Tài liệu Bài giải mạch P1 ppt

Chapter 1, Solution 1

(a) q = 6.482x10
17

x [-1.602x10
-19

C] = -0.10384 C

(b) q = 1. 24x10
18

x [-1.602x10
-19

C] = -0.19865 C

(c) q = 2.46x10
19

x [-1.602x10
-19

C] = -3.941 C

(d) q = 1.628x10
20

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 =
1200 120
π
π
cos
t
pA
(e) i =dq/dt =

−+
−
et
t4
80 50 1000 50
(cos sin )
A

µ
t


Chapter 1, Solution 3

(a)
C 1)(3t +=+=
∫
q(0)i(t)dt q(t)
(b) mC 5t)(t
2
+=++=
∫
q(v)dt s)(2tq(t)
(c)
()
q(t) 20 cos 10t / 6 q(0) (2sin(10 / 6) 1) Ct
π
πµ
=++=++
∫


(d)

C 40t) sin 0.12t(0.16cos40e
30t-
+−=
−

+
=+=
∫
t)cos 40-t40sin30(
1600900
e10
q(0)t40sin10eq(t)
-30t
30t-




Chapter 1, Solution 4

()
mC 4.698=−=
−
===
∫∫
π
π
π
06.0cos1
6
5
tπ6cos
6
5
dt t π 6 5sinidtq

10
0




Chapter 1, Solution 5

µCmC )e1(
2
1
e
2
1
-mC dteidtq
4
2
0
2t-2t-
490=−=
===
∫∫



Chapter 1, Solution 6

(a) At t = 1ms,
mA 40===
2

80
dt
dq
i


(b) At t = 6ms,
mA 0==
dt
dq
i


(c) At t = 10ms,
mA 20-===
4
80
dt
dq
i




Chapter 1, Solution 7







<<
<<
<<
==
8t6 25A,
6t2 25A,-
2t0 A,25
dt
dq
i


which is sketched below:







Chapter 1, Solution 8

C 15 µ110
2
110
idtq =×+
×
==
∫

Chapter 1, Solution 9

(a)
C 10===
∫∫
1
0
dt 10idtq

(b)
C 22.5=−+=
×+






×
−+×==
∫
251015
15
2
15
10110idtq
3

0


(c)
C 30=++==
∫
101010idt
5
0
q



Chapter 1, Solution 10


qixt x xx
== =
−
8 10 15 10 120
36
C
µ




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,

qt idt q tdt t
tt
() ( ) .
=+= +=
∫∫
03015
0
2
0

At t=6, q(6) = 1.5(6)
2
= 54
For 6 < t < 10s,

qt idt q dt t
tt
() ( )

=+= +=−
∫∫
61854185
66
4
66

At t=10, q(10) = 180 – 54 = 126
For 10<t<15s,

qt idt q dt t
tt
() ( ) ( )
=+ =−+=−+
∫∫
10 12 126 12 246
10 10


At t=15, q(15) = -12x15 + 246 = 66
For 15<t<20s,

qt dt q
t
() ( )
=+=
∫
015
15

Thus,

qt
t
t
t
()
.
,
=
−
−+







15
18 54
12 246
66
2
C, 0 < t < 6s
C, 6 < t < 10s
C, 10 < t < 15s
C 15 < t < 20s

The plot of the charge is shown below.
0 5 10 15 20
0
20
40
60
80
100
120
140
t
q(t)




Chapter 1, Solution 13

kJ 2.486-=






−=







−=
==
==
∫
∫∫
216sin
4
1
12008sin
8
2
1200
1)-2x cos 2xcos (since,1)dt -t8cos2(1200
dtt 4cos1200vidtw
2
0
2
0
2
2
0
2
2
0
tt

Chapter 1, Solution 14

(a)
(
)
(
)
()
C 2.131=−+=
+===
∫∫
2e2110
2et10dte-110idtq
0.5-
1
0
0.5t-
1
0
0.5t-


(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

(a)
()
C 1.297=−−=
−
===
∫∫
1e5.1
e
2
3
dt3eidtq
2-
2
0
2t
2
0
2t-


(b)
We90
)(
t4−
−==
−=−==

vip
e305e6
dt
di5
v
2t-2t


(c) J 22.5−=
−
−
===
∫∫
3
0
4t-
3
0
4t-
e
4
90
dt e-90pdtw






Chapter 1, Solution 16

mJ 916.7
)(
(
(((
,
=








++






+−−++=
−+









++=
−+−+++==










<<
<<
<<
=
<<
<<
=
∫
∫∫∫∫∫
4
3
2
2
4
3
2

3
2
2
2
1
1
0
3
4
3
3
2
2
1
1
0
3
t
4t-t1625028
2
9
122503
2
250
3
250
dtt)4250
2
t
-t4250

2
250
t
3
250
25t)mJ-t)(1001040t)dt2510010t)dt2510(25t)dt10v(t)i(t)dtw
4t3 V10t -40
3t1 V 10
1t0 V10t
v(t)
4t2 mA25t -100
2t0 mA25t
i(t)




Chapter 1, Solution 17

Σ
p = 0 → -205 + 60 + 45 + 30 + p
3
= 0

p
3
= 205 – 135 = 70 W

Thus element 3 receives
70 W.

