| Teach Yourself Electricity and Electronics, 5th edition |
| Stan Gibilisco |
| Explanations for Quiz Answers in Chapter 10 |
| 1. The correct choice is (a). To perform permeability tuning, we must move a ferromagnetic core in and out of a coil, varying the number of coil turns that "enclose" the core. We can do that easily with a rod-shaped (solenoidal) core and a helical coil. A toroidal core, with its donut shape, does not lend itself to permeability tuning; it remains "enclosed" by all of the coil turns no matter how we manipulate it. |
| 2. The net inductance of a series combination equals the sum of the individual inductances. In this case, each inductor exhibits 88.0 mH, so the net series inductance is four times that, or 352 mH. The correct answer is (d). |
| 3. The net inductance of a parallel combination of identical inductances equals the inductance of any one of them, divided by the number of inductors. (Remember that inductances combine just like resistances do.) In this case, each inductor exhibits 88.0 mH and we have four of them in parallel, so the net inductance equals 88.0 / 4 mH, or 22.0 mH. The correct answer is (a). |
| 4. One of the inductances (8.8 H) exceeds the other two (88 nH and 88 mH) by a large factor. If we connect the three inductors in series, then the net inductance differs from the largest one to an extent so small that we can ignore it. Even if we don't ignore the difference, it won't show up in the final answer when we round off the net inductance to the two significant digits that we're allowed. The correct choice is (d). |
| 5. This scenario works "inside-out" from the series connection we encountered in the previous problem. One of the inductors has a far smaller value (88 nH) than the other two (88 mH and 8.8 H). If we connect the three inductors in parallel, then the net inductance will essentially equal the smallest one; although it will be a little less than 88 nH, the difference won't show up when we round the final answer off to two significant digits. In this case, (a) represents the correct choice. |
| 6. Pot cores provide a lot of inductance in a small package -- more than any other known configuration. The correct choice is (c). |
| 7. In a transmission line intended for use as an inductor, the frequency affects the inductance because a change in the frequency produces an effective change in the line's electrical length (in wavelengths). Therefore, we can rule out (a). The physical line length and the velocity factor also affect the electrical length if we hold all other factors constant, so we can rule out (c) and (d). The signal strength, under normal circumstances, has no effect on the inductance as long as we keep all other factors constant. The correct answer is (b). |
| 8. As we increase the inductance of a coil, it takes longer and longer for it to store or release a given amount of energy in the form of a magnetic field, as long as external factors (such as any resistance that might appear in parallel with the coil) do not change. The energy-storage and energy-release rates both decrease as the inductance goes up. The correct answer is (c). |
| 9. The inductance of an air-core coil depends on its dimensions and the number of turns it contains, but not on the frequency of the applied AC signal. The correct choice is (b). |
| 10. If we keep the number of turns in a solenoidal coil constant, we can increase the inductance by making the solenoid shorter (a), by increasing the permeability of the core material (b), or by increasing the diameter (c). The correct choice is therefore (d), "Any of the above." |
| 11. When we connect two inductors in series and they exhibit reinforcing mutual
inductance, the net inductance L of the combination equals L = L1 + L2 + 2M where L1 and L2 represent the values of the individual inductors, and M represents the mutual inductance (with all inductances expressed in the same units). In this case, we have L1 = 400 µH L2 = 600 µH M = 100 µH When we plug these values into the formula for the net inductance, we obtain L = 400 + 600 + 2 x 100 which equals 1.20 mH, rounded to three significant figures. The correct answer is (d). |
| 12. When we connect two inductors in series and they exhibit opposing mutual
inductance, the net inductance L of the combination equals L = L1 + L2 - 2M where L1 and L2 represent the values of the individual inductors, and M represents the mutual inductance (with all inductances expressed in the same units). In this case, as in Question 11, we have L1 = 400 µH L2 = 600 µH M = 100 µH When we plug these values into the formula for the net inductance, we obtain L = 400 + 600 - 2 x 100 The correct answer is (a). |
