Teach Yourself Electricity and Electronics, 5th edition
Stan Gibilisco
Explanations for Quiz Answers in Chapter 17
1. For a transmission line to operate at peak efficiency (minimum possible loss), we must make sure that the load impedance is purely resistive, and that the load resistance equals the characteristic impedance of the line. The correct choice is (a). You might think that choice (d) would also work, but that's true only in the special case where the load happens to contain no reactance.
2. The ninth harmonic of a signal occurs at a frequency equal to 9 times the fundamental. In this case, the fundamental frequency equals 900 kHz, or 0.900 MHz. When we multiply that figure by 9, we get 8.10 MHz. The correct answer is (d).
3. A pure resistance radiates or dissipates true power. The answer is (c). Imaginary power (b) exists only in reactances, and apparent power (d) expresses a mathematical combination of true and imaginary power. We haven't defined "complex power" (a) at all.
4. We want to build a half-wave dipole antenna for a fundamental frequency of 14.3 MHz. If we specify the end-to-end length of the antenna in meters as Lm, then

Lm = 143 / fo

for fo in megahertz. When we plug 14.3 for fo, we get

Lm = 143 / 14.3
= 10.0 m

The correct choice is (b).

5. In a transmission line that has standing waves, the voltage maxima (loops) exist at the same points as the current minima (nodes). The correct choice is (b).
6. When a transmission line operates with an impedance mismatch, standing waves occur, producing higher currents and voltages at certain points along the line compared with the perfectly-matched condition (assuming that the transmitter generates the same amount of signal output in either case). At the current maxima, the loss increases in the wire conductors, so the correct answer to this question is (a). At the voltage maxima, the loss increases in the dielectric material between the conductors -- but actually decreases somewhat in the wire conductors because voltage maxima correspond to current minima!
7. In an AC circuit containing both resistance and reactance, the cosine of the phase angle equals the power factor by definition. The correct choice is (d).
8. True power always shows up as dissipation, radiation, or a change in form. Of the four choices given here, only (c) meets that criterion, so it's the right answer.
9. If we know the apparent power PVA and the true power PT, we can calculate the power factor PF using the formula

PF = PT / PVA

Plugging in the numbers PT = 30 and PVA = 40, we obtain

PF = 30/40
= 0.75
= 75%

The correct answer is (b).

10. To determine the power factor in this situation, we must know the apparent or volt-ampere power PVA. We've been told the true power PT and the imaginary power PX, so we can use the formula

PVA = (PT2 + PX2)1/2

When we plug in the values PT = 40 and PX = 30, we get

PVA = (402 + 302)1/2
= (1600 + 900)1/2
= 25001/2
= 50

Now we can calculate the power factor PF using the formula

PF = PT / PVA

Plugging in the numbers PT = 40 and PVA = 50 yields the result

PF = 40/50
= 0.80
= 80%

The correct choice is (c).

11. We've been told that R = 24 ohms and X = 10 ohms. First, let's find the absolute-value impedance Z using the formula for series circuits:

Z = (R 2 + X 2)1/2
= (242 + 102)1/2
= (576 + 100)1/2
= 6761/2
= 26 ohms

Now we can calculate the power factor as

PF = R / Z
= 24/26
= 0.92
= 92%

The correct choice is (c).

12. The calculation of PF yields the same result here as it did in the solution to Question 11. We have R = 24 ohms and X = -10 ohms. The calculations for Z proceed as follows:

Z = (R 2 + X 2)1/2
= [242 + (-10)2]1/2
= (576 + 100)1/2
= 6761/2
= 26 ohms

so we end up with

PF = R / Z
= 24/26
= 0.92
= 92%

Once again, the correct answer is (c).

13. When we solved Question 11, we found that PF = 92%, or 0.92. The formula for PF in terms of true power PT and VA power PVA is

PF = PT / PVA

We can rearrange this formula to

PT = PF x PVA

When we plug in PF = 0.92 and PVA = 100, we get

PT = 0.92 x 100
= 92 W

The correct choice is (d).

14. We've been told that R = 60.0 ohms and X = 80.0 ohms. We also know that the true power PT equals 100 W. First, let's calculate the phase angle as

f = Arctan (X / R)
= Arctan (80.0 / 60.0)
= 53.1301

We take the Arctangent to a few extra digits because it doesn't come out as a "clean value." The power factor equals the cosine of the phase angle, so

PF = cos 53.1301
= 0.600000

The formula for the power factor in terms of true and VA power is

PF = PT / PVA

We can rearrange this to get

PVA = PT / PF

When we plug in PF = 0.600000 and PT = 100 and round off to three significant figures, we obtain

PVA = 100 / 0.600000
= 167 W

The correct choice is (a).

15. In this case, we have R = 80.0 ohms and X = 60.0 ohms. As in the situation described by Question 14, the true power, PT, equals 100 W. We calculate the phase angle as

f = Arctan (X / R)
= Arctan (60.0 / 80.0)
= 36.8699

The power factor equals the cosine of the phase angle, so

PF = cos 36.8699
= 0.800000

We determined in the solution to Question 14 that

PVA = PT / PF

When we plug in PF = 0.800 and PT = 100 and round off to three significant figures, we obtain

PVA = 100 / 0.800000
= 125 W

The correct choice is (c).

16. We begin by calculating the product of the inductance L in microhenrys and the capacitance C in microfarads, as follows:

LC = 36 x 0.0010
= 0.036

We can use the given values directly, because microhenrys and microfarads go together in the formula. When we take the square root of 0.036, we get 0.18974. Finally, we divide 0.15915 by 0.18974 and round off to two significant figures, getting a resonant frequency of 0.84 MHz. The correct choice is (b).

17. If we place a pure resistance of any value in series with the coil and capacitor, the resonant frequency will remain the same. A pure resistance has no effect on the resonant frequency of any LC circuit, regardless of where we insert that resistance. (The resistance will, however, affect the "sharpness" of the resonant response.) The correct choice is (b).
18. First, let’s convert the capacitance to microfarads; 150 pF = 0.000150 F. Then we find the product

LC = 75 x 0.000150
= 0.01125

When we take the square root of 0.01125, we get 0.10607. Finally, we divide 0.15915 by 0.10607 and then round off to two significant figures, getting fo = 1.5 MHz. The correct answer is (a).

19. If we connect a 22-pF capacitor in parallel with the circuit described in Question 18, we'll end up with a net capacitance of 150 pF + 22 pF = 172 pF. Because this capacitance exceeds the original capacitance, the product LC will increase, and its square root will likewise increase. When we divide 0.15915 by this increased quantity, we'll get a lower resonant frequency than the one we had before. The correct choice is (c).
20. We have an operating frequency of fo = 18.1 MHz and a transmission line with a velocity factor of v = 0.667. The length Lm of a quarter-wave section in meters is

Lm = 75.0 v / fo

When we plug in the numbers, we get

Lm = 75.0 x 0.667 / 18.1
= 2.76 m

The correct choice is (d).