Chapter 1, Solution 18

p
1
= 30(-10) = -300 W
p
2
= 10(10) = 100 W
p
3
= 20(14) = 280 W
p
4
= 8(-4) = -32 W
p
5
= 12(-4) = -48 W



Chapter 1, Solution 19


pIxxx
ss
=→−−−+=→=
∑
0 4 2 6 13 2 5 10 0 3

A
I




Chapter 1, Solution 20

Since Σ
p = 0

-30×6 + 6×12 + 3V
0
+ 28 + 28×2 - 3×10 = 0

72 + 84 + 3V
0
= 210 or 3V
0
= 54

V
0
= 18 V



Chapter 1, Solution 21

nA8.

(.
C/s100.8 C/s106110
8
4
electron) / C1061
photon
electron
8
1
sec
photon
104


8-1911
1911
=×=×××=
×⋅








⋅







×=
∆
∆
=
−
t
q
i




Chapter 1, Solution 22

It should be noted that these are only typical answers.

(a) Light bulb 60 W, 100 W
(b)
Radio set 4 W
(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

(a) W12.5===
120
1500
v
p
i
(b) kWh 1.125. =×=⋅×××== kWh
60
45
1.5J60451051
3
ptw
(c) Cost = 1.125 × 10 = 11.25 cents


Chapter 1, Solution 24


p = vi = 110 x 8 = 880 W

Chapter 1, Solution 25

cents 21.6 cents/kWh 930hr
6
4
kW 1.2 Cost =×××=




Chapter 1, Solution 26

(a)
mA 80
.
=
⋅
=
10h
hA80
i

(b)
p = vi = 6 × 0.08 = 0.48 W
(c)
w = pt = 0.48 × 10 Wh = 0.0048 kWh



Chapter 1, Solution 27

∫∫
=××====
×==
kC 43.2 36004 3 T33dt idt q
36005 4 4h T Let (a)
T
0



[]
kJ 475.2
.
.
.
)((
=
××+×=








+=







+===
×
∫∫∫
3600162503600403
3600
250
103
dt
3600
t50
103vidtpdtW b)
36004
0
2
0
T
0
t
t
T


cents 1.188
(
=×=
==

cent 9kWh
3600
475.2
Cost
Ws)(J kWs, 475.2
W
c)




Chapter 1, Solution 28

A 0.25===
120
30
(a)
V
P
i


$31.54
(
=×=
=
××==
262.8 $0.12Cost
kWh 262.8Wh 2436530ptW b)

Chapter 1, Solution 29

cents 39.6
.
=×=
=+=






+
+
++
==
3.3 cents 12Cost
kWh 3.30.92.4
hr
60
30
kW 1.8hr
60
45)1540(20
kW21
ptw

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

cents 39.6
.
=×=
=+=






+

+
++
==
3.3 cents 12Cost
kWh 3.30.92.4
hr
60
30
kW 1.8hr
60
45)1540(20
kW21
ptw




Chapter 1, Solution 33

C 6=××==→=
∫
3
1032000idtq
dt
dq
i



Chapter 1, Solution 34

(b)
Energy =
∑
= 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2
pt
= 10,000 kWh

(c)
Average power = 10,000/24 = 416.67 W



Chapter 1, Solution 35

kWh 10.4
)((
=+=
×+×××+×+×==
∫
28007200
24002120012200210006400W a) dttp


W/h 433.3( =
h 24
kW 10.4
b)

Chapter 1, Solution 36

days 6,667
,
(
A 4
===
=
⋅
=
day / 24h
h000160
0.001A
160Ah
tb)
40
hA160
i (a)




Chapter 1, Solution 37

(
)
J 901.2.

−=×−==

−=×−×=
−
12180W
C 180106021105q
1920
qv




Chapter 1, Solution 38

P = 10 hp = 7460 W

W = pt = 7460 × 30 × 60 J =
13.43 × 10
6
J



Chapter 1, Solution 39

A 16.667=
×
==→=
120
102
v
p

ivip
3