15.10: Alternating-Current Circuits (Answers)
Check Your Understanding
15.1. 10 ms
15.2. a. \(\displaystyle (20V)sin200πt,(0.20A)sin200πt\);
b. \(\displaystyle (20V)sin200πt,(0.13A)sin(200πt+π/2)\);
c. \(\displaystyle (20V)sin200πt,(2.1A)sin(200πt−π/2)\)
15.3. \(\displaystyle v_R=(V_0R/Z)sin(ωt−ϕ);v_C=(V_0X_C/Z)sin(ωt−ϕ+π/2)=−(V_0X_C/Z)cos(ωt−ϕ);v_L=(V_0X_L/Z)sin(ωt−ϕ-π/2)=(V_0X_L/Z)cos(ωt−ϕ)\)
15.4. \(\displaystyle v(t)=(10.0V)sin90πt\)
15.5. 2.00 V; 10.01 V; 8.01 V
15.6. a. 160 Hz;
b. \(\displaystyle 40Ω\);
c. \(\displaystyle (0.25A)sin10^3t\);
d. 0.023 rad
15.7. a. halved;
b. halved;
c. same
15.8. \(\displaystyle v(t)=(0.14V)sin(4.0×10^2t)\)
15.9. a. 12:1;
b. 0.042 A;
c. \(\displaystyle 2.6×10^3Ω\)
Conceptual Questions
1. Angular frequency is \(\displaystyle 2π\) times frequency.
3. yes for both
5. The instantaneous power is the power at a given instant. The average power is the power averaged over a cycle or number of cycles.
7. The instantaneous power can be negative, but the power output can’t be negative.
9. There is less thermal loss if the transmission lines operate at low currents and high voltages.
11. The adapter has a step-down transformer to have a lower voltage and possibly higher current at which the device can operate.
13. so each loop can experience the same changing magnetic flux
Problems
15. a. \(\displaystyle 530Ω\);
b. \(\displaystyle 53Ω\);
c. \(\displaystyle 5.3Ω\)
17. a. \(\displaystyle 1.9Ω\);
b. \(\displaystyle 19Ω\);
c. \(\displaystyle 190Ω\)
19. 360 Hz
21. \(\displaystyle i(t)=(3.2A)sin(120πt)\)
23. a. \(\displaystyle 38Ω\);
b. \(\displaystyle i(t)=(4.24A)sin(120πt−π/2)\)
25. a. \(\displaystyle 770Ω\);
b. 0.16 A;
c. \(\displaystyle I=(0.16A)cos(120πt)\);
d. \(\displaystyle v_R=120cos(120πt); v_C=120cos(120πt−π/2)\)
27. a. \(\displaystyle 690Ω\);
b. 0.15 A;
c. \(\displaystyle I=(0.15A)sin(1000πt−0.753)\);
d. \(\displaystyle 1100Ω\), 0.092 A, \(\displaystyle I=(0.092A)sin(1000πt+1.09)\)
29. a. \(\displaystyle 5.7Ω\);
b. \(\displaystyle 29°\);
c. \(\displaystyle I=(30.A)cos(120πt)\)
31. a. 0.89 A;
b. 5.6A;
c. 1.4 A
33. a. 5.3 W;
b. 2.1 W
35. a. inductor;
b. \(\displaystyle X_{L}=52Ω\)
37. \(\displaystyle 1.3×10^{−7}F\)
39. a. 820 Hz;
b. 7.8
41. a. 50 Hz;
b. 50 W;
c. 13;
d. 25 rad/s
43. The reactance of the capacitor is larger than the reactance of the inductor because the current leads the voltage. The power usage is 30 W.
45. a. 45:1;
b. 0.68 A, 0.015 A;
c. \(\displaystyle 160Ω\)
47. a. 41 turns;
b. 40.9 mA
Additional Problems
49. a. \(\displaystyle i(t)=(1.26A)sin(200πt+π/2)\);
b. \(\displaystyle i(t)=(12.6A)sin(200πt−π/2)\);
c. \(\displaystyle i(t)=(2A)sin(200πt)\)
51. a. \(\displaystyle 2.5×10^3Ω,3.6×10^{−3}A\);
b. \(\displaystyle 7.5Ω,1.2A\)
53. a. 19 A;
b. inductor leads by \(\displaystyle 90°\)
55. \(\displaystyle 11.7Ω\)
57. 36 W
59. a. \(\displaystyle 5.9×10^4W\);
b. \(\displaystyle 1.64×10^{11}W\)
Challenge Problems
61. a. 335 MV;
b. the result is way too high, well beyond the breakdown voltage of air over reasonable distances;
c. the input voltage is too high
63. a. \(\displaystyle 20Ω\);
b. 0.5 A;
c. \(\displaystyle 5.4°\), lagging;
d. \(\displaystyle V_R=(9.96V)cos(250πt+5.4°),V_C=(12.7V)cos(250πt+5.4°−90°),V_L=(11.8V)cos(250πt+5.4°+90°),V_{source}=(10.0V)cos(250πt);\);
e. 0.995;
f. 6.25 J
65. a. \(\displaystyle 0.75Ω\);
b. \(\displaystyle 7.5Ω\);
c. \(\displaystyle 0.75Ω\);
d. \(\displaystyle 7.5Ω\);
e. \(\displaystyle 1.3Ω\);
f. \(\displaystyle 0.13Ω\)
67. The units as written for inductive reactance Equation 15.8 are \(\displaystyle \frac{rad}{s}\)H. Radians can be ignored in unit analysis. The Henry can be defined as \(\displaystyle H=\frac{V⋅s}{A}=Ω⋅s\). Combining these together results in a unit of \(\displaystyle Ω\) for reactance.
69. a. 156 V;
b. 42 V;
c. 154 V
71. a. \(\displaystyle \frac{v_{out}}{v_{in}}=\frac{1}{\sqrt{1+1/ω^2R^2C^2}}\) and \(\displaystyle \frac{v_{out}}{v_{in}}=\frac{ωL}{\sqrt{R^2+ω^2L^2}};
b. \(\displaystyle v_{out}≈v_{in}\) and \(\displaystyle v_{out}≈0\)