Physics Demystified, 2nd edition

Stan Gibilisco

Explanations for Quiz Answers in Chapter 8

1. By definition, a ramp is a sawtooth wave with a steady, finite rise and an instantaneous decay. The correct choice is B.

2. We're told that the vectors for two waves of equal frequency point at an angle of π/4 rad with respect to each other (where π represents the ratio of a circle's circumference to its diameter, roughly equal to 3.14159). A full cycle comprises 2π rad. We know that π/4 rad represents 1/8 of a full circle because (π/4) x 8 = 2π. The correct choice is C.

3. We've never learned how to calculate the root-mean-square (RMS) voltage of a sawtooth wave based only on its peak-to-peak voltage. The correct choice is A.

4. To answer this question, we must assume that the voltage-versus-time waveform of our AC utility electricity constitutes a perfect sinusoid with no DC component. That's true "for all practical intents and purposes." In that case, the peak-to-peak voltage equals approximately 2.828 times the RMS voltage. When we multiply 117 V RMS by 2.828, we get 331 V pk-pk. The answer is D.

5. Remember that a full 360º cycle contains 2π rad of phase. If two waves have the same frequency but differ in phase by 30º, that's the equivalent of 30/360, or 1/12, of a cycle. The radian equivalent is therefore (1/12) x 2π = π/6 rad. The correct choice is D.

6. The period of an AC wave varies inversely in proportion to the frequency. If we double the frequency, we cut the period in half. The correct choice is D.

7. We're told that in Fig. 8-18, each horizontal increment represents 1.00 ms = 0.00100 s. When we examine the waveform, we can see that any pair of adjacent decay instants (shown as heavy vertical lines) is separated by two divisions, representing 0.00200 s. Because any two successive decay instants occur exactly one cycle apart, we know that the period equals 0.00200 s. To calculate the frequency (in hertz), we must take the reciprocal of the period (in seconds), getting 1 / 0.00200 = 500 Hz. The answer is D.

8. In Fig. 8-18, the uppermost (presumably positive) peaks of the waveform reach five divisions above the horizontal axis, representing +50 V, while the lowermost (presumably negative) peaks reach one division below the horizontal axis, representing -10 V. The peak-to-peak voltage of an AC wave, by definition, equals the positive peak voltage minus the negative peak voltage. In this case that's +50 - (-10) = 50 + 10 = 60 V. The correct choice is B.

9. This problem constitutes nothing more or less than an "arithmetic mess"! Let's start by calculating the number of seconds in a year. We're told that one earth year equals 365.24 days of 24.000 hours each. We know that an hour contains precisely 60 minutes, and a minute contains precisely 60 seconds. Therefore, for our purposes, we can consider a year as 365.24 x 24.000 x 60.000 x 60.000 = 3.1556736 x 107 s. (We carry out this result to a few extra digits to avoid cumulative rounding error, an annoying phenomenon that can sometimes occur if we round off our values before we finish a calculation sequence.) Our distant star's "sunspot cycle" has a duration of 25.000 earth years, representing 3.1556736 x 107 x 25.000 = 7.88918400 x 108 s. The frequency (in hertz) equals the reciprocal of this unwieldy quantity, or 1.2676 x 10-9 Hz, rounded off to five significant figures. The correct choice is B.

10. The peak-to-peak voltage of any AC wave, no matter how irregular, equals the maximum instantaneous voltage minus the minimum instantaneous voltage. In this situation, that's +1.35 - (-0.35) = 1.35 + 0.35 = 1.70 V. The answer is A.