电路基础