Teach Yourself Electricity and Electronics, 5th edition |

Stan Gibilisco |

Explanations for Quiz Answers in Chapter 9 |

1. The text tells us that when we have a perfect sine wave, the positive peak voltage equals approximately 1.414 times the RMS voltage. The correct choice is (c). |

2. A complete wave cycle represents 2p radians (rad) of phase, where p represents the number of diameters in a circle's circumference (approximately 3.1416). Therefore, 1 rad equals 1/(2p) of a full cycle, and 2 rad represents 2/(2p) of the cycle, which simplifies to 1/p of the cycle. The correct answer is (a). |

3. As defined in the text, a ramp wave has a finite, nonzero, defined rise time, but the amplitude decays instantly. Therefore, the correct answer is (b). If you get confused by the distinction between the ramp and the opposite type of sawtooth (instantaneous rise, finite decay), you might try a little memory trick: In electricity and electronics, ramps always slant upward as you move toward the right (forward in time). |

4. In order to determine the frequency of an AC wave in hertz when we know its angular frequency in radians per second, we must divide the angular frequency by 2p. That's because a complete wave cycle equals 2p radians. If we let p = 3.1416, then 2p = 6.2832. When we divide 94.25 by 6.2832, we get 15.00 Hz. The correct answer is (d). |

5. We define the period of a waveform as the length of time between two adjacent
points on a wave that occur at the same place on the cycle. In Fig. 9-15, let's reference
the abrupt decays (vertical lines) as these points. They're easy to identify, and they
cross the horizontal axis at clear locations -- every other division. We know that
each horizontal division represents 2.0 milliseconds (ms). Therefore, the wave's period
equals twice that time interval, or 4.0 ms. The correct answer is (c). |

6. We define the peak-to-peak (pk-pk) voltage of a waveform as the difference between the maximum instantaneous voltage and the minimum instantaneous voltage. In Fig. 9-15, that difference equals the vertical displacement between the highest and lowest points on the wave. By visual inspection, we can see that this displacement amounts to roughly 5.2 divisions (approximately 2.6 divisions going up from the horizontal axis, and approximately 2.6 divisions going down from the horizontal axis). Each vertical division represents 10 millivolts (mV). Therefore, the peak-to-peak voltage equals approximately 5.2 x 10 mV, or 52 mV. The correct answer is (d). |

7. Our main challenge lies in deciding which point on the wave cycle to use as the reference for determining the period. Let's use the points where the wave crosses the horizontal axis going downward. Any two such points appear to lie exactly four divisions apart on the horizontal axis. Each horizontal division represents 2.0 ms. Therefore, the period equals 4.0 x 2.0 ms, or 8.0 ms. The correct answer is (b). |

8. We define the positive peak voltage of a waveform as the displacement going upward from the horizontal axis to the maximum instantaneous voltage point. In Fig. 9-16, that displacement appears to be exactly two divisions. Because each vertical division represents 10 mV, and because the maximum instantaneous voltage point lies above the horizontal axis "in positive territory," we can conclude that the positive peak voltage equals twice 10 mV in the positive direction, or +20 mV pk+. The correct answer is (b). |

9. As with Question 7, we must decide which point on the wave cycle to use as the reference for determining the period. Let's use the points where the wave crosses the horizontal axis going downward, just as we did in Question 7. Any two such points appear separated by approximately four divisions. Each horizontal division represents 2.0 ms, so the period equals about 4.0 x 2.0 ms, or 8.0 ms. The correct answer is (a). |

10. The frequency of any wave in hertz equals the reciprocal of the period in seconds. We have determined the period as roughly 8.0 ms, or 0.0080 s. Therefore, the approximate frequency equals 1 divided by 0.0080, or 125 Hz. The correct answer is (b). |

11. In Fig. 9-17, the displacement going upward from the horizontal axis to the maximum instantaneous voltage point looks like about 2.4 divisions. Because each vertical division represents 10 mV, we can estimate that the positive peak voltage equals approximately 2.4 x 10 mV in the positive direction, or +24 mV pk+. The correct answer is (d). |

