| 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 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 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 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 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 Now we can calculate the power factor as PF = R / Z 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 so we end up with PF = R / Z 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 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) = 53.1301º 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 53.1301º 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 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) = 36.8699º The power factor equals the cosine of the phase angle, so PF = cos 36.8699º 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 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 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 µF.
Then we find the product LC = 75 x 0.000150 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 The correct choice is (d). |