|Teach Yourself Electricity and Electronics, 5th edition|
|Explanations for Quiz Answers in Chapter 6|
|1. If we want to minimize the danger of electric shock when we work on a power supply, we must install a bleeder resistor in parallel with each filter capacitor. We must always switch off the power supply before starting any repair or other internal work. Once the supply has been powered-down, we can give ourselves some extra "safety insurance" by shorting out each individual filter capacitor using a tool with an insulated handle, and while wearing rubber gloves. The correct answer is (b). Answer (a) might work with some power supplies, but it's no guarantee of safety. Action (c) would likely blow the supply's fuse or trip its breaker, and could also destroy one or more of the rectifier diodes. Answer (d) certainly won't do. Series resistors would render the filter capacitors ineffectual, and wouldn't bleed off any charge they might acquire.|
|2. To determine the range of resistances we should expect according to the manufacturer's specifications, we should first find 5% of 220 ohms. That's 220 x 0.05, or 11 ohms. The expected minimum resistance is therefore 220 - 11, or 209 ohms; the expected maximum is 220 + 11, or 231 ohms. All three of the values given by answers (a), (b), and (c) lie within this range. No value lies outside the range, so the correct answer is (d), "None of the above.".|
|3. In an RF circuit, high-frequency AC flows. We want all resistors for use in such applications to have a minimum of reactance; that means they should exhibit little or no capacitance or inductance. Carbon composition (a), metal film (b), and nonreactive (d) resistors all meet this requirement, but wirewound resistors (c) usually present significant inductive reactance. They're a poor choice for use in RF systems. The correct answer is (c).|
|4. We define the threshold of hearing as a sound level of 0 dB. If we hear a musical
note and someone tells us that its intensity level equals "63 dB," that person
means to say that its level equals 63 dB above the threshold of hearing. Plugging 63 dB
into the formula for decibel gain or loss in terms of the ratio Q / P of the
output power Q to the input power P, we get the equation
63 = 10 log (Q / P)
where "log" represents the base-10 logarithm. We can divide each side by 10 to get
6.3 = log (Q / P)
When we take the base-10 antilogarithm of both sides, we obtain
antilog 6.3 = Q / P
A calculator tells us that the base-10 antilogarithm of 6.3 equals a little more than 1,995,262, which we can round off to 2,000,000. The sound power of our musical note is about 2,000,000 times greater than the sound power of a whisper at the threshold of hearing! The correct answer is (c).
|5. If we have a sound that starts out at a level of P watts and then it changes
by x dB, we can find the final sound power Q using the formula
Q = P antilog (x / 10)
In this situation, P = 7.70 W and x = -20.0. We can plug in the numbers to obtain
Q = 7.70 antilog (-20.0 / 10)
A calculator tells us that the base-10 antilogarithm of -2.00 equals 0.0100, so
Q = 7.70 x 0.0100
The correct choice is (b).
|6. A resistor between the emitter of an NPN bipolar transistor and the negative power supply terminal limits the current that can flow through the transistor. The correct answer is (a).|
|7. An orange band indicates the numeral 3, so you know that the first two digits in the resistance value both equal 3. A red band in the third position tells you to add two zeros to the numeral. Therefore, the resistor is rated at 3300 ohms. The correct choice is (d).|
|8. To get the optimum base bias for a bipolar transistor, we can connect a pair of resistors in series between the ground and the power supply terminal that provides the voltage for the collector, and then connect the base to the point between the resistors. In a PNP transistor, the collector receives a negative voltage relative to ground. The correct answer is therefore (a).|
|9. Let's go through each of the choices in order. Choice (a) won't work because, in a high-power amplifier, a 1/2-W resistor probably won't have a sufficient power rating to handle the current in the emitter circuit. Choice (b) is plausible, so let's keep it in mind while we examine the last two possibilities. Choice (c) won't work; we would never have reason to connect a resistor in series with a power-supply filter capacitor! We can rule out choice (d) because we've found a decent answer, namely (b). So that's it.|
|10. If we have a 4700-ohm resistor that will draw up to 12 mA of current, we can
calculate the maximum power that it will have to handle using the formula
P = I 2 R
where P represents the dissipated power in watts, I represents the current through the resistor in amperes, and R represents the resistance in ohms. A current of 12 mA equals 0.012 A. Therefore, we can plug in the numbers to get
P = 0.012 x 0.012 x 4700
If we allow for a 10% "safety margin," we should multiply this power rating by 1.1, getting 0.75 W. Choices (a) and (b) are obviously too low. Choice (c) will work; 1 W is plenty. Choice (d) would do the job, but the question specifically asks for the power rating that's high enough but not needlessly high. Therefore, the best answer is (c).
