Teach Yourself Electricity and Electronics, 5th edition
Stan Gibilisco
Explanations for Quiz Answers in Chapter 12
1. An AC wave with a frequency of 10 kHz goes through 10,000 complete cycles every second. Therefore, one complete cycle takes 1/10,000 of a second, or 0.00010 s, accurate to two significant figures. The correct answer is (a).
2. In a pure AC sine wave having a constant period, all the energy exists at a single frequency. The correct choice is (a). The wavelength doesn't vary periodically, so (b) is wrong. No wave disturbance has a wavelength of zero, because in that case it wouldn't be a wave at all; choice (c) is false. Transitions never occur instantaneously in a sine wave, so (d) is wrong.
3. In order to define a phase relationship between two sine waves, they must have the same frequency. The correct answer is (c).
4. One-sixth of a cycle equals 360║ of phase (a full cycle) times 1/6, or 60║. If a particular wave X leads another wave Y by 1/6 of a cycle, then X leads Y by 60║ of phase. We can also say that wave Y lags wave X by 60║, so the correct answer is (c).
5. If we shift the phase of a continuous sine wave to the left (earlier) or to the right (later) by any whole number of cycles, we get the original wave. A complete cycle equals 360║ of phase. The correct choice is therefore (b).
6. If we change the phase of a pure sine wave having a constant frequency and no DC component by 1/2 cycle, we end up with the same sine wave "upside-down." A half cycle constitutes 180║ of phase. The correct choice is (d).
7. We represent a full cycle for a sine wave as 360║ of phase. Therefore, 45║ represents 45/360, or 1/8, of a cycle. The correct choice is (b).
8. If two waves coincide in phase, then their positive and negative peak voltages add. In this case, our waves have ▒7 V pk and ▒3 V pk, so the composite wave will exhibit  ▒(7+3) V pk, or ▒10 V pk. The peak-to-peak voltage of a sine wave equals the difference between the positive and negative peak voltages, in this case +10 V pk+ minus -10 V pk-, which works out to [10 - (-10)] V pk-pk, or 20 V pk-pk. The correct choice is (c).
9. We're told that our wave has a frequency of 5000 Hz. The period in seconds (s) equals the 1/5000 s, or 0.0002000 s. To obtain the period in microseconds, we must multiply this figure by 1,000,000, getting 200.0 Ás. A phase angle of 45║ represents 1/8 of a cycle (because 45║ x 8 = 360║), which occurs in 1/8 of the time it takes for a full cycle. Therefore, our wave goes through 45║ in 1/8 x 200.0 Ás, or 25.00 Ás. The correct answer is (c).
10. A cosine wave has the same general shape as a sine wave, so choice (a) works. A cosine wave constitutes the derivative of a sine wave, so (b) works. A cosine wave portrays the instantaneous rate of change in a sine wave at the same frequency, so (c) also works. The correct answer is (d), "All of the above."
11. If two pure sine waves have identical frequency and no DC components are offset by 360║ of phase, then they line up precisely, just as if they were not offset at all. Therefore, we can consider them coincident in phase. The correct answer is (d).
12. The term phase quadrature refers to two waves having the same frequency but differing in phase by 1/4 of a cycle or 90║. The correct answer is (b).
13. In a polar-coordinate vector diagram showing a pure sine wave having no DC component, the vector length represents the peak positive amplitude, or the peak negative amplitude without the minus sign. That's half the peak-to-peak amplitude, so the correct choice is (a).
14. If wave X lags wave Y by 135║, then wave X lags wave Y by 135/360, or 3/8, of a cycle. That's equivalent to saying that wave Y leads wave X by 3/8 of a cycle, or that wave Y is 3/8 of a cycle ahead of wave X. The correct answer is (c).
15. Because vector X lies 120║ counterclockwise from vector Y in the graph of Fig. 12-12, we know that wave X leads wave Y by 120║ of phase. (Remember that in diagrams of this sort, we should imagine that the vectors rotate counterclockwise as time passes, as shown by the dashed circle with arrows in the figure). A phase difference of 120║ represents 1/3 of a cycle, because 120/360 = 1/3. Therefore, the correct answer is (d).
16. Because vector X is longer than vector Y, and because we know that waves X and Y both constitute pure sine waves having no DC components, we can conclude that wave X has greater amplitude than wave Y. This fact holds true for amplitude however we define it (as positive peak, negative peak, or peak-to-peak). The correct answer is therefore (b).
17. The two waves shown in Fig. 12-13 have the same frequency but different amplitudes; they "line up" exactly (the positive and negative peaks coincide). Therefore, we can see that the two waves exist in phase coincidence. The correct answer is (a).
18. If we invert either one of the waves in Fig. 12-13, we'll end up with waves in phase opposition because neither wave has any superimposed DC component on it (as we were told in the statement of Question 17). The correct choice is (b).
19. The two waves have identical amplitude, but they oppose in phase. Neither has a DC component. They have identical frequency. Under these conditions, the two wave signals will completely cancel each other, leaving us with no signal at all; it will manifest an amplitude equal to 0 V pk-pk. The correct answer is (c).
20. If we shift a wave by 1/10 of a cycle to the left or right on a graph with time plotted horizontally and instantaneous amplitude plotted vertically, we in effect shift the wave's phase by 1/10 x 360║, or 36║. The correct answer is (c).