Teach Yourself Electricity and Electronics, 5th edition |

Stan Gibilisco |

Explanations for Quiz Answers in Chapter 13 |

1. If the frequency of an AC source equals f (in hertz) and the inductance of a
coil equals L (in henrys), then we can calculate the inductive reactance X
(in ohms) using the approximate formula_{L}
fLThis formula also works for values of f = 2.55 kHz. When we plug those values into the above formula, we get100 = 6.2832 x 2.55 x Dividing this equation through by the quantity (6.2832 x 2.55), we obtain
The correct choice is (c). |

2. As we raise the frequency of an AC signal through a component that exhibits a fixed
inductance, the component's inductive reactance increases. The formula
fLtells us that the inductive reactance increases goes up in direct proportion to the frequency. The correct answer is (c). |

3. Once again, let's use the formula for inductive reactance in terms of frequency and
inductance, specifying L in microhenrys (µH) and f in megahertz:
fLWe know that f = 1.50. Plugging in those
values yields200 = 6.2832 x 1.50 x Solving for
The correct choice is (a). |

4. If we have an RL circuit containing finite, nonzero inductive reactance and
finite, nonzero resistance, then the phase angle exceeds 0º but is smaller than 90º. The
current lags the voltage by some amount less than 1/4 of a cycle. The correct answer is
(b). |

5. Let's use the formula for the phase angle and then plug in the ratio directly. We
can do this trick even though we don't know the exact values of the inductive reactance or
the resistance. The formula is
R)where R represents the resistance (in the same
units as the inductive reactance). We've been informed that the ratio X / _{L}R
equals 2, so we have
A scientific calculator tells us that Arctan 2 = 63º (rounded off to the nearest degree). That's more than 45º but less than 90º, so the correct choice is (c). |

6. We can solve this problem in the same fashion as we solved the previous one, except
that now the ratio X / _{L}R equals 1/2, so we get the formula
A scientific calculator reveals that Arctan (1/2) = 27º (rounded off to the nearest degree). That's more than 0º but less than 45º, so the correct choice is (a). |

7. We can see from the diagram that the inductive reactance is approximately 60 ohms
and the resistance is approximately 14 ohms. Therefore, X / _{L}R
equals approximately 60/14, or 4.29. The correct answer is (d). |

8. We can find the phase angle by taking the Arctangent of the ratio X / _{L}R.
In this case, that ratio equals approximately 4.29. When we carry out the operation with
our calculator, we get approximately 77º. The correct choice is (a). |

9. If we take away turns from a coil, the inductance goes down. If we keep the AC frequency constant through a coil as its inductance decreases, the inductive reactance also decreases. The correct answer is (c). |

10. We're told (in the statement of Question 9) that the coil contains no resistance, and we're also assured that no external resistance exists either. Therefore, we have a theoretically pure inductance, no matter how many turns the applied AC passes through. In a pure inductance carrying AC, the current always lags the voltage by 90º. Therefore, in theory, the phase angle does not change as we adjust the coil to have fewer turns. The correct choice is(a). |

11. Mathematically, each point in the RX quarter-plane corresponds
to a unique (one and only one) complex impedance, and vice-versa. Those two conditions,
taken together, constitute a _{L}one-to-one correspondence, so the correct answer is
(d). A given inductance can correspond to infinitely many different points in the
quarter-plane, so (a) won't work. A given value of inductive reactance can correspond to
infinitely many different points in the quarter-plane, so (b) is wrong. The same thing
happens with specific values of resistance, so (c) doesn't work either. |

12. We're told that a coil exhibits an inductance of 75.00 mH, and we apply an AC
signal of 4.400 kHz. We can find the inductive reactance X in ohms by
plugging _{L}L = 75.00 and f = 4.400 directly into the formula
fLthereby obtaining
= 2073 ohms The correct answer is (b). |

13. We're given the inductance in microhenrys and the reactance in ohms. If we input
the given values for L and X directly into the formula_{L}
fLwe will get an equation for 1800 = 6.2832 x When we divide through by the quantity (6.2832 x 150), we get
The correct answer is (c). |

14. If we allow both the resistance R and the inductive reactance X
in an _{L}RL circuit to vary from zero to unlimited values while always remaining equal
to each other, the ratio X / _{L}R always equals 1. That means the
phase angle f equals 45º because
R)= Arctan 1 = 45º The actual values of |

15. This problem is tricky if you don't remember the equation for a circle in
rectangular coordinates. You should have learned that equation in algebra or pre-calculus
courses, but if you never got that far or if you forgot it, don't worry. Even if you can't
remember the equation, you can figure out by elimination that the right choice here is
(d). Just plug in a few numbers for R and X that satisfy the
equation_{L}
^{2} = 100and see how the points come out! If you remember your algebra or pre-calculus, you'll recognize that the above equation produces a circle with a radius of 10 units in rectangular coordinates. Because we only have a quarter-plane to work with when dealing with resistance and inductive reactance, the set of points satisfying the above equation will turn out as a quarter-circle (therefore an arc) centered at the origin. |

16. If the X / _{L}R ratio is tiny, then the phase angle (Arctan X / _{L}R)
is close to 0º. As the ratio X / _{L}R increases, the phase angle
also increases. If X / _{L}R increases without limit, its Arctangent
approaches 90º. The phase angle can approach (but never reach) 90º, no matter how large
the ratio gets. The correct answer is (a). |

17. If the ratio X / _{L}R is very large, then the phase angle
(Arctan X / _{L}R) is close to 90º. As the ratio X / _{L}R
decreases, the phase angle also decreases. If X / _{L}R attains the
value of 1:1 (which equals 1), its Arctangent approaches 45º (because Arctan 1 = 45º).
The correct answer is (b). |

18. We're told that a coil has an inductance of L = 250 nH. In microhenrys
(µH), that's equivalent to 0.250, because 1 nH = 0.001 µH. The frequency f equals
144 MHz. We can plug the values L = 0.250 and f = 144 directly into the
formula for inductive reactance X, obtaining_{L}
= 226 ohms accurate to three significant figures. The correct choice is (d). |

19. We must take two steps to solve this problem. First, let's find the inductive
reactance X at the frequency _{L}f = 300 kHz. We're told that L
= 0.050 mH. We have units that agree for our formula (kilohertz and millihenrys, expressed
in thousands and thousandths, respectively), so we can plug the numbers f = 300 and
L = 0.050 straight in, getting
= 94.248 ohms We have a resistor of
R = 94.248 / 200= 0.47124 (Let's leave in the extra digits to minimize the effect of cumulative rounding errors
in our calculation; we'll round off to the nearest phase-angle degree at the end of the
calculation process.) Now we're ready for the second step: Calculate the phase angle
R)= Arctan 0.47124 = 25º The correct answer is (b). |

20. As in the previous problem, we need to go through two steps. First, let's find the
inductive reactance X at _{L}f = 880 Hz. We're told that L =
88 mH, which equals 0.088 H. Now that we have units that agree (hertz and henrys), we can
plug in f = 880 and L = 0.088 to obtain
= 486.57 ohms The circuit contains a resistance of
R = 486.57 / 88= 5.5292 We calculate the phase angle as
R)= Arctan 5.5292 = 80º The correct answer is (a). |