capacitors as we had in Problem 15, but this time we connect them in series. When we series-connect two capacitors and one of them has a value many times that of the other, we can consider the net capacitance equal to the smaller of the two values "for all intents and purposes." In this case that's 10 pF, so the correct answer is (d).|
|17. Let's calculate how far above or below 470 pF we can go and still not exceed the
tolerance limit of ±5%. Five percent of 470 pF equals 470 x 0.05, or 23.5 pF, so the
smallest allowable capacitance is
470 - 23.5 = 446.5 pF
and the largest allowable capacitance is
470 + 23.5 = 493.5 pF
All of the capacitance values listed in choices (a), (b), and (c) fall within the acceptable range of 446.5 pF to 493.5 pF. Therefore, the correct answer is (d).
|18. For the component in question here, the actual capacitance is 20 pF lower than the rated, or quoted, value of 470 pF. We can figure out the fractional difference by dividing 20 by 470, getting 0.0426 (rounded to four decimal places). That's 4.26%. Because the actual value is smaller than the quoted value, we consider the percentage discrepancy as a negative quantity, in this case -4.26%. The correct choice is (a).|
|19. We can construct a 4 x 4 series-parallel matrix of components in two different
ways: as a parallel combination of four sets of four series-connected components, or as a
series combination of four sets of four parallel-connected components. Either way, if all
the components are 100-pF capacitors, we'll end up with a net capacitance of 100 pF. The
correct choice is (c).
Using the first method described above, we construct four 25-pF capacitors by connecting sets of four 100-pF capacitors in series; when we parallel-connect these four series sets, we get 25 pF + 25 pF + 25 pF + 25 pF = 100 pF.
Using the second method described above, we construct four 400-pF capacitors by connecting sets of four of 100-pF capacitors in parallel; when we series-connect these four parallel sets, we get 400 pF / 4 = 100 pF.
|20. If we increase the spacing between the plates in a capacitor while leaving all other factors constant, we reduce the capacitance of the component. The correct answer is (b). Choice (d), by the way, has no theoretical meaning because, in an ideal capacitor, the resistance is "infinite" regardless of the spacing between the plates.|