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15.1

10 ms

15.2

a. (20V)sin200πt,(0.20A)sin200πt(20V)sin200πt,(0.20A)sin200πt; b. (20V)sin200πt,(0.13A)sin(200πt+π/2)(20V)sin200πt,(0.13A)sin(200πt+π/2); c. (20V)sin200πt,(2.1A)sin(200πtπ/2)(20V)sin200πt,(2.1A)sin(200πtπ/2)

15.3

v R = ( V 0 R / Z ) sin ( ω t ϕ ) ; v C = ( V 0 X C / Z ) sin ( ω t ϕ + π / 2 ) = ( V 0 X C / Z ) cos ( ω t ϕ ) ; v L = ( V 0 X L / Z ) sin ( ω t ϕ π / 2 ) = ( V 0 X L / Z ) cos ( ω t ϕ ) v R = ( V 0 R / Z ) sin ( ω t ϕ ) ; v C = ( V 0 X C / Z ) sin ( ω t ϕ + π / 2 ) = ( V 0 X C / Z ) cos ( ω t ϕ ) ; v L = ( V 0 X L / Z ) sin ( ω t ϕ π / 2 ) = ( V 0 X L / Z ) cos ( ω t ϕ )

15.4

v ( t ) = ( 10.0 V ) sin 90 π t v ( t ) = ( 10.0 V ) sin 90 π t

15.5

2.00 V; 10.01 V; 8.01 V

15.6

a. 160 Hz; b. 40Ω40Ω; c. (0.25A)sin103t(0.25A)sin103t; d. 0.023 rad

15.7

a. halved; b. halved; c. same

15.8

v ( t ) = ( 0.14 V ) sin ( 4.0 × 10 2 t ) v ( t ) = ( 0.14 V ) sin ( 4.0 × 10 2 t )

15.9

a. 12:1; b. 0.042 A; c. 2.6×103Ω2.6×103Ω

Conceptual Questions

1.

Angular frequency is 2π2π 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. 530Ω530Ω; b. 53Ω53Ω; c. 5.3Ω5.3Ω

17.

a. 1.9Ω1.9Ω; b. 19Ω19Ω; c. 190Ω190Ω

19.

360 Hz

21.

i ( t ) = ( 3.2 A ) sin ( 120 π t ) i ( t ) = ( 3.2 A ) sin ( 120 π t )

23.

a. 38Ω38Ω; b. i(t)=(4.24A)sin(120πtπ/2)i(t)=(4.24A)sin(120πtπ/2)

25.

a. 770Ω770Ω; b. 0.16 A; c. I=(0.16A)cos(120πt0.33π)I=(0.16A)cos(120πt0.33π); d. vR=62cos(120πt0.33π)vR=62cos(120πt0.33π); vC=103cos(120πt0.83π)vC=103cos(120πt0.83π)

27.

a. 690Ω690Ω; b. 0.15 A; c. I=(0.15A)sin(1000πt0.753)I=(0.15A)sin(1000πt0.753); d. 1100Ω1100Ω, 0.092 A, I=(0.092A)sin(1000πt+1.09)I=(0.092A)sin(1000πt+1.09)

29.

a. 5.7Ω5.7Ω; b. 29°29°; c. I=(30.A)cos(120πt+0.51)I=(30.A)cos(120πt+0.51)

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. XL=52ΩXL=52Ω

37.

1.3 × 10 −6 F 1.3 × 10 −6 F

39.

a. 820 Hz; b. 7.8

41.

a. 50 Hz; b. 50 W; c. 6.32; d. 50 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. 160Ω160Ω

47.

a. 41 turns; b. 40.9 mA

Additional Problems

49.

a. i(t)=(1.26A)sin(200πt+π/2)i(t)=(1.26A)sin(200πt+π/2); b. i(t)=(7.96A)sin(200πtπ/2)i(t)=(7.96A)sin(200πtπ/2); c. i(t)=(2A)sin(200πt)i(t)=(2A)sin(200πt)

51.

a. 2.5×103Ω,3.6×10−3A2.5×103Ω,3.6×10−3A; b. 7.5Ω,1.2A7.5Ω,1.2A

53.

a. 19 A; b. inductor leads by 90°90°

55.

11.7 Ω 11.7 Ω

57.

14 W

59.

a. 5.9×104W5.9×104W; b. 1.64×1011W1.64×1011W

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. 20Ω20Ω; b. 0.5 A; c. 5.4°5.4°, lagging;
d. VR=(9.96V)cos(250πt+5.4°),VC=(12.7V)cos(250πt+5.4°90°),VL=(11.8V)cos(250πt+5.4°+90°),Vsource=(10.0V)cos(250πt);VR=(9.96V)cos(250πt+5.4°),VC=(12.7V)cos(250πt+5.4°90°),VL=(11.8V)cos(250πt+5.4°+90°),Vsource=(10.0V)cos(250πt); e. 0.995; f. 6.25 J

65.

a. 0.75Ω0.75Ω; b. 7.5Ω7.5Ω; c. 0.75Ω0.75Ω; d. 7.5Ω7.5Ω; e. 1.3Ω1.3Ω; f. 0.13Ω0.13Ω

67.

The units as written for inductive reactance Equation 15.8 are radsHradsH. Radians can be ignored in unit analysis. The Henry can be defined as H=V·sA=Ω·sH=V·sA=Ω·s. Combining these together results in a unit of ΩΩ for reactance.

69.

a. 156 V; b. 42 V; c. 154 V

71.

a. voutvin=11+1/ω2R2C2voutvin=11+1/ω2R2C2 and voutvin=ωLR2+ω2L2voutvin=ωLR2+ω2L2; b. voutvinvoutvin and vout0vout0

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