Physics Demystified, 2nd edition
Stan Gibilisco
Explanations for Quiz Answers in Chapter 10
1. The complex-impedance vector for a pure inductance always points straight up in the resistance-reactance (RX) half-plane along the inductive-reactance (XL) axis, as long as the inductance and frequency are both nonzero. The complex-impedance vector for a pure capacitance always points straight down along the capacitive-reactance (XC) axis, as long as the frequency and capacitance are both nonzero. Therefore, the vectors for a pure inductance and a pure capacitance always point in opposite directions in the RX half-plane. The correct choice is B.
2. In order to determine the complex-number impedance of an inductor, we must know the frequency, and we must also know the amount of internal resistance (if any) that the component has. We haven't been given either of those two parameters, so we can't calculate an answer here. The correct choice is D, "We need more information to determine it."
3. The schematic diagram of Fig. 10-13 shows two pairs of series-connected capacitors wired in parallel with each other. All four of the capacitors have the same value; let's call it C. The capacitance of the upper pair of series-connected components equals C/2; the capacitance of the lower pair also equals C/2. When we connect these two series combinations in parallel with each other, we get 2 x (C/2) = C. The capacitance of this so-called two-by-two series-parallel matrix is the same as the capacitance of any one of the four components taken alone. The correct choice is B.

4. Let's recall the formula for capacitive reactance in terms of capacitance and frequency. If the capacitance equals C (in microfarads) and the frequency equals f (in megahertz), then we can calculate the capacitive reactance XC (in ohms) as

XC = -(6.2832 f C)-1
= -1 / (6.2832 f C)

When we examine this formula, we can see that if we double the value of f (from 5 MHz to 10 MHz, for example), we double the denominator so that XC becomes 1/2 of its original value. The correct choice is B. (Let's keep in mind that both before and after the frequency change, XC is negative. So, when we say that the value becomes 1/2 as great, we mean that it becomes 1/2 as great negatively.)
5. A complex-number impedance figure of 50 + j0 indicates no reactance whatsoever. The component in question must therefore exhibit a pure resistance of 50 ohms at the specified frequency (in this case 75 MHz). The correct choice is C.
6. For any fixed value of inductance, the inductive reactance varies directly in proportion to the AC frequency. The reactance value XL always equals a positive number of ohms (as compared with negative ohmic values for capacitive reactances). If we pass extremely low-frequency AC through an inductor made from perfectly conducting wire (so that it has no resistance at all), we'll observe an extremely small, positive reactance. The correct choice is C.
7. For any fixed capacitance, the capacitive reactance varies inversely in proportion to the negative of the AC frequency. The reactance XC always equals a negative number of ohms. If we pass extremely low-frequency AC through a capacitor made from perfectly conducting plates separated by dry air (an excellent dielectric medium), we'll observe an extremely large, negative reactance. The correct choice is A.
8. Once again, let's recall that for any fixed inductance, the reactance varies in direct proportion to the frequency of applied AC. If we reduce the frequency of an AC signal passing through a component having a constant inductance, the reactance goes down, telling us that the component exhibits diminishing opposition to the AC. The correct choice is A.
9. If we insert a high-permeability ferromagnetic core inside an inductor that contains mostly air to begin with, we'll increase the inductance a lot. At a fixed AC frequency, the inductive reactance XL of a coil varies in direct proportion to its inductance. Therefore, the addition of the ferromagnetic core will cause XL to increase dramatically. The correct choice is A.
10. This question asks us to start with a fixed capacitor in a vacuum, and then insert a sample of material with a dielectric constant of 1.5 between the plates. This action will increase the capacitance. At a fixed AC frequency, we'll observe a reactance figure that decreases in the negative sense. When a number decreases negatively, its value remains negative but gets closer to zero. The correct choice is therefore B.