Electronics Demystified, 2nd edition
Stan Gibilisco
Explanations for Quiz Answers in Chapter 2
1. We're told that each horizontal-axis division represents exactly one microsecond (1.0 Ás) in the graph of Fig. 2-18. The waveform crosses the horizontal axis going downward at regular intervals corresponding to three divisions, representing 3.0 Ás. The distance between any adjacent pair of such points represents a period of the waveform. Therefore, the period equals 3.0 Ás. The correct choice is B.
2. The frequency in hertz (Hz) equals the reciprocal of the period in seconds (s). We've determined that the period equals 3.0 Ás, or 0.0000030 s. When we take the reciprocal of this value, we obtain 333,333 Hz, which rounds off to 333 kHz. The correct choice is B.
3. We're told that each vertical-axis division in Fig. 2-18 represents exactly 10 volts (10 V). The positive waveform peaks are clarified with a dashed line running horizontally near the top of the graph. That line passes through the vertical axis roughly 3.9 divisions above the origin (the point where the graph's axes intersect each other). Therefore, the positive peak voltage is about +3.9 x 10 V, or +39 V pk+. The correct answer is C.
4. In Fig. 2-18, the negative waveform peaks are clarified with a dashed line running horizontally, somewhat below the horizontal graph axis. That line passes through the vertical axis about 2.1 divisions below the origin. Therefore, the negative peak voltage equals approximately -2.1 x 10 V, or -21 V pk-. The correct choice is B.
5. The peak-to-peak voltage of a waveform (let's call it Epk-pk) equals its positive peak voltage minus its negative peak voltage. In this case, we have

Epk-pk = 39 - (-21)
= 39 + 21
= 60 V pk-pk

We don't use polarity signs when expressing peak-to-peak voltages, because polarity lacks relevance in the peak-to-peak context. The correct choice is D.

6. The root-mean-square (RMS) voltage of a pure AC sine wave with no DC component equals approximately 0.3536 times its peak-to-peak voltage. From this fact, we can deduce that the peak-to-peak voltage equals approximately 1 / 0.3536, or 2.828, times the RMS voltage. In this situation, we measure 24.00 V RMS for a pure AC wave with no DC component. We can calculate the peak-to-peak voltage, Epk-pk, as

Epk-pk = 24.00 x 2.828
= 67.87 V pk-pk

The correct choice is A.

7. If we measure the positive peak voltage of a pure AC sine wave as +24.00 pk+ and its negative peak voltage as -24.00 V pk-, then the peak-to-peak voltage equals 48.00 V pk-pk. That's the positive peak voltage minus the negative peak voltage. The RMS voltage (let's call it ERMS) equals approximately 0.3536 times the peak-to-peak voltage. Therefore, in this case,

ERMS = 48.00 x 0.3536
= 16.97 V RMS

The correct choice is C.

8. To convert "regular frequency" (in hertz) to angular frequency (in radians per second), we multiply by 6.2832. In this case, we have a 60.0-Hz wave, so the angular frequency equals 60.0 x 6.2832, or 377 rad/s. The correct choice is D.
9. We're told that the loss resistance Rloss in the span of high-tension line is 10 ohms. The line carries a current I of 100 A. This is a bit of a trick question, because the voltage in the line doesn't matter here. We don't need to know the voltage in order to calculate the power loss; the current and the line resistance are enough! Remember power equals the current squared, multiplied by the resistance. We can therefore calculate the power loss (let's call it Ploss) as

Ploss = I 2 Rloss
= 1002 x 10
= 10,000 x 10
= 100,000 W
= 100 kW

The correct choice is C.

10. The loss resistance Rloss in the span of line still equals 10 ohms. The line now carries a current I of 50 A. We can calculate the new power loss as

Ploss = I 2 Rloss
= 502 x 10
= 2500 x 10
= 25,000 W
= 25 kW

The correct choice is D.