Teach Yourself Electricity and Electronics, 5th edition
Stan Gibilisco
Explanations for Quiz Answers in Chapter 4
1.We want to build a component that can dissipate at least 16 W of power (14 W plus 2 W as a "safety margin"). We have an unlimited supply of 20-ohm resistors rated at 1 W each. We can combine 16 resistors in series, 16 resistors in parallel, or a 4 x 4 series-parallel matrix; any of these arrangements will provide 16 W of power-handling capacity because the resistors all have identical ohmic values. Remember, the net resistance makes no difference! (If the ohmic values weren't all the same, this simple rule would not apply.) Any of the answers (a), (b), or (c) will work. Therefore, the correct answer is (d).
2.Recall the formula for the dissipated power P (in watts) in terms of the voltage E (in volts) and the resistance R (in ohms):

P = E 2 / R

In this situation, E = 1.50 V and R = 100 ohms. Therefore, we have

P = (1.50)2 / 100
= 2.25 / 100
= 0.0225 W
= 22.5 mW

The correct answer is (c).

3. Recall the formula for converting conductance G (in siemens) into resistance R (in ohms):

G = 1 / R

Each of our components exhibits a conductance of 0.10 S. When we take the reciprocal of this figure to obtain a resistance value, we get 10 ohms. When we connect four 10-ohm resistances in series, we obtain a net resistance of 10 + 10 + 10 + 10 = 4 x 10 = 40 ohms. The correct answer is therefore (c).

4. Each 10-ohm resistor can safely handle up to 1/2 W. We want to get a 10-ohm resistor that can handle 5.5 W, which is 11 times 1/2 W. We want the net resistance of the combination to equal the ohmic value of any one of the original resistors; this fact tells us that we need to build an n x n series-parallel matrix. If we build a 3 x 3 network, we'll get nine times the original power-handling capacity, or 4.5 W. That's not enough, so answer (a) won't work. If we build a 4 x 4 network, we'll get 16 times the original power-handling capacity, or 8 W. That's more than enough, but it's the smallest n x n series-parallel matrix that we can use for this purpose. Therefore, (b) is the right answer. The networks cited in answers (c) and (d) will give us workable networks, but much larger than necessary; the question specifically asks us which network constitutes the smallest functional arrangement.
5. We can solve this problem using the Ohm's Law formula for determining current as a function of voltage and resistance. We have E = 18 V and R = 2.2 k = 2200 ohms. Therefore, the resistor draws a current of

I = E / R
= 18 / 2200
= 0.0082 A
= 8.2 mA

The correct answer is (d).

6. To solve this problem, we can use the Ohm's Law formula for finding the resistance when we know the voltage and the current. In this situation, we have E = 20 mV = 0.020 V and I = 20 mA = 0.020 A. When we plug these values into the formula for resistance, we get

R = E / I
= 0.020 / 0.020
= 1.0 ohm

The correct answer is (a).

7. We have four resistors in parallel, and their values are all the same (1.60 M). Whenever we connect a bunch of identical resistances in parallel, the net resistance of the combination equals the resistance of any one component divided by the number of components. Therefore, the net resistance equals 1.60 M / 4, which is 0.400 M. Most engineers would express this value as 400 k, so the correct answer is (b).
8. Let's find the net resistance R of the parallel combination. Using the formula for resistances in parallel, we have

R = 1 / (1/51 + 1/68 + 1/82)
= 1 / (0.0196 + 0.0147 + 0.0122)
= 1 / 0.0465
= 21.5 ohms

This resistance appears across a 12 V battery. Letting E = 12 V, we can calculate the total current I through the combination using Ohm's Law as

I = E / R
= 12 / 21.5
= 0.56 A

That's choice (b).

9. If we increase the conductance of a component by a factor of 16, then we make its resistance 1/16 as great. Recall the formula for the wattage P in terms of the voltage E (in volts) and the resistance R (in ohms):

P = E 2 / R

We're told that the voltage remains constant. The resistance becomes 1/16 as great. The quotient E 2 / R, and therefore the dissipated power P, increases by a factor of 16. The correct answer is (b).