| 13. At audio frequencies (AF), pot cores represent a good choice for inductors. A frequency of 6 kHz lies within the AF range. Pot cores don't work very well at frequencies much above a few hundred kilohertz. The correct choice is (a). |
| 14. This problem requires some careful thinking! We've been told that two inductors
have identical values and "the maximum possible amount of mutual inductance." To
figure out how much inductance that is, we must do some mathematics with the formula for M,
the mutual inductance, in terms of k, the coefficient of coupling. If two coils exhibit as much mutual inductance as they can, then the coefficient of coupling k equals exactly 1 because, by definition that's as large as it can ever get! The formula relating M and k is M = k (L1L2)1/2 where L1 and L2 represent the individual inductances. In this case, we know that they're equal; let's call them both L1. Then the above formula becomes M = k (L1L1)1/2 We know that k = 1, so by substitution we know that M = L1 Now we can use the formula for the net inductance L of a series combination in which mutual inductance exists, and the fields from the inductors reinforce each other: L = L1 + L2 + 2M Because L1 = L2 = M, we can simplify this formula to L = L1 + L1 + 2L1 We've agreed to call the values of the individual inductors both L1. The foregoing tidbit of trickery demonstrates that, when we connect two identical inductors in series and they exhibit the maximum possible amount of mutual inductance, the net inductance of the combination equals four times the inductance of either component all by itself (assuming, of course, that we stick to the same unit, such as millihenrys or microhenrys, for all inductance values). The correct answer is (c). |
| 15. As in the previous problem, we've been told that two inductors have identical
values and "the maximum possible amount of mutual inductance." If, as before, we
call both inductances L1 and the mutual inductance M, then once
again we have M = L1 The formula for the net inductance L of a series combination in which mutual inductance exists, and the fields from the inductors oppose each other, is L = L1 + L2 - 2M Because L1 = L2 = M, we can simplify this formula to L = L1 + L1 - 2L1 The correct answer is (a), as long as we use the same unit to express all inductances. |
| 16. Any of the three actions (a), (b), or (c), performed all by itself, will reduce the inductance of a permeability-tuned solenoidal coil. Remember that if we hold all other factors constant and make a coil longer, the inductance goes down, so choice (a) will work. Pulling the core farther out from inside the solenoid reduces the proportion of the coil turns affected by the core, so (b) will work. If we reduce the core permeability but don't change anything else, the inductance goes down, so choice (c) will work. The correct answer is therefore (d), "Any of the above." |
| 17. Solid plastic has approximately the same permeability as air or a vacuum, that is, 1. Plastic won't do anything for us if we want to get a lot of inductance in a small physical space, so we can throw out choice (a). Choice (c) is also wrong. Powdered iron has permeability much larger than 1, so a coil with a plastic core would have far less inductance than the same coil with a powdered-iron core. Choice (d) borders on the ridiculous! Putting a plastic pot-core "shell" around a coil would do little or nothing to change the behavior of the coil compared with leaving it out in the air. However, solid plastic can function as an inductor core in place of air to enhance the structural rigidity of the coil windings. Choice (b) is the one we want. |
| 18. If f represents the frequency in megahertz and v represents the
velocity factor as a fractional value between 0 and 1, then we can calculate 1/4
electrical wavelength in centimeters (scm) using the formula scm = 7500v / f Plugging the values from the question directly in, we get scm = 7500 x 0.800 / 144 If we want a transmission line to work as an inductor when we connect the conductors at the far end together, then the line must measure less than 1/4 electrical wavelength from end to end. Choice (c) is the only one that ensures this criterion. |
| 19. If we cut a transmission-line section so that it measures exactly 1/4 electrical wavelength from end to end, and if we keep the far end shorted out, then the non-shorted (input) end will "look like" an open circuit. The correct answer is (d). |
| 20. If we make a transmission-line section longer than 1/4 electrical wavelength but shorter than 1/2 electrical wavelength, and if we short out the line's far end, then the non-shorted (input) end will "look like" a capacitor. The correct choice is (a). |