| 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 XL
(in ohms) using the approximate formula XL = 6.2832 fL This formula also works for values of f in kilohertz and values of L in millihenrys. We've been told that XL = 100 ohms and f = 2.55 kHz. When we plug those values into the above formula, we get 100 = 6.2832 x 2.55 x L Dividing this equation through by the quantity (6.2832 x 2.55), we obtain L = 100 / (6.2832 x 2.55) 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 XL = 6.2832 fL tells 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: XL = 6.2832 fL We know that XL = 200 and f = 1.50. Plugging in those values yields 200 = 6.2832 x 1.50 x L Solving for L, we obtain L = 200 / (6.2832 x 1.50) 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 f = Arctan (XL / R)where f represents the phase angle in degrees, XL represents the inductive reactance, and R represents the resistance (in the same units as the inductive reactance). We've been informed that the ratio XL / R equals 2, so we have f = Arctan 2A 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 XL / R equals 1/2, so we get the formula f = Arctan (1/2)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, XL / 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 XL / 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 RXL quarter-plane corresponds to a unique (one and only one) complex impedance, and vice-versa. Those two conditions, taken together, constitute a 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 XL in ohms by
plugging L = 75.00 and f = 4.400 directly into the formula XL = 6.2832 fL thereby obtaining XL = 6.2832 x 4.400 x 75.00 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 XL directly into the formula XL = 6.2832 fL we will get an equation for f in megahertz. Let's do it! The equation becomes 1800 = 6.2832 x f x 150 When we divide through by the quantity (6.2832 x 150), we get f = 1800 / (6.2832 x 150) The correct answer is (c). |
| 14. If we allow both the resistance R and the inductive reactance XL
in an RL circuit to vary from zero to unlimited values while always remaining equal
to each other, the ratio XL / R always equals 1. That means the
phase angle f equals 45º because f = Arctan (XL / R)= Arctan 1 = 45º The actual values of R and XL can range from "zero to infinity"; the only constraint is that they always have to be the same. Geometrically, then, we can get points anywhere along a ray starting at the origin and ramping up as we move toward the right. The correct choice is (c). |
| 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 XL that satisfy the
equation R2 + XL2 = 100 and 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 XL / R ratio is tiny, then the phase angle (Arctan XL / R) is close to 0º. As the ratio XL / R increases, the phase angle also increases. If XL / 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 XL / R is very large, then the phase angle (Arctan XL / R) is close to 90º. As the ratio XL / R decreases, the phase angle also decreases. If XL / 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 XL, obtaining XL
= 6.2832 x 144 x 0.250 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 XL at the frequency 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 XL = 6.2832 x 300
x 0.050 We have a resistor of R = 200 ohms in the RL circuit. When we divide out the ratio of the inductive reactance to the resistance, we obtain XL / R = 94.248 / 200 (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 f. We use the formula f = Arctan (XL / 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 XL at 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 XL
= 6.2832 x 880 x 0.088 The circuit contains a resistance of R = 88 ohms. When we divide out the ratio of the inductive reactance to the resistance, we get XL / R = 486.57 / 88 We calculate the phase angle as f = Arctan (XL / R)= Arctan 5.5292 = 80º The correct answer is (a). |