|11. We have a resistor whose value R equals 4700 ohms, and we expect that it
will draw a current I of up to 0.012 A. We can use Ohm's Law to calculate the
potential difference E at that current level, which is the largest voltage we
should expect to observe:
E = IR
The correct choice is (c).
|12. You measure a resistor's value as 795 ohms, but the manufacturer rates it at 820
ohms. You want to find out the error, as a percentage, between your measured value of 795
ohms and the quoted value of 820 ohms. If you call this percentage error Err%,
then you can calculate its value as
Err% = 100 x (795 -
820) / 820
The correct answer is (c). You should use the manufacturer's rated resistance, not your measured resistance, as the basis for determining the error. You get a negative result because the measured value is less than the quoted value.
|13. A linear-taper potentiometer's resistance varies in direct proportion to the extent of control-shaft rotation. The correct answer is (a).|
|14. You expect the component to dissipate up to 850 mW, which is 0.85 W. You should
add a 10% "safety margin" to this rating to obtain 0.85 x 1.10 = 0.935 W. That's
the minimum power dissipation capability that your component should have. As for the
resistance, you have room for an error of up to 20% either way from 680-ohm value. That's
a range of 680 ± 136 ohms, or anywhere between 544 and 816 ohms.
You can go through the choices one by one, and find the answer by elimination. If you connect four 680-ohm, 1/2-W resistors in series-parallel as described in choice (a), you'll get a 2-by-2 matrix that has a resistance of 680 ohms and that can dissipate up to 4 x 1/2 W, or 2 W. That will work, and it'll cost you 40 cents. If you go with choice (b) and connect two 1500-ohm, 1/2-W resistors in parallel, you'll get a 750-ohm, 1-W resistor. That will work, and it'll cost you 20 cents. Now you can rule out choice (a) because it costs too much; the same goes for choice (c) without your having to do any calculation because it would cost you 30 cents no matter what. That leaves choice (d). If you connect two 240-ohm, 1/2-W resistors in series, you'll get a 480-ohm, 1-W component. That's enough power-handling capacity, but the resistance falls outside the acceptable range of 544 to 816 ohms. By a process of elimination, you've found that choice (b) will do the job for the least expense.
|15. You've connected two 1500-ohm, 1/2-W resistors in parallel to solve the problem posed in Question 14. According to the manufacturer's claimed resistance values, therefore, you should have a 750-ohm, 1-W component. However, you discover that all of your resistors have values 20% above the rated amounts. Therefore, your "1500-ohm" resistors actually have values of 1500 + 300, or 1800 ohms. Two of these in parallel give you 900 ohms. The correct choice is (c).|
|16. The first band is green, telling us that the first digit equals 5. The second band is blue, so the second digit equals 6. The third band is orange, which indicates three zeros. That means the component has a rated resistance of 56,000 ohms or 56 k. The gold band tells us that the tolerance equals ±5%. The correct answer is (d).|
|17. A potentiometer's resistance changes smoothly and continuously as we rotate its shaft, but the resistance of a rheostat varies in discrete "jumps" as the moving contact passes from one wire turn to the next. The correct answer is therefore (d). Choice (a) is wrong because a potentiometer lacks inductive reactance while a rheostat has it. Choice (b) is wrong because a rheostat lacks capacitive reactance. Choice (c) is wrong because the ideal roles are exactly reversed!|
|18. The correct answer is (a). Potentiometers and metal-film resistors lack the power-handling capability to function in high-voltage, high-current applications. A heavy-duty rheostat can do the job, assuming it's rugged enough to handle the high current that flows in the transformer primary.|
|19. We're told that the power level increases by a factor of 40. We calculate the
decibel change (dB) using the formula
dB = 10 log (Q / P)
where Q represents the final power and P represents the initial power, both in watts (or in the same units, at least). In this situation, we have a power ratio Q / P of 40, so
dB = 10 log 40
The correct answer is (c).
|20. The first band is violet, telling us that the first digit is 7. The second band is green, telling us that the second digit is 5. The third band is brown, which indicates one zero. That means the rated resistance is 750 ohms. The silver band tells us that the tolerance is ±10%. The resistance can therefore vary up to 75 ohms either way from 750 ohms: a range of 675 to 825 ohms. That's approximately 680 to 830 ohms, so choice (b) is correct.|