Electronics Demystified, 2nd edition |

Stan Gibilisco |

Explanations for Quiz Answers in Chapter 3 |

1. When we want to add two complex numbers, we should add the real parts to each
other, add the imaginary parts to each other, and then combine the results as a sum. In
this case, we have (5 + The real parts add to give us 5 + (-3) = 2 and the imaginary parts add to produce
The sum of the original two complex numbers is therefore (5 + The correct choice is C. |

2. To find the reciprocal of an imaginary number, we reverse the sign of the j
operator and take the reciprocal of the real coefficient. If we start with j2, as
in this case, we end up with -j(1/2), which equals -j0.5. The correct choice
is B. |

3. Let's start by calculating the reactance of the 100-µH inductor at 318.31 kHz. We
can change the frequency to 0.31831 MHz to go with microhenrys in the inductive-reactance
formula. That formula is
f Lwhere f represents the frequency in megahertz, and L represents the
inductance in microhenrys. In this case, f = 0.31831 MHz and L = 100 µH, so
= 200 ohms The resistance
j200The correct choice is A. |

4. First, let's find the reactance of a 1000-pF capacitor. We're told that the
frequency is 3.1831 MHz. We can express the capacitance as 0.001000 µF to go with
megahertz in the formula
f C)where f
represents the frequency in megahertz, and C represents the capacitance in
microfarads. In this case, f = 3.1831 MHz and C = 0.001000 µF, so
= -50 ohms The resistance
j50The correct choice is A. |

5. Whenever we find resistance and reactance together in an AC circuit, we must know the frequency before we can determine the reactance. Because complex impedance contains components of resistance and reactance, we can't calculate the complex impedance unless we know the frequency. We aren't given any value for the frequency, so the correct choice is D, "We need more information to say." |

6. The resistance has no effect on the resonant frequency of this circuit. Only the
inductance L and the capacitance C have significance. If we express the
inductance in microhenrys, we have L = 100. If we express the capacitance in
microfarads, we have C = 0.001000. The formula is
where
That's the equivalent of 503 kHz, so the correct choice is C. |

7. Because the resistance makes no difference in the resonant frequency of an LC
circuit, shorting out the resistor won't change it. The correct answer is C. The
resistance does influence the "sharpness" of the resonant response,
however. In a series RLC circuit, the "sharpness" increases as the
resistance goes down; the "sharpest" possible resonant response occurs with the
resistor shorted out. In a parallel RLC circuit, the opposite holds true; the
"sharpness" increases as the resistance goes up, reaching its maximum when the
resistor is removed altogether. |

8. First, let's find the susceptance of a 100-µH inductor at 318.31 kHz. Because
we're given the inductance in microhenrys, it goes with megahertz in the formula
f L)where f
represents the frequency in megahertz, and L represents the inductance in
microhenrys. In this case, f = 0.31831 MHz and L = 100 µH, so
= -0.00500 S The resistance
j0.00500The correct choice is B. |

9. Let's start by calculating the susceptance of the 1000-pF capacitor at 3.1831 MHz.
That capacitance equals 0.001000 µF, a value that goes with megahertz in the
capacitive-susceptance formula
f Cwhere f
represents the frequency in megahertz, and C represents the capacitance in
microfarads. In this case, f = 3.1831 MHz and C = 0.001000 µF, so
= 0.0200 S The resistance
j0.0200The correct choice is A. |

10. The component values in this example coincide exactly with those in the situation
of Question 6, except this time, they're all connected in parallel rather than in series.
In an RLC circuit, the resonant-frequency formulas for the series and parallel
cases are identical. Therefore, the resonant frequency of this parallel combination is the
same as the value that we got when we solved Question 6, that is, 0.503 MHz or 503 kHz. The correct
choice is C. |