# Bài giải phần giải mạch P1

Chia sẻ: Tran Long | Ngày: | Loại File: PDF | Số trang:11

0
61
lượt xem
16

## Bài giải phần giải mạch P1

Mô tả tài liệu

Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C (b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C (c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C (d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C Chapter 1, Solution 2 (a) (b) (c) (d) (e) i = dq/dt = 3 mA i = dq/dt = (16t + 4) A i = dq/dt = (-3e-t + 10e-2t) nA i=dq/dt = 1200π cos 120π t pA i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ...

Chủ đề:

Bình luận(0)

Lưu

## Nội dung Text: Bài giải phần giải mạch P1

1. Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C (b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C (c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C (d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C Chapter 1, Solution 2 (a) i = dq/dt = 3 mA (b) i = dq/dt = (16t + 4) A (c) i = dq/dt = (-3e-t + 10e-2t) nA (d) i=dq/dt = 1200π cos 120π t pA (e) i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ A Chapter 1, Solution 3 (a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C (b) q(t) = ∫ (2t + s) dt + q(v) = (t 2 + 5t) mC (c) q(t) = ∫ 20 cos (10t + π / 6 ) + q(0) = (2sin(10t + π / 6) + 1) µ C 10e -30t q(t) = ∫ 10e -30t sin 40t + q(0) = ( −30 sin 40t - 40 cos t) (d) 900 + 1600 = − e - 30t (0.16cos40 t + 0.12 sin 40t) C Chapter 1, Solution 4 10 −5 q = ∫ idt = ∫ 5sin 6 π t dt = cos 6π t 6π 0 5 = (1 − cos 0.06π ) = 4.698 mC 6π
2. Chapter 1, Solution 5 2 1 q = ∫ idt = ∫ e dt mC = - e -2t -2t 2 0 1 = (1 − e 4 ) mC = 490 µC 2 Chapter 1, Solution 6 dq 80 (a) At t = 1ms, i = = = 40 mA dt 2 dq (b) At t = 6ms, i = = 0 mA dt dq 80 (c) At t = 10ms, i = = = - 20 mA dt 4 Chapter 1, Solution 7 25A, 0
3. Chapter 1, Solution 8 10 × 1 q = ∫ idt = + 10 × 1 = 15 µC 2 Chapter 1, Solution 9 1 (a) q = ∫ idt = ∫ 10 dt = 10 C 0 3  5 ×1 q = ∫ idt = 10 × 1 + 10 −  + 5 ×1 (b) 0  2  = 15 + 10 − 25 = 22.5 C 5 (c) q = ∫ idt = 10 + 10 + 10 = 30 C 0 Chapter 1, Solution 10 q = ixt = 8 x10 3 x15 x10 − 6 = 120 µ C Chapter 1, Solution 11 q = it = 85 x10-3 x 12 x 60 x 60 = 3,672 C E = pt = ivt = qv = 3672 x1.2 = 4406.4 J Chapter 1, Solution 12 For 0 < t < 6s, assuming q(0) = 0, t t ∫ ∫ q (t ) = idt + q (0 ) = 3tdt + 0 = 1.5t 2 0 0 At t=6, q(6) = 1.5(6)2 = 54 For 6 < t < 10s,
4. t t ∫ ∫ q (t ) = idt + q (6 ) = 18 dt + 54 = 18 t − 54 6 6 At t=10, q(10) = 180 – 54 = 126 For 10
5. Chapter 1, Solution 13 2 2 w = ∫ vidt = ∫ 1200 cos 2 4 t dt 0 0 2 = 1200 ∫ ( 2 cos 8t - 1)dt (since, cos 2 x = 2 cos 2x - 1) 0 2 2  1  = 1200 sin 8t − t  = 1200 sin 16 − 2  8 0 4  = - 2.486 kJ Chapter 1, Solution 14 q = ∫ idt = ∫ 10(1 - e -0.5t )dt = 10(t + 2e -0.5t ) 1 1 (a) 0 0 = 10(1 + 2e -0.5 − 2 ) = 2.131 C (b) p(t) = v(t)i(t) p(1) = 5cos2 ⋅ 10(1- e-0.5) = (-2.081)(3.935) = -8.188 W Chapter 1, Solution 15 2 2 − 3 2t q = ∫ idt = ∫ 3e dt =-2t e (a) 0 2 0 = −1.5(e − 1) = 1.297 C -2 5di v= = −6e 2t ( 5) = −30e -2t (b) dt p = vi = − 90 e − 4 t W 3 3 − 90 -4t (c) w = ∫ pdt = -90∫ e -4t dt = e = − 22.5 J 0 −4 0
