| Teach Yourself Electricity and Electronics, 5th edition |
| Stan Gibilisco |
| Explanations for Quiz Answers in Chapter 18 |
| 1. An autotransformer consists of a single coil with a tap somewhere along the winding. Usually, we apply the AC input to the entire winding (across the two end terminals) and take the output from between the tap and either end terminal. That gives us a step-down transformer. In some cases we can step up a voltage or impedance slightly if we apply the input signal to most of the coil and take the output from the end terminals. However, we can't easily get a large step-up voltage or impedance ratio that way. The best answer to this question is (b). Neither (a) nor (b) applies because an autotransformer doesn't do anything "automatically." Choice (d) is false; an autotransformer does not employ a transmission line. |
| 2. The value 35 nH represents a tiny amount of inductance, which we would best obtain by using a core material having the lowest possible permeability. Of the four choices given here, that material is air, so the correct choice is (a). |
| 3. The correct choice is (d). At the highest frequencies, transformers perform best when the core material has the lowest possible permeability. Choice (a) is wrong because air has lower permeability than powdered iron, concentrating magnetic lines of flux less (not more) than powdered iron. Choice (b) is wrong because air has the lowest loss of any known core material. Choice (c) is simply false. |
| 4. If we have a load that does not contain any reactance, then that load represents a pure resistance. Assuming that the load does not constitute a short circuit (zero resistance) or open circuit (infinite resistance) -- a reasonable supposition considering that short or open circuits can't dissipate or radiate any power -- we can use a transformer to match that load to a transmission line having any characteristic impedance within reason. The best choice here is (a). |
| 5. By definition, a step-down transformer has a higher primary (or input) voltage than secondary (or output) voltage, and therefore a higher primary impedance than secondary impedance. The correct answer is (c). Autotransformers and balanced-to-unbalanced transformers can function in step-up mode or step-down mode, so (a) and (d) won't work. Choice (b) is totally wrong; in a step-up transformer, the secondary voltage or impedance exceeds the primary voltage or impedance. |
| 6. When we design or build an AC utility transformer, we can keep eddy currents to a minimum by winding the primary and secondary around a laminated-iron core (as opposed to a solid-iron core). The correct answer is (b). We would not find air satisfactory as a core material for utility AC because we need far more permeability than air can provide, so (a) is wrong. We don't have to restrict ourselves to the core winding method, and we don't have to wind the primary right over the secondary, so choices (c) and (d) won't work here. |
| 7. If we wind the coils in a transformer directly over one another, we always end up with a considerable amount of capacitance between the windings. Choice (c) serves best. The results described in choices (a), (b), and (d) are possible, but we don't necessarily have to expect them (unless we deliberately design a transformer to have a large impedance-transfer ratio, or to work at high frequencies, or to exhibit a lot of hysteresis loss). |
| 8. Let’s use the formula that relates the turns ratio to the voltage ratio to
solve for the secondary voltage. Remember that Epri / Esec = Tpri / Tsec We can plug in Epri = 20.0 and Tpri / Tsec = 4.00 to obtain 20.0 / Esec = 4.00 Dividing through by 20.0 gives us 1 / Esec = 4.00 / 20.0 Taking the reciprocals of both sides, we obtain Esec = 20.0 / 4.00 The correct choice is (d). |
| 9. We can use the same formula as we did in the previous problem, and plug in the new
numbers. We start with Epri / Esec = Tpri / Tsec Letting Epri = 20.0 and Tpri / Tsec = 1 / 4.00, we get 20.0 / Esec = 1 / 4.00 Dividing through by 20.0, we obtain 1 / Esec = 1 / (4.00 x 20.0) Taking the reciprocals of both sides, we obtain Esec = 80.0 V RMS The correct choice is (a). |
| 10. In this case, we're told that the secondary-to-primary turns ratio equals 2.00:1.
The primary-to-secondary turns ratio is therefore the reciprocal of this value, or 1:2.00.
