Teach Yourself Electricity and Electronics, 5th edition
Stan Gibilisco
Explanations for Quiz Answers in Chapter 3
1. An electroscope consists of two foil leaves suspended inside an insulating container, and connected to an electrode that extends through the top of the container to the outside. When an electrically charged object comes near, or touches, the outside electrode, the foil leaves become charged with the same polarity. The resulting electrostatic force pushes the leaves apart. The correct answer is (b). If we want to analyze an AC wave, we might employ an oscilloscope, but never an electroscope, so (a) is wrong. Answers (c) and (d) are likewise wrong; we might use an ohmmeter in either of these applications, but an electroscope would do us no good at all.
2. A current-carrying wire produces a magnetic field, which exerts force on the needle of a magnetic compass. The needle is actually a small permanent magnet with its own magnetic field. The interaction between the magnetic fields around the wire and the compass needle produce electromagnetic deflection of the needle. The correct answer is (a). The force and deflection have nothing to do with electrostatic effects, so (b) and (c) are wrong. No such term as "electroscopic force" even exists, at least not among engineers I've encountered, so (d) is wrong.
3. A mischievous person might apply a large AC voltage to a solar cell, thermocouple, or illuminometer, destroying it and causing it to emit a visible flash along with a puff of smoke; but a competent, sane engineer would never do such a thing. None of those devices are designed to produce visible light from electricity of any kind. Answers (a), (b), and (c) are all wrong. That leaves us with choice (d), "None of the above," as the correct answer.
4. The meter indicates 60 mV, because we see six of the 10 rectangles illuminated, and each rectangle indicates 10 mV. Remember that 1 mV equals 1/1000 V or 0.001 V. Our meter shows (6 x 10) / 1000 V, which equals 0.06 V. The correct answer is (c).
5. Analog meters, such as the D'Arsonval type, give the observer a good idea of how a quantity fluctuates from moment to moment. Most people find it easy to visually follow along with the motions of a needle on a scale, but difficult or impossible to intuit the moment-to-moment variations in a direct numeric reading that doesn't hold to a constant value. The best answer is (d). Analog meters don't offer any advantage for energy measurement as opposed to power measurement, so (a) is no good. Analog meters aren't any better than direct numeric displays for measurement of large currents, so (b) is wrong. Answer (c) doesn't make sense; it implies that numeric digital displays have needles and graduated scales. They don't.
6. Oscilloscopes work pretty well for waveform analysis, so (a) does not apply here. We can use an oscilloscope to get a good idea of the frequency of an AC wave, so we can throw out (b). We can measure an AC signal's peak level with an oscilloscope, so (d) does not apply. That leaves us with only answer (c), and that's definitely right in this context! Knowledgeable engineers or technicians wouldn't try to use an oscilloscope to measure DC resistance (although, in theory, they could "rig up a scope" to make it do that indirectly); they'd have ohmmeters available for resistance determinations.
7. The correct answer is (c). Of the four units mentioned here, only kilowatt-hours express energy. The units given in (a) and (b) quantify power, not energy. Ampere-hours, specified in (d), represent neither power nor energy, but you'll occasionally see battery capacity quoted in those terms.
8. If you want to reverse the direction of needle deflection, you must reverse the direction of the current in the wire coil surrounding the compass. You can do that by reversing the battery polarity. You can therefore rule out answers (c) and (d). You want to reduce the needle deflection from 20 degrees to 10 degrees, so you must reduce the current in the coil. You can do that by increasing the resistance. The correct answer is (b).
9. Electromagnetic deflection involves the interaction of electric currents and magnetic fields. None of the three answers (a), (b), and (c) mention magnetism in any way, so they're all wrong. Answer (d) is correct; if we bring a current-carrying wire near a magnetic compass, the magnetic field surrounding the wire can cause the compass needle to veer away from geomagnetic north.