6. Chapter 1, Solution 16   10t V 0< t
7. Chapter 1, Solution 20 Since Σ p = 0 -30×6 + 6×12 + 3V0 + 28 + 28×2 - 3×10 = 0 72 + 84 + 3V0 = 210 or 3V0 = 54 V0 = 18 V Chapter 1, Solution 21 ∆q  photon  1  electron  i= = 4 × 1011  ⋅  photon  ⋅ 1. 6 × 10 ( C / electron)  19 ∆t  sec  8   4 = × 1011 × 1. 6 × 10 −19 C/s = 0.8 × 10 -8 C/s = 8 nA 8 Chapter 1, Solution 22 It should be noted that these are only typical answers. (a) Light bulb 60 W, 100 W (b) Radio set 4W (c) TV set 110 W (d) Refrigerator 700 W (e) PC 120 W (f) PC printer 18 W (g) Microwave oven 1000 W (h) Blender 350 W Chapter 1, Solution 23 p 1500 (a) i = = = 12.5 W v 120 45 (b) w = pt = 1. 5 × 103 × 45 × 60 ⋅ J = 1.5 × kWh = 1.125 kWh 60 (c) Cost = 1.125 × 10 = 11.25 cents
8. Chapter 1, Solution 24 p = vi = 110 x 8 = 880 W Chapter 1, Solution 25 4 Cost = 1.2 kW × hr × 30 × 9 cents/kWh = 21.6 cents 6 Chapter 1, Solution 26 0. 8A ⋅ h (a) i = = 80 mA 10h (b) p = vi = 6 × 0.08 = 0.48 W (c) w = pt = 0.48 × 10 Wh = 0.0048 kWh Chapter 1, Solution 27 (a) Let T = 4h = 4 × 36005 T q = ∫ idt = ∫ 3dt = 3T = 3 × 4 × 3600 = 43.2 kC 0 T T  0 . 5t  ( b) W = ∫ pdt = ∫ vidt = ∫ ( 3) 10 + dt 0 0  3600  4×3600  0. 25t 2  = 310t +   = 3[40 × 3600 + 0. 25 × 16 × 3600]  3600  0 = 475.2 kJ ( c) W = 475.2 kWs, (J = Ws) 475.2 Cost = kWh × 9 cent = 1.188 cents 3600
9. Chapter 1, Solution 28 P 30 (a) i = = = 0.25 A V 120 ( b) W = pt = 30 × 365 × 24 Wh = 262.8 kWh Cost = $0.12 × 262.8 =$31.54 Chapter 1, Solution 29 (20 + 40 + 15 + 45)  30  w = pt = 1. 2kW hr + 1.8 kW  hr 60  60  = 2.4 + 0.9 = 3.3 kWh Cost = 12 cents × 3.3 = 39.6 cents Chapter 1, Solution 30 Energy = (52.75 – 5.23)/0.11 = 432 kWh Chapter 1, Solution 31 Total energy consumed = 365(4 +8) W Cost = $0.12 x 365 x 12 =$526.60 Chapter 1, Solution 32 (20 + 40 + 15 + 45)  30  w = pt = 1. 2kW hr + 1.8 kW  hr 60  60  = 2.4 + 0.9 = 3.3 kWh Cost = 12 cents × 3.3 = 39.6 cents
10. Chapter 1, Solution 33 dq i= → q = ∫ idt = 2000 × 3 × 10 3 = 6 C dt Chapter 1, Solution 34 (b) Energy = ∑ pt = 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2 = 10,000 kWh (c) Average power = 10,000/24 = 416.67 W Chapter 1, Solution 35 ( a) W = ∫ p( t ) dt = 400 × 6 + 1000 × 2 + 200 × 12 × 1200 × 2 + 400 × 2 = 7200 + 2800 = 10.4 kWh 10.4 kW ( b) = 433.3 W/h 24 h Chapter 1, Solution 36 160A ⋅ h (a) i= =4A 40 160Ah 160, 000h ( b) t = = = 6,667 days 0.001A 24h / day Chapter 1, Solution 37 q = 5 × 10 20 (− 1. 602 × 10 −19 ) = −80. 1 C W = qv = −80. 1 × 12 = − 901.2 J
11. Chapter 1, Solution 38 P = 10 hp = 7460 W W = pt = 7460 × 30 × 60 J = 13.43 × 106 J Chapter 1, Solution 39 p 2 × 10 3 p = vi → i = = = 16.667 A v 120