电路分析基础(英文版)课后答案第一章

DE 1.8 [a] 24 = v2 + v5 ¡ v1 = 3i5 + 7i5 ¡ (¡2i5) = 12i5 Therefore i5 = 24=12 = 2 A
[b] v1 = ¡2i5 = ¡4 V [c] v2 = 3i5 = 6 V [d] v5 = 7i5 = 14 V
Problems 3 [e] p24 = ¡(24)(2) = ¡48 W; therefore 24 V source is delivering 48 W. DE 1.9
imax
=
1 ®e
=
1 0:03679e
»=
10
A
DE 1.3 [a]
when
t
=
1 ;
®
Therefore (a) v = ¡20 V;
i = ¡4 A; (b) v = ¡20 V, i = 4 A
(c) v = 20 V, i = ¡4 A; (d) v = 20 V, i = 4 A
[b] Using the reference system in Fig. 1.3(a), p = vi = (¡20)(¡4) = 80 W, so the box is absorbing power.
0
0
w = 360t = 360(1 £ 60) = 21:6 kJ
6 CHAPTER 1. Circuit Variables and Circuit Elements
P 1.7
p = vi;
Zt
w = p dx
0
Since the energy is the area under the power vs. time plot, let us plot p vs. t.
p = (6)(100) £ 10¡3 = 0:6 W; w = (0:6)(3)(60)(60) = 6480 J
Assume we are standing at box A looking toward box B, then p = vi.
[a] p = (120)(5) = 600 W
from A to B
= 21:67 ¡ 60e¡500t + 20e¡1500t + 40e¡1000t¡
25e¡2000t + 3:33e¡3000t¹J
w(1 ms) = 1:24¹J
[c] wtotal = 21:67¹J
[a] v(20 ms) = 100e¡1 sin 3 = 5:19 V i(20 ms) = 20e¡1 sin 3 = 1:04 A
90 · t · 95 பைடு நூலகம்s:
v = 180 ¡ 2000t V; i = 0:9 mA; p = 162 ¡ 1800t mW
95 · t · 100 ms:
v = ¡200 + 2000t V; i = 0:9 mA; p = ¡180 + 1800t mW
8 CHAPTER 1. Circuit Variables and Circuit Elements
q(0)
0
q(t) ¡ q(0) =
sin 4000y 24
4000
¯¯¯¯t
0
P 1.2 P 1.3
P 1.4
But q(0) = 0 by hypothesis, i.e., the current passes through its maximum value at t = 0, so q(t) = 6 £ 10¡3 sin 4000t C = 6 sin 4000t mC
i = 0:6 mA; p = 0:6t mW
10 · t · 25 ms:
v = 10 V;
i = 0:6 mA; p = 6 mW
25 · t · 35 ms:
v = 75 ¡ 2500t V; i = 0 mA; p = 0 mW
35 · t · 60 ms:
v = ¡50 + 1000t V; i = ¡0:4 mA; p = 20 ¡ 400t mW
across the 12 V battery, therefore p = vi = ¡12(30) = ¡360 W.
Thus the power °ow is from B to A, and Car A has the \dead" battery.
Zt
Zt
[b] w = p dx = 360 dx
[b] p = (250)(¡8) = ¡2000 W
from B to A
[c] p = (¡150)(16) = ¡2400 W
from B to A
[d] p = (¡480)(¡10) = 4800 W
from A to B
[a]
P 1.5
p = vi = (40)(¡10) = ¡400 W Power is being delivered by the box. [b] Entering [c] Gain
p(20 ms) = vi = 5:39 W
Problems 7
[b]
p = vi = 2000e¡100t sin:2 150t
=
2000e¡100t
·1
¡
1
cos
¸
300t
22
= 1000e¡100t ¡ 1000e¡100t cos 300t
Z1
Z1
w=
1000e¡100t dt ¡ 1000e¡100t cos 300t dt
[a] p = vi = 30e¡500t ¡ 30e¡1500t ¡ 40e¡1000t + 50e¡2000t ¡ 10e¡3000t
p(1 ms) = 3:1 mW
Zt
[b]
w(t) =
(30e¡500x ¡ 30e¡1500x ¡ 40e¡1000x+
0
50e¡2000x ¡ 10e¡3000x)dx
dp = ¡1000e¡500t + 2000e¡1000t = 0 at t = 1.4 ms dt
pmax = p(1:4 ms) = 0:5 W
[b]
¡0:25 + vt + vt = 0; 100 25
p = vt2 = 1 W: 25
5vt = 25;
vt = 5 V
Problems
P 1.1
i
=
dq dt
=
24 cos 4000t
Therefore, dq = 24 cos 4000t dt
Problems 5
Z q(t)
Zt
dx = 24 cos 4000y dy
p(0) = (6)(15 £ 10¡3) = 90 £ 10¡3 W
p(216 ks) = (4)(15 £ 10¡3) = 60 £ 10¡3 W
w
=
(60
£
10¡3)(216
£
103)
+
1 (216)(30)
=
16:2
kJ
2
P 1.8 P 1.9
Note: 60 hr ´ 216;000 s = 216 ks
60 · t · 70 ms:
v = ¡50 + 1000t V; i = 0 mA; p = 0 mW
70 · t · 80 ms:
v = 20 V;
i = ¡0:5 mA; p = ¡10 mW
80 · t · 90 ms:
v = 180 ¡ 2000t V; i = 0 mA; p = 0 mW
[c] The box is absorbing 80 W.
