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 L_{m},
then
for
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 P_{VA} and the true power P_{T},
we can calculate the power factor PF using the formula
Plugging in the numbers
The correct answer is (b). |

10. To determine the power factor in this situation, we must know the apparent or
volt-ampere power P_{VA}. We've been told the true power P_{T}
and the imaginary power P_{X}, so we can use the formula
When we plug in the values
Now we can calculate the power factor
Plugging in the numbers
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:
Now we can calculate the power factor as
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:
so we end up with
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 P_{T} and VA power P_{VA}
is
We can rearrange this formula to
When we plug in
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 P_{T} equals 100 W. First, let's calculate the phase
angle as
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
The formula for the power factor in terms of true and VA power is
We can rearrange this to get
When we plug in
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, P_{T}, equals 100 W. We
calculate the phase angle as
The power factor equals the cosine of the phase angle, so
We determined in the solution to Question 14 that
When we plug in
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:
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
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 |

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 f_{o} = 18.1 MHz and a
transmission line with a velocity factor of v = 0.667. The length L_{m}
of a quarter-wave section in meters is
When we plug in the numbers, we get
The correct choice is (d). |