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 = jXL + jXC
= j50 + (-j100)
= j (50 - 100)
= j (-50)
= -j50

The correct choice is (c).

9. First, let's calculate the reactance jXL of the inductor L = 3.00 H 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

jXL = j6.2832fL
= j (6.2832 x 6.00 x 3.00)
= j113.0976

Next, we calculate the reactance jXC of the capacitor C = 100 pF at f = 6.00 MHz. We should convert picofarads to microfarads, getting C = 0.000100 F, so that the numbers will work directly in our formula. Calculating, we obtain

jXC = -j [1 / (6.2832fC)]
= -j [1 / (6.2832 x 6.00 x 0.000100)]
= -j265.2576

Finally, we calculate the net reactance jX as

jX = jXL + jXC
= j113.0976 + (-j265.2576)
= -j152.16

which rounds off to -j152. The correct answer is (a). Note that we extend our intermediate calculations to a few extra digits, so as to avoid cumulative rounding errors. We should always round our answer off when we've finished our calculations. We'll repeat this precaution several more times in this quiz!

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 XL. Remember that

XL = 6.2832fL

and that megahertz and microhenrys go together in the formula. We can therefore plug L = 44.00 and f = 5.650 directly into the formula to get

jXL = j (6.2832 x 5.650 x 44.00)
= j1562.0035

Next, let's calculate the capacitive reactance using the formula

XC = -1 / (6.2832fC)

We should convert 500.00 pF to microfarads to go with megahertz in the formula, obtaining C = 0.00050000 F. Then we have

jXC = -j [1 / (6.2832 x 5.650 x 0.00050000)]
= -j56.3379

We know that R = 220.0 ohms. Let's put the resistance and the inductive reactance together, so one of the impedances becomes 220.0 + j1562.0035. The other impedance equals 0.000 - j56.3379. When we add these two complex numbers, we get the complex impedance Z as

Z = (220.0 + j1562.0035) + (0.000 - j56.3379)
= 220.0 + j1562.0035 - j56.3379
= 220.0 + j1505.6656

We should round this result off to Z = 220.0 + j1506, because we're allowed four significant figures. The correct answer is therefore (a).

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 H, so the inductive reactance XL is given by the formula

XL = 6.2832fL

Therefore

jXL = j (6.2832 x 1.340 x 8.88)
= j74.7651

Next, let's calculate the capacitive reactance using the formula

XC = -1 / (6.2832fC)

We note that C = 980 pF = 0.000980 F, so we calculate

jXC = -j [1 / (6.2832 x 1.340 x 0.000980)]
= -j121.1960

We've been told that R = 75.3 ohms. Putting the resistance and the inductive reactance together, we get a complex value of 75.3 + j74.7651. The other impedance equals 0.00 - j121.1960. When we add these two complex numbers, we obtain

Z = (75.3 + j74.7651) + (0.00 - j121.1960)
= 75.3 + j74.7651 - j121.1960
= 75.3 - j46.4309

We can round this result off to Z = 75.3 - j46.4. The correct answer is (c).

13. We can add the values jBL = -j0.32 and jBC = j0.20 arithmetically to get the net susceptance

jB = jBL + jBC
= -j0.32 + j0.20
= -j0.12

The correct choice is (d).

14. We're given the inductance L = 8.5 H and the frequency f = 7.10 MHz, so we can calculate the susceptance jBL of the coil as

jBL = -j [1 / (6.2832fL)]
= -j [1 / (6.2832 x 7.10 x 8.5)]
= -j0.00263719

Next, we can calculate the susceptance jBC of the capacitor whose value is C = 100 pF, after converting it to 0.000100 F, as

jBC = j (6.2832fC)
= j (6.2832 x 7.10 x 0.000100)
= j0.00446107

Finally, we add the inductive and capacitive susceptances to obtain the net susceptance

jB = jBL + jBC
= -j0.00263719 + j0.00446107
= j0.00182388

We should round this off to j0.0018, accurate to two significant figures. The correct choice is (b).

15. As before, we have L = 8.5 H, but now f = 14.2 MHz. We calculate the susceptance jBL of the coil as

jBL = -j [1 / (6.2832fL)]
= -j [1 / (6.2832 x 14.2 x 8.5)]
= -j0.00131860

We calculate the susceptance jBC of the capacitor whose value is C = 100 pF, after converting it to 0.000100 F, as

jBC = j (6.2832fC)
= j (6.2832 x 14.2 x 0.000100)
= j0.00892214

Finally, we add the inductive and capacitive susceptances to obtain the net susceptance

jB = jBL + jBC
= -j0.00131860 + j0.00892214
= j0.00760354

which we should round off to j0.0076. The correct choice is (d), "None of the above."

16. First, let's calculate the inductive susceptance. We know that R = 7.50 ohms, L = 22.0 H, C = 100 pF = 0.000100 F, and f = 5.33 MHz. We can plug in the numbers to the formula for inductive susceptance to get

jBL = -j [1 / (6.2832fL)]
= -j [1 / (6.2832 x 5.33 x 22.0)]
= -j0.00135728

Next, we calculate the capacitive susceptance using the appropriate formula, obtaining

jBC = j (6.2832fC)
= j (6.2832 x 5.33 x 0.000100)
= j0.00334895

Finally, 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 - j0.00135728, while the other admittance equals 0 + j0.00334895. Adding these two complex numbers and rounding the imaginary coefficient to three significant figures at the end of the process, we obtain the net admittance as

Y = 0.133 - j0.00135728 + j0.00334895
= 0.133 + j0.00199

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

G 2 + B 2 = 0.3332 + 0.6672
= 0.110889 + 0.444889
= 0.555778

Now that we know this value, we can calculate and round off the resistive component of the complex impedance as

R = G / 0.555778
= 0.333 / 0.555778
= 0.599

and the reactive component as

X = -B / 0.555778
= -0.667 / 0.555778
= -1.20

When we put these two components together into a single complex number, we obtain an impedance of Z = 0.599 - j1.20. The correct choice is (c).

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 BL = -1 / (6.2832fL) = -1 / (6.2832 x 2.0 x 7.7) = -0.0103347.
  • Calculate BC = 6.2832fC = 6.2832 x 2.0 x 0.0020 = 0.0251328.
  • Calculate B = BL + BC = -0.0103347 + 0.0251328 = 0.0147981.
  • Define G 2 + B 2 = 0.0402 + 0.01479812 = 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 - j8.1.

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

Z = (R 2 + X 2)1/2
= [202 + (-20)2]1/2
= (400 + 400)1/2
= 8001/2
= 28.28427 ohms

Now we can calculate the current as

I = E / Z
= 42 / 28.28427
= 1.5 A RMS

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

Z = [R 2 X 2 / (R 2 + X 2)]1/2
= [502 x 402 / (502 + 402)]1/2
= [2500 x 1600 / (2500 + 1600)]1/2
= (4,000,000 / 4100)1/2
= 975.60981/2
= 31.23475 ohms

The total current is therefore

I = E / Z
= 155 / 31.23475
= 5.0 A RMS

The correct answer is (a).