10. Figure 3-18 shows a volume-unit (VU) meter, of the sort used in high-fidelity audio systems. The needle is in the "heavy-line" part of the scale, indicating a high risk of audio distortion, which can result from setting the gain (or volume control) too high. The correct answer is (c). Answer (a) is wrong; VU meters have nothing to do with UV (ultraviolet rays)! Answer (b) is no good, because VU meters do not quantify resistance. Answer (d) is likewise wrong; VU meters have nothing to do with measurement of utility voltage.
11. If you want to get the meter to register higher currents than its full-scale maximum of 50 mA, you must connect something in parallel with the meter to take away part of the "current burden." A meter shunt will do this, so the correct answer is (b). Answer (a) is obviously wrong; you don't have to buy a new meter. Answer (c) is wrong; if you connect a resistance in series with a milliammeter, you'll reduce the current through the entire circuit, not only through the meter. A meter should never alter the behavior of the circuit under test. Answer (d) is likewise wrong. If you connect a battery in series with a meter, you'll probably change the behavior of the circuit.
12. Photovoltaic cells can produce DC electricity from infrared (IR), visible light, and ultraviolet (UV) rays. An illuminometer takes advantage of this property, making it possible to measure IR, visible light, or UV intensity. The correct answer is (a). A volume-unit (VU) meter indicates the loudness of sound, so (b) is wrong. A utility meter tells us the amount of electrical energy consumed over a period of time, so (c) is wrong. We use an ohmmeter to measure resistance, so (d) is wrong.
13. Engineers and technicians use ammeters to directly measure current, not voltage. The correct answer is therefore (a). Answers (b) and (c) describe techniques for measuring the current that a circuit draws from a battery. Performing the action described by (d) would short out the battery, causing the circuit to shut down.
14. A typical ohmmeter can quantify resistances that range from a fraction of an ohm to several megohms, so we can rule out answers (a) and (b). If a voltage exists between the points under test, then that voltage will appear as a "phantom battery" in series with the ohmmeter's own internal battery, throwing off the meter reading. (We might actually get a reading of "more than infinity ohms" if the external voltage is large enough, and if its polarity opposes that of the ohmmeter's internal battery.) The correct answer is (c).
15. An ammeter should have the lowest possible internal resistance, not the highest. We must therefore choose (d).
16. Answer (c) is correct. A FET voltmeter (FETVM) is designed specifically to measure small DC voltages, and/or to minimize the effect of the meter on the circuit under test. Answer (a) won't work; a microammeter quantifies current, not voltage. Answer (b) is wrong; a galvanometer is a crude current meter, not a voltmeter. Answer (d) won't work, because VU meters are designed to quantify sound volume in audio systems, not to determine DC voltages.
17. We commonly find frequency counters in electronics labs. A device of this type literally counts the number of wave pulses or cycles in a fixed period of time, thereby determining the frequency. The correct answer is (b). An engineer would never use a pulse counter to measure DC voltage, DC power, or sound volume, so answers (a), (c), and (d) are all wrong.
18. The correct answer is (a). The instantaneous motor speed varies in direct proportion to the amount of electrical power consumed by the household. The voltage in a properly functioning utility system varies little or not at all, so (b) is wrong. The energy consumed by the household constantly increases with time; if the motor speed were proportional to it, the motor would run faster and faster as time went by, eventually reaching impossible speeds! Therefore, (c) is wrong. The AC frequency remains constant with time; it's irrelevant to the utility usage anyway, so (d) is wrong.
19. Take note of the word directly here, because it's critical to understanding the question. In a pointer-type ohmmeter, the D'Arsonval movement deflects in proportion to the amount of direct current flowing through its coil(s). This current is identical to the current flowing through the resistance under test. Therefore, the correct answer is (a). The meter reading may vary with changes in the voltage across the resistance, the power dissipated by the resistance, or the energy consumed by the resistance, but the D'Arsonval movement itself does not directly respond to any of those parameters.
20. The meter needle rests at a point corresponding to roughly 13 on the scale. The range switch position tells us to multiply the meter reading by 10 k or 10,000 ohms. The meter therefore indicates approximately 13 x 10,000, or 130,000 ohms. The correct answer is (c).