1
2 CHAPTER 1. Circuit Variables and Circuit Elements
DE 1.4 p = vi = 20 £ 104e¡10;000t W;
w = Z 1 20 £ 104e¡10;000t dt = 20 J 0
i2 = 120=24 = 5 A i3 = 120=8 = 15 A i1 = i2 + i3 = 20 A ¡200 + 20R + 120 = 0 R = 80=20 = 4 − DE 1.10 [a] Plotting a graph of vt versus it gives
Note that when it = 0, vt = 25 V; therefore the voltage source must be 25 V. When vt is zero, it = 0:25 A, hence the resistor must be 25=0:25 or 100−. A circuit model having the same v ¡ i characteristic is a 25 V source in series with a 100− resistor.
DE 1.5 p = 800 £ 103 £ 1:8 £ 103 = 1440 £ 106 = 1440 MW
from Oregon to California
DE 1.6
The interconnection is valid:
is = 10 + 15 = 25 A
p100V = 100is = 2500 W (absorbing)
[a] p = vi = (¡60)(¡10) = 600 W, so power is being absorbed by the box. [b] Entering [c] Lose
P 1.6
[a] Looking from A to B the current i is in the direction of the voltage rise
1
Circuit Variables and Circuit Elements
Drill Exercises
DE 1.1 q = Z 1 20e¡5000t dt = 4000 ¹C
0
DE 1.2 i = dq = te¡®t; dt
di = (1 ¡ ®t)e¡®t; dt
di =0
dt
Therefore
p10A = ¡100(10) = ¡1000 W (generating)
¡100 + vs ¡ 40 = 0
so vs = 140 V
p15A = ¡15(140) = ¡2100 W (generating)
p40V = 15(40) = 600 W (absorbing)
X
pdev = p10A + p15A = 3100 W
[b]
w(25)
=
1 2
(6)(10)
+
(6)(15)
=
120
¹J
w(60)
=
120
+
1 2
(15)(6)
¡
1 2
(10)(4)
=
145
¹J
w(90) = 145 ¡ (10)(10) = 45 ¹J
w(100)
=
45
¡
1 2
(10)(9)
=
0
¹J
P 1.11 [a] p = vi = (2e¡500t ¡ 2e¡1000t) W
4 CHAPTER 1. Circuit Variables and Circuit Elements [b]
25 it = 125 = 0:2 A;
p = (0:2)2(25) = 1 W:
DE 1.11 [a] Since we are constructing the model from two elements, we have two choices on interconnecting them|series or parallel. From the v ¡ i characteristic we require vt = 25 V when it = 0. The only way we can satisfy this requirement is with a parallel connection. The constraint that vt = 0 when it = 0:25 A tells us the ideal current source must produce 0:25 A. Therefore the parallel resistor must be 25=0:25 or 100−.
X
pabs = p100V + p40V = 3100 W
X
X
pdev = pabs = 3100 W
DE 1.7 [a] vl ¡ vc + v1 ¡ vs = 0; ilRl ¡ icRc + i1R1 ¡ vs = 0
isRl + isRc + isR1 ¡ vs = 0
[b] is = vs=(Rl + Rc + R1)
= =
0
1000
10 ¡
1e¡0¡10100000·t ¯¯¯¯¯11 0 £¡1100410+00009(£(110004)e2¸¡+=100(1t3000¡)21[¡100
cos
300t
+
300
sin
300t])¯¯¯¯¯1 0
w = 9J
P 1.10 [a] 0 · t · 10 ms:
v = 1000t V;