12. We define the negative peak voltage as the negative of the displacement going downward from the horizontal axis to the minimum instantaneous voltage point. In Fig. 9-17, it looks like one division. Because each vertical division represents 10 mV, and because the minimum instantaneous voltage point lies below the horizontal axis "in negative territory," we can conclude that the negative peak voltage equals -10 mV pk-. The correct answer is (a). |

13. In Fig. 9-17, the difference between the maximum instantaneous voltage and the minimum instantaneous voltage equals roughly +24 mV minus -10 mV. Arithmetically, that difference works out as the sum 24 + 10 mV, which gives us 34 mV pk-pk. The correct answer is (c). We can also determine the peak-to-peak voltage by observing how many vertical divisions exist between the wave's maximum and the wave's minimum. It appears as about 3.4 divisions. Because each vertical division represents 10 mV, we can estimate that the peak-to-peak voltage equals approximately 10 x 3.4, or 34 mV pk-pk. |

14. If two pure AC sine waves have the same frequency but different amplitudes, and if they coincide in phase, and if neither of them has any DC component, then the composite wave's amplitude equals the sum of the amplitudes of the two original waves. We know that the waves measure 21 V pk-pk and 17 V pk-pk. The sum of these values equals 38 V pk-pk. The correct answer is (a). |

15. If two pure AC sine waves have the same frequency but different amplitudes, they differ in phase by 180º, and neither one has a DC component, then the composite wave's amplitude equals the difference between the amplitudes of the originals. In this case, that's 21 V pk-pk minus 17 V pk-pk, which equals 4 V pk-pk. The correct answer is (c). |

16. If two pure AC sine waves have the same frequency but differ in phase by some odd amount such as an angle between 30º and 60º, and assuming that neither wave has a DC component, then the resulting signal amplitude depends on the strength of the two waves and also on the difference in phase. Because we don't know the exact phase difference, we can't answer this question. We must choose (d). |

17. We know that a full wave cycle represents 360º of phase. Therefore, 1/8 of a cycle represents 360º/8, or 45º. The correct answer is (d). |

18. Remember that the frequency of an AC wave in hertz equals the reciprocal of its period in seconds. If we want to know the frequency in kilohertz (thousands of hertz), therefore, we must take the reciprocal of the period in milliseconds (thousandths of seconds). That method works because when we divide 1 by thousandths, we get thousands! The correct choice is (b). |

19. Let's find the frequency of the fundamental wave, whose period equals 4.096 milliseconds (ms), and then derive the fourth harmonic from that. The wave's time period equals 0.004096 seconds; when we take the reciprocal of this number, we get 244.140625 Hz (including all the digits that our calculator can produce). The fourth harmonic frequency therefore equals 4 x 244.140625 Hz. When we multiply these numbers and round off to four significant figures, we get 976.6 Hz. The correct answer is (c). |

20. To work out this problem, we must start by figuring out the positive and negative peak voltages of the waveform appearing at the AC wall outlet. Then, we can see whether or not the series DC voltage is high enough to overcome the peaks. We've been told that the AC voltage equals 117 V RMS. We can safely assume that standard utility outlets don't provide any DC component along with the AC wave (in any case, they had better not!). We can assume that the utility AC waveform constitutes a pure sine wave for all practical purposes. Recall from the chapter text that the positive and negative peak amplitudes of a sine wave equal approximately 1.414 and -1.414 times the RMS amplitude, respectively. Therefore, the positive peak voltage of our utility waveform equals 117 x 1.414 or +165 V pk+, while the negative peak voltage equals 117 x (-1.414) or -165 V pk-. If we connect a DC source of 135 V in series with this AC wave, the entire sine wave will displace towards the positive or negative pole (depending on the polarity of the DC source) by 135 V. That's not enough to overcome the +165 V positive peak or the -165 V negative peak. We'll end up with a wave that constitutes AC (because its polarity alternates) but appears "lopsided," with one peak voltage much greater than the other. Therefore, our correct choice is (c). |