Electricity Demystified, 2nd edition Stan Gibilisco Explanations for Quiz Answers in Chapter 6 1. In Fig. 6-7, each vertical division represents 5.00 mV. The positive peak of the waveform lies exactly four divisions above the horizontal axis (which represents 0.00 V). Therefore, the positive peak voltage equals +4.00 x 5.00 = +20.0 mV pk+. The correct choice is C. 2. In Fig. 6-7, the negative peak of the waveform lies exactly two divisions below the horizontal axis. Therefore, the negative peak voltage equals -2.00 x 5.00 = -10.0 mV pk-. The correct choice is C. 3. The peak-to-peak voltage of the waveform of Fig. 6-7 equals the positive peak voltage minus the negative peak voltage. In this case, if we let Epk-pk represent the peak-to-peak voltage, we calculateEpk-pk =  +20.0 - (-10.0) = +20.0 + 10.0 = 30.0 mV pk-pk The correct choice is C. Remember, we don't need to put a polarity sign in front of a peak-to-peak voltage (and in fact we shouldn't), because polarity lacks relevance in the peak-to-peak context. 4. We're told that the waveform of Fig. 6-7 is a perfect sinusoid, even though the positive and negative peak voltages differ. Therefore, we can find the average voltage Eavg by adding the positive peak voltage and the negative peak voltage, and then dividing by exactly 2. We getEavg = [+20.0 + (-10.0)] / 2.00 = +10.0 / 2.00 = +5.00 mV avg The correct choice is B. 5. In Fig. 6-7, each horizontal division represents 100 ns. The illustration clearly shows us that successive positive peaks lie exactly four divisions apart. The period therefore equals 4.00 x 100 = 400 ns. The correct choice is A. 6. We can calculate the frequency of an AC wave (in hertz) as the reciprocal of its period (in seconds). We've found that the period of the wave in question equals 400 ns, which we can write as 0.000000400 s (or 4.00 x 10-7 s, for those of you who prefer scientific notation). The reciprocal of that number is 2,500,000 (or 2.50 x 106), so the frequency equals 2.50 x 106 Hz or 2.50 MHz. The correct choice is D. 7. The period of an AC wave varies inversely in proportion to its frequency. Neither the period nor the frequency have anything to do with the wave amplitude in any form (peak-to-peak, RMS, average, or anything else). The correct choice is D, "None of the above." 8. The frequency of an AC wave varies inversely in proportion to its period. The voltage, current, or waveform have nothing to do with either the frequency or the period. The correct choice is B. 9. The efficiency of a generator, in percent, equals precisely 100 times the useful electrical output power divided by the mechanical input power. In this situation, we drive a generator with 2100 W of mechanical power, and we obtain 1500 W of electricity at the output. The efficiency in percent, Eff%, can be calculated asEff% = 100 x (1500 / 2100) = 100 x 0.7143 = 71.43% The correct choice is C. 10. Assuming that the utility AC constitutes a perfect sine wave with no DC component, the peak-to-peak voltage, Epk-pk, equals approximately 2.828 times the RMS voltage. If a utility wave arrives at our wall outlets at 117 V RMS, thenEpk-pk = 2.828 x 117 = 331 V pk-pk The correct choice is A.