|Teach Yourself Electricity and Electronics, 5th edition|
|Explanations for Quiz Answers in Chapter 2|
|1. An electric dipole consists of a pair of charge centers in proximity, and having
different quantities of charge. Electrons carry negative charge, so a dipole's negative
charge center has relatively more electrons than the positive charge center. The
correct answer is (a).
The word relatively holds significance here. Not all electric dipoles have one negative pole and one positive pole in an absolute sense. Both poles might have electron surpluses relative to the neutral state (and therefore both have negative absolute charge); both poles might have electron shortages relative to the neutral state (and therefore both have positive absolute charge). However, the negative charge center in an electric dipole always has relatively more electrons than the positive charge center does, even if the negative charge center is less deficient in electrons.
The key phrase constant polarity also warrants attention. If the dipole's polarity alternates or fluctuates, answer (d) could work -- but in this example, we're told that the polarity remains constant.
|2. The correct answer is (c). A high voltage can drive more current through a fixed resistance than a low voltage can -- and high current kills! High-voltage sources don't always exist along with large quantities of charge, while low-voltage sources can exist with huge charge quantities, so choice (a) is wrong. Magnetic fields arise from current, not from voltage, so (b) is wrong. Electrical conductance has nothing to do with voltage, so (d) is wrong.|
|3. We calculate a component's consumed power by taking the product of the voltage across it and the current through it. If we quadruple the current while leaving the voltage unchanged, we quadruple the power, so the correct answer is (a).|
|4. A millisiemens (mS) represents 0.001 S, so 10 mS represents 10 x 0.001 S, or 0.01
S. The resistance R of a component (in ohms) equals the reciprocal of its
conductance G (in siemens). Here, we have G = 0.01, so we can calculate
R = 1 / G
The correct answer is (d).
|5. The conductance G of a component (in siemens) equals the reciprocal of its
resistance R (in ohms). In this case, we have a resistor with R = 470, so we
G = 1 / R
rounded off to three significant figures. To calculate the conductance in millisiemens, we multiply siemens by 1000, getting G = 2.13 mS. The correct answer is (a).
|6. To calculate the power level P (in watts) at which this circuit breaker will
"trip," we multiply the battery voltage E (in volts) by the breaker's
rated threshold current I (in amperes). In this case, E = 6 V and I =
10 A, so we have
P = EI
The correct answer is (d).
|7. This question has an embedded trick, but hopefully it will make you remember an important distinction. The British thermal unit (Btu) expresses energy, not power. The correct answer is (d). Lay people (and some professionals) have a bad habit of saying "Btu" or "Btu's" when they really mean "British thermal units per hour (Btu/hr)." If someone tells you that a furnace is rated at "90,000 Btu," for example, you can be reasonably sure that they mean 90,000 Btu/hr.|
|8. Let's recall the formula for calculating the power P (in watts) as a
function of the voltage E (in volts) and the current I (in amperes):
P = EI
We can rearrange this formula to express the current in terms of the power and the voltage, getting
I = P / E
In this case, we're told that E = 12 and P = 6. Therefore
I = P / E
The correct answer is (b).
|9. Unless we're careful, we might wander astray when comparing conductance values for
variable wire lengths. We can "stay safe" by converting conductance to
resistance at the outset, and then converting resistance back to conductance at the end.
Our 2-km length of wire has a conductance (in siemens) of G = 800 mS = 0.8 S. The
resistance R (in ohms) is therefore
R = 1 / G
If we look at a span 1 km long, that's half the original wire length. We see only half the total wire resistance, or 0.625 ohms, in that span. When we take the reciprocal of 0.625 ohms, we get a conductance figure of 1.6 S. The correct answer is therefore (c).
|10. The kilowatt expresses power. All of the other units listed in this question express energy. The right answer is (b).|
|11. One full cycle every five seconds represents 1/5 cycle per second, which translates to a frequency of 1/5 Hz or 0.2 Hz. Therefore, (c) is the correct answer.|
|12. In the United States, all (or nearly all) utility companies supply AC electricity at a frequency of 60 Hz. The correct answer is (a).|
|13. A full-wave rectifier takes advantage of the entire AC cycle, inverting one half to obtain pulsating DC output while leaving the other half unchanged. Answers (a), (b), and (c) are all wrong, although answer (b) would work if we were dealing with a half-wave rectifier. The correct answer is (d).|
|14. Voltage has nothing to do with conductance, so (a) does not apply here. A low voltage can produce high current if the potential difference appears across a low resistance, so (c) is wrong. Voltage does not necessarily produce a magnetic field; that only happens if the EMF drives current through something. Therefore, (d) is wrong. The correct answer is (b). A low voltage can occur across a high resistance (or any nonzero resistance, for that matter).|
|15. We can express magnetomotive force (in ampere-turns) as the product of current and the number of turns in a coil or loop. The radius of the loop makes no difference (as strange as that fact might seem). Therefore, if we double the radius of a loop while leaving the number of turns and the current the same, we don't change the magnetomotive force. The correct answer is (b).|
|16. If we double the number of turns in a current-carrying wire loop while leaving the current unchanged, we double the number of ampere-turns. Therefore, we double the magnetomotive force. The correct answer is (c).|
|17. We can express magnetic flux density in teslas or gauss. The gauss represents a maxwell per square centimeter. The correct answer is therefore (c).|
|18. Effective voltage can never exceed peak voltage in any AC circuit, so we can rule out (c). All rectifiers produce DC at their outputs (never AC), so we can rule out (d). If we supply pure sine-wave AC input to a half-wave rectifier, we get an effective output voltage of about 0.354 times the peak output voltage, so (b) obviously won't do. The correct answer is (a).|
|19. The situation here almost duplicates that of the previous question. Effective voltage can never exceed peak voltage, so (c) is wrong. All rectifiers have DC output, so (d) is wrong. If we supply pure sine-wave AC input to a full-wave rectifier, we get an effective output voltage of about 0.707 times the peak output voltage, so (b) is wrong. As before, (a) is correct.|
|20. The chapter text tells us that we can express magnetomotive force in gilberts. Therefore, the correct answer is (d).|