10. The formula stated in the solution to Question 9 applies, but now we leave the conductance (and therefore the resistance) constant,and increase the voltage by a factor of 16. This voltage increase will cause the quotient E 2 / R, and therefore the dissipated power P, to grow larger by a factor of 162, or 256. The correct answer is (a).
11. Ohm's Law provides a formula for the current that a circuit draws, in terms of the voltage it receives and its overall resistance. That formula is

I = E / R

where we specify I in amperes, E in volts, and R in ohms. By looking at this formula, we can see that if we triple the voltage and leave the resistance the same, the current will triple. Therefore, the correct answer is (c).

12. Recall the formula for wattage P in terms of voltage E (in volts) and resistance R (in ohms):

P = E 2 / R

If we triple the voltage E but leave the resistance R the same, the numerator in the quotient increases by a factor of of 32, which is 9. The denominator doesn't change. Therefore, the power must increase by a factor of 9. The correct answer is (d).

13. The current is 20 mA, which equals 0.020 A. The resistance is 35.35 K, which equals 35,350 ohms. Using the Ohm's Law formula for voltage in terms of current and resistance, we obtain

E = IR
= 0.020 x 35,350
= 707 V

We know the current to an accuracy of two significant figures, while we know the resistance to four significant figures. We must round our answer off to the lesser degree of accuracy, which is two significant figures. In order to accomplish that task with the voltage as we've derived it, we can convert it to kilovolts so we can express it as 0.71 kV. Therefore, the best expression for this voltage is given by answer (c). Answers (a) and (b) claim too much accuracy, while answer (d) doesn't claim enough accuracy.

14. The potentiometer carries 2.00 mA, which equals 0.00200 A. The resistance is 240 ohms. We can calculate the dissipated power using the formula

P = I 2R

where we express P in watts, I in amperes, and R in ohms. When we plug in the numbers and do the arithmetic, we obtain

P = 0.002002 x 240
= 0.00000400 x 240
= 0.000960 W
= 960 W

that is, 960 microwatts, choice (d).

15. When we connect multiple resistors in parallel, all having identical ohmic values, then the net resistance of the combination equals the value of any single resistor divided by the number of resistors. We have seven 70-ohm resistors in parallel, so the net resistance is 70/7, or 10 ohms. (The question contains a slight distraction, because the battery voltage has nothing to do with the resistance of the circuit.) The correct answer is (c).
16. If we remove one of the resistors from the circuit described in Problem 15, we will have a net resistance of 70/6, which is between 11 and 12 ohms. Remember the Ohm's Law formula for current in terms of voltage and resistance:

I = E / R

Because the resistance of the combination has increased slightly, the current that it draws must go down slightly. The correct answer is therefore (b).

17. Let's begin by calculating the net resistance R, in ohms. That's easy, because we have a series combination:

R = 10 + 33 + 100
= 143 ohms

Our power supply provides 50 V DC. Using the formula for wattage P in terms of voltage E (in volts) and resistance R (in ohms), we have

P = E 2 / R
= 502 / 143
= 2500/143
= 17 W

That's choice (a).

18. Once again, recall the Ohm's Law formula for current in terms of voltage and resistance:

I = E / R

We have E = 750 mV = 0.750 V and R = 750 k = 750,000 ohms. When we plug these values into the formula and calculate, we obtain

I = 0.750 / 750,000
= 0.00000100 A
= 1.00 A

or 1.00 microampere, which is choice (c).

19. The voltage is 13.52 V. The current is 4.6 mA, which equals 0.0046 A. Using the Ohm's Law formula for resistance in terms of voltage and current, we obtain

R = E / I
= 13.52 / 0.0046
= 2939 ohms

We know the voltage to four significant figures. We know the current to two significant figures. We must therefore round our result off to two significant figures. To do that here, we can convert the resistance to kilohms and express it as 2.9 k. Answer (c) offers the best expression for the resistance. Answers (a) and (b) claim too much accuracy, while answer (d) doesn't claim enough.

20. Let's convert the conductance to resistance by taking its reciprocal. The reciprocal of 0.25 is 4.0, so we know that the component has a resistance of 4.0 ohms. It carries 20 A of DC. Using the formula for voltage in terms of current and resistance, we obtain

E = IR
= 20 x 4.0
= 80 V

The correct answer is (c).