Now we can use the same formula as we have with the previous two problems. We start with Epri / Esec = Tpri / Tsec Letting Epri = 20.0 and Tpri / Tsec = 1 / 2.00, we get 20.0 / Esec = 1 / 2.00 Dividing through by 20.0, we obtain 1 / Esec = 1 / (2.00 x 20.0) Taking the reciprocals of both sides, we conclude that Esec = 40.0 V RMS The correct choice is (b). |
| 11. We know that the secondary-to-primary turns ratio equals 1:2:00, so the
primary-to-secondary turns ratio is 2.00:1. Using the familiar formula again, we start
with Epri / Esec = Tpri / Tsec Plugging in Epri = 20.0 and Tpri / Tsec = 2.00, we get 20.0 / Esec = 2.00 Dividing through by 20.0 gives us 1 / Esec = 2.00 / 20.0 Taking the reciprocals of both sides, we obtain Esec = 20.0 / 2.00 The correct choice is (c). |
| 12. The transformer must have a step-down impedance ratio of Zpri / Zsec
= 300 / 50.0 From the above formulas, we can calculate Tpri / Tsec = (Zpri / Zsec)1/2 The correct answer is (c). |
| 13. The impedance-transfer ratio equals the square of the turns ratio. Therefore,
according to the formula that relates these ratios, we have Zpri / Zsec
= (Tpri / Tsec)2 We know that the secondary impedance, Zsec, equals 50 ohms, so we can calculate Zpri = 16 x Zsec The correct answer is (b). |
| 14. Let's begin with the general formula that relates impedance-transfer ratio to
turns ratio. We know that Zpri / Zsec = (Tpri / Tsec)2 We recall that the voltage-transfer ratio in a transformer always equals the turns ratio. Therefore, we can revise the above formula to Zpri / Zsec = (Epri / Esec)2 Plugging in the known values, we have 4.00 / 1 = (200 / Esec)2 Taking the positive square roots of both sides, we get 2.00 = 200 / Esec We can use simple algebra to solve this equation, getting Esec = 100 V RMS. The correct choice is (d). |
| 15. We're told that the antenna (or output) impedance Rout equals
600 ohms, while the characteristic impedance Zo of the line equals 92
ohms. We can use the following formula to calculate the resistive input impedance, Rin: Rin = Zo2 / Rout When we plug in the values, grind out the arithmetic, and round the result off to two significant figures, we get Rin = 922 / 600 The correct choice is (a). |
| 16. We can use the formula for the length of a 1/4-wave section in terms of the
frequency fo and the velocity factor v. All of our choices are
specified in meters, so we use Lm = 75v / fo Plugging in the known values yields Lm = 75 x 0.75 / 14 The correct choice is (d). |
| 17. Let's use the formula that relates a transformer's turns ratio to its
impedance-transfer ratio, plug in the known values, and then solve for the turns ratio.
Our input impedance is 50 ohms and our output impedance is 800 ohms, so the
input-to-output (primary-to-secondary) impedance-transfer ratio is (Zpri / Zsec)
= 50 / 800 Inputting the known values and solving, we get Tpri / Tsec = (1/16)1/2 The correct choice is (c). |
| 18. We can calculate the characteristic impedance of the section as follows: Zo = (RinRout)1/2
The correct choice is (b). |
| 19. We need a reactance of -35 ohms to cancel out the reactive part of the antenna impedance. That means we need a capacitor. The correct location for that component is where the feed system meets the load; in other words, where the transformer secondary or 1/4-wave section output connects to the antenna. The correct choice is (a). |
| 20. If we want to obtain a good impedance match between a feed line and an antenna over a band of frequencies such as 10 to 20 MHz, we can use a transmatch at the feed point. The correct choice is (c). A variable inductor or variable capacitor, all by itself, will sometimes work in a situation like this, but usually not, so choices (a) and (b) aren't the best here. Choice (d) is, of course, false. |