Baixe solution of fundamentals of electric circuits 2nd ed sadiku e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity!
(a) q = 6.482x
17 x [-1.602x
(b) q = 1. 24x
18 x [-1.602x
(c) q = 2.46x
19 x [-1.602x
(d) q = 1.628x
20 x [-1.602x
Chapter 1, Solution 2
(a) i = dq/dt = 3 mA
(b) i = dq/dt = (16t + 4) A
(c) i = dq/dt = (-3e
-t
-2t ) nA
(d) i=dq/dt = 1200 πcos 120 π t pA
(e) i =dq/dt = − +
− e t
4 t ( 80 cos 50 1000 sin 50 t ) μ A
Chapter 1, Solution 3
(a) q(t)= ∫ i(t)dt+q(0)=(3t +1) C
(b) (t 5t) mC
2
q(t)= ∫ (2t+s)dt+q(v)= +
(c) q(t) = ∫20 cos (10t + π/ 6) + q(0) = (2sin(10 t + π/ 6) +1) μC
(d)
e (0.16cos40t 0.12sin40t) C
= ∫ + = ( 30 sin 40 t- 40 cost)
10 e q(t) 10e sin 40 t q(0)
-30t
Chapter 1, Solution 4
=^ (^ − )^ =4.698mC
1 cos 0. 06 6
cos 6 πt 6
q idt 5sin 6 πtdt
10
0
( 1 e )mC μC 2
e 2
q idt e dtmC -
4
2
0
Chapter 1, Solution 6
(a) At t = 1ms, = = = 40 mA 2
dt
dq i
(b) At t = 6ms, = = 0 mA dt
dq i
(c) At t = 10ms, = = =- 20 mA 4
dt
dq i
Chapter 1, Solution 7
25A, 6 t 8
25 A, 0 t 2
dt
dq i
which is sketched below:
q t idt q dt t
t t
6 6
4
66
At t=10, q(10) = 180 – 54 = 126
For 10<t<15s,
q t idt q dt t
t t ( ) = + ( ) = ( − ) + = − +
10 12 126 12 246
10 10
At t=15, q(15) = -12x15 + 246 = 66
For 15<t<20s,
q t dt q
t ( ) = + ( )=
0 15
15
Thus,
q t
t
t
t
( )
.
,
=
−
− +
1 5
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)
=-2.486 kJ
sin 16 2 4
sin 8 1200 8
1200 ( 2 cos 8 t-1)dt(since,cosx 2 cos2x-1)
w vidt 1200 cos 4 tdt
2
0
2
0
2
2
0
(^22)
0
t t
Chapter 1, Solution 14
(a)
( ) ( )
= ( + − ) =2.131C
101 2 e 2
q idt 101 - e dt 10 t 2e
1
0
(^1) - 0.5t
0
(b) p(t) = v(t)i(t)
p(1) = 5cos2 ⋅ 10(1- e
-0. ) = (-2.081)(3.935)
= -8.188 W
Chapter 1, Solution 15
(a)
=− ( − ) =1.297C
- 5 e 1
e 2
q idt 3e dt
2
0
(^2) 2t
0
(b)
90 e W
− 4 t = =−
p vi
6 e 5 30 e dt
5 di v
2t -2t
(c) =−22.5J −
3
0
(^3) - 4t
0
w pdt -90 e dt
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
. 8 nA
10 16 10 C/s 0.8 10 C/s 8
16 10 C/electron) photon
electron
8
sec
photon 4 10
11 19 - 8
11 19
= × × × = × =
⋅ ×
= ×
−
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) = = =12.5W 120
v
p i
(b) = =. × × × ⋅ = × kWh=1.125kWh 60
1 5 10 45 60 J 1.
3 w pt
(c) Cost = 1.125 × 10 = 11.25 cents
p = vi = 110 x 8 = 880 W
Chapter 1, Solution 25
hr 30 9 cents/kWh 21.6 cents 6
Cost =1.2kW× × × =
Chapter 1, Solution 26
(a) 80 mA
10h
08 A h i
(b) p = vi = 6 × 0.08 = 0.48 W
(c) w = pt = 0.48 × 10 Wh = 0.0048 kWh
Chapter 1, Solution 27
= ∫ =∫ = = × × =
= = ×
q idt 3dt 3 T 3 4 3600 43.2 kC
(a) LetT 4h 4 36005
T
0
[ ]
475.2 kJ
= × + × ×
×
dt 3600
05 t b) W pdt vidt 3 10
43600
0
2
0
T
0
t t
T
1.188 cents
= × =
kWh 9 cent 3600
Cost
c) W 475.2kWs, (J Ws)
= → =∫ = × × = 6 C
3 q idt 2000 3 10 dt
dq i
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
10.4 kWh
= = × + × + × × × + ×
a) W pt dt 400 6 1000 2 200 12 1200 2 400 2
( =433.3 W/h 24 h
10.4kW b)
Chapter 1, Solution 36
6,667 days
4 A
24h/ day
160000 h
0.001A
160Ah b) t
160 A h (a) i
Chapter 1, Solution 37
( )
. 901.2J
= =− × = −
= × − × =−
−
W 801 12
q 5 10 1602 10 801 C
20 19
qv
P = 10 hp = 7460 W
W = pt = 7460 × 30 × 60 J = 13.43 × 10
6 J
Chapter 1, Solution 39
=16.667 A
×
v
p p vi i
3
c d
b
a
9 A
i (^3) i (^2)
12 A
12 A
i (^1)
8 A
At node a, 8 = 12 + i 1 i 1 = - 4A
At node c, 9 = 8 + i 2 i 2 = 1A
At node d, 9 = 12 + i 3 i 3 = -3A
Chapter 2, Solution 9
Applying KCL,
i 1 + 1 = 10 + 2 i 1 = 11A 1 + i 2 = 2 + 3 i 2 = 4A
i 2 = i 3 + 3 i 3 = 1A
Chapter 2, Solution 10
At node 1, 4 + 3 = i 1 i 1 = 7A
At node 3, 3 + i 2 = -2 i 2 = -5A
-2A
3A
4A
i (^2) i (^1)
Applying KVL to each loop gives
-8 + v 1 + 12 = 0 v 1 = 4v
-12 - v 2 + 6 = 0 v 2 = -6v
10 - 6 - v 3 = 0 v 3 = 4v
-v 4 + 8 - 10 = 0 v 4 = -2v
Chapter 2, Solution 12
For loop 1, -20 -25 +10 + v 1 = 0 v 1 = 35v
For loop 2, -10 +15 -v 2 = 0 v 2 = 5v
For loop 3, -v 1 +v 2 +v 3 = 0 v 3 = 30v
+ 15v -
loop 3
loop 2
loop 1
20v
- - 25v + + 10v^ - + v^2 -
v
-
v
-
Chapter 2, Solution 13
2A
I 2 7A I 4
4A
I 1
3A I 3
v 3 10V
For loop 1,
8 − 12 + v 2 = 0 → v 2 = 4 V
V
V
For loop 2,
− v 3 − 8 − 10 = 0 → v 3 = − 18
For loop 3,
− v 1 + 12 + v 3 = 0 → v 1 = − 6
Thus,
v 1 = − 6 V , v 2 = 4 V , v 3 = − 18 V
Chapter 2, Solution 16
+ v 1 -
6V
+ loop 1
loop 2
12V
10V
v 1
-
+ v 2 -
Applying KVL around loop 1,
–6 + v 1 + v 1 – 10 – 12 = 0 v 1 = 14V
Applying KVL around loop 2,
12 + 10 – v 2 = 0 v 2 = 22V
+ v 1 -
+ 10V
12V
24V
loop 2
v
-
v
loop 1
+
It is evident that v 3 = 10V
Applying KVL to loop 2,
v 2 + v 3 + 12 = 0 v 2 = -22V
Applying KVL to loop 1,
-24 + v 1 - v 2 = 0 v 1 = 2V
Thus,
v 1 = 2V , v 2 = -22V , v 3 = 10V
Chapter 2, Solution 18
Applying KVL,
-30 -10 +8 + I(3+5) = 0
8I = 32 I = 4A
-Vab + 5I + 8 = 0 Vab = 28V
+ v 0 -
6 Ω 10A 2v
At the node, KCL requires that
0
0 10 2 v 4
v
The current through the controlled source is
i = 2V 0 = -8.888A
and the voltage across it is
v = (6 + 4) i 0 = 10 11. 111 4
v (^0) =−
Hence,
p 2 vi = (-8.888)(-11.111) = 98.75 W
Chapter 2, Solution 23
The circuit is reduced to that shown below.
ix 1 Ω
6A 2 Ω 3 Ω
Applying current division,
i (^) x = A A v (^) x i
= =
2
2 1 3
( 6 ) 2 , (^1) x = 2 V
The current through the 1.2- Ω resistor is 0.5ix = 1A. The voltage across the 12-
resistor is 1 x 4.8 = 4.8 V. Hence the power is
Ω
p
v
R
= = = W
2 2 4 8
12
1 92
. .
Chapter 2, Solution 24
(a) I 0 = R 1 R 2
Vs
V 0 =− αI 0 ( R 3 R 4 )=
3 4
3 4
(^12)
0
R R
RR
R R
V
( 1 2 )( 3 4 )
0 3 4
R R R R
RR
Vs + +
V −
(b) If R 1 = R 2 = R 3 = R 4 = R,
R
V 2 R
V
S
0
α ⋅ =
α = α = 40
Chapter 2, Solution 25
V 0 = 5 x 10
3 = 50V
Using current division,
I 20 =
= x
0.1 A
V 20 = 20 x 0.1 kV = 2 kV
p 20 = I 20 V 20 = 0.2 kW
