Teach Yourself Electricity and Electronics, 5th edition |

Stan Gibilisco |

Explanations for Quiz Answers in Chapter 16 |

1. When we have resistance in a series RLC circuit but no reactance (either
because the inductive and capacitive reactances are equal and opposite, or because the
circuit contains no reactive components whatsoever), the complex-impedance vector points
directly toward the right along the resistance axis. The correct choice is (c). |

2. In a parallel RLC circuit where the conductance and susceptance values are
comparable, the admittance vectors point either upward toward the right (for capacitive
susceptance, which has positive imaginary values) or downward toward the right (for
inductive susceptance, which has negative imaginary values). In this example, we have a
negative imaginary susceptance whose absolute value equals the conductance, so the
complex-admittance vector points downward and toward the right. The correct answer is (d). |

3. In this situation, we have a resistance of 10 ohms, so the conductance equals 0.10
S (the reciprocal of 10 ohms). We have a reactance of -j10, so the susceptance
equals j0.10 (the reciprocal of -j10). Note that the sign changes when we
convert from reactance to susceptance. In the GB half-plane, we have the complex
admittance vector 0.10 + j0.10, which points upward and to the right. The correct
choice is (c). |

4. A vector pointing upward and toward the right in the GB half-plane indicates
conductance along with positive susceptance. Positive susceptance values indicate
capacitance, so the correct choice is (c). |

5. In the RX half-plane, we'll never see a vector pointing toward the left of
the vertical axis, either upward or downward. The correct choice here is (d), "None
of the above"! |

6. We have a coil with a reactance of j20 ohms. The susceptance in siemens
equals the reciprocal of the reactance in ohms. When we want to find the reciprocal of an
imaginary number, we must take the reciprocal of the real-number coefficient and then
change the sign. The reciprocal of 20 is 0.050, so when we "attach" 0.050 to the
j operator and reverse the sign, we get a susceptance of -j0.050 S. The
correct answer is (b). |

7. We have a capacitor with a susceptance of j0.040 S. The reactance in ohms
equals the reciprocal of the susceptance in siemens. As in the previous problem, we take
the reciprocal of the real-number coefficient and then change the sign. The reciprocal of
0.040 equals 25. When we "attach" 25 to the j operator and reverse the
sign, we get a reactance of -j25 ohms. The correct answer is (d). |

8. In series, reactances add arithmetically. When we take the sum of the two imaginary
values given here, we get
jX_{C}= j50 + (-j100)= j (50 - 100)= j (-50)= - j50The correct choice is (c). |

9. First, let's calculate the reactance jX of the inductor _{L}L
= 3.00 µH at f = 6.00 MHz. We have L in microhenrys and f in
megahertz, so we can plug the values into the formula directly to get
j6.2832fL= j (6.2832 x 6.00 x 3.00)= j113.0976Next, we calculate the reactance C = 100
pF at f = 6.00 MHz. We should convert picofarads to microfarads, getting C =
0.000100 µF, so that the numbers will work directly in our formula. Calculating, we
obtain
j [1 / (6.2832fC)]= - j [1 / (6.2832 x 6.00 x 0.000100)]= - j265.2576Finally, we calculate the net reactance
jX_{C}= j113.0976 + (-j265.2576)= - j152.16which rounds off to - |

10. Let's consider the resistor as part of the inductor, obtaining the two complex
numbers 10 + j72 and 0 - j83. Adding these complex numbers gives us a
resistive component of 10 + 0 = 10 and a reactive component of j72 - j83 = -j11.
Therefore, Z = 10 - j11. The correct answer is (b). |

11. First, let's calculate the inductive reactance X. Remember
that_{L}
fLand that megahertz and microhenrys go together in the formula. We can therefore plug
j (6.2832 x 5.650 x 44.00)= j1562.0035Next, let's calculate the capacitive reactance using the formula
fC)We should convert 500.00 pF to microfarads to go with megahertz in the formula, obtaining
j [1 / (6.2832
x 5.650 x 0.00050000)]= - j56.3379We know that
We should round this result off to |

12. Let's convert the frequency to megahertz, so that we can use it directly in the
formula for inductive reactance. We have f = 1340 kHz = 1.340 MHz and L =
8.88 µH, so the inductive reactance X is given by the formula _{L}
fLTherefore
j (6.2832 x 1.340 x 8.88)= j74.7651Next, let's calculate the capacitive reactance using the formula
fC)We note that
j [1 / (6.2832
x 1.340 x 0.000980)]= - j121.1960We've been told that
We can round this result off to |

13. We can add the values jB = -_{L}j0.32 and jB
= _{C}j0.20 arithmetically to get the net susceptance
jB_{C}= - j0.32 + j0.20= - j0.12The correct choice is (d). |

14. We're given the inductance L = 8.5 µH and the frequency f =
7.10 MHz, so we can calculate the susceptance jB of the coil as_{L}
j [1 / (6.2832fL)]= - j [1 / (6.2832 x 7.10 x 8.5)]= - j0.00263719Next, we can calculate the susceptance C = 100 pF, after converting it to 0.000100 µF, as
j (6.2832fC)= j (6.2832 x 7.10 x 0.000100)= j0.00446107Finally, we add the inductive and capacitive susceptances to obtain the net susceptance
jB_{C}= - j0.00263719 + j0.00446107= j0.00182388We should round this off to |

15. As before, we have L = 8.5 µH, but now f = 14.2 MHz. We calculate
the susceptance jB of the coil as_{L}
j [1 / (6.2832fL)]= - j [1 / (6.2832 x 14.2 x 8.5)]= - j0.00131860We calculate the susceptance C
= 100 pF, after converting it to 0.000100 µF, as
j (6.2832fC)= j (6.2832 x 14.2 x 0.000100)= j0.00892214Finally, we add the inductive and capacitive susceptances to obtain the net susceptance
jB_{C}= - j0.00131860 + j0.00892214= j0.00760354which we should round off to |

16. First, let's calculate the inductive susceptance. We know that R =
7.50 ohms, L = 22.0 µH, C = 100 pF = 0.000100 µF, and f = 5.33 MHz.
We can plug in the numbers to the formula for inductive susceptance to get
j [1 / (6.2832fL)]= - j [1 / (6.2832 x 5.33 x 22.0)]= - j0.00135728Next, we calculate the capacitive susceptance using the appropriate formula, obtaining
j (6.2832fC)= j (6.2832 x 5.33 x 0.000100)= j0.00334895Finally, we consider the conductance, which is 1/7.50 = 0.133 S, and the inductive
susceptance as existing together in a single component. Working that way, we see that one
of the parallel-connected admittances equals 0.133 -
The correct choice is (a). |

17. From the statement of the question, we can see that G = 0.333 S and B = 0.667 S. First, let's
calculate the sum of the squares of the conductance and susceptance to get
Now that we know this value, we can calculate and round off the resistive component of the complex impedance as
and the reactive component as
When we put these two components together into a single complex number, we obtain an
impedance of |

18. Let's follow the eight-step procedure described in the chapter text for solving
problems of this type, going out to seven decimal places in our intermediate calculations, and then rounding off to two significant figures at the end. It goes as follows:- Calculate
*G*= 1/*R*= 1/25 = 0.0400000. - Calculate
*B*= -1 / (6.2832_{L}*fL*) = -1 / (6.2832 x 2.0 x 7.7) = -0.0103347. - Calculate
*B*= 6.2832_{C}*fC*= 6.2832 x 2.0 x 0.0020 = 0.0251328. - Calculate
*B*=*B*+_{L}*B*= -0.0103347 + 0.0251328 = 0.0147981._{C} - Define
*G*^{ 2}+*B*^{ 2}= 0.040^{2}+ 0.0147981^{2}= 0.0018190. - Calculate
*R*=*G*/ 0.0018190 = 0.0400000 / 0.0018190 = 22. - Calculate
*X*= -*B*/ 0.0018190 = -0.0147981 / 0.0018190 = -8.1. - The complex impedance is
*R*+*jX*= 22 -*j*8.1.
The correct choice is (d). |

19. Let's calculate the complex impedance using the formula in the text for series
circuits, taking the square root of 800 to seven significant figures and then rounding the final answer off to two significant figures. We have
Now we can calculate the current as
The correct choice is (b). |

20. Let's calculate the absolute-value impedance, remembering the formula for parallel
circuits, and then calculate the current on the basis of that result. As before, we'll carry our inexact intermediate results to a few extra significant figures, and round off our answer at the end of the calculation process. We get
The total current is therefore
The correct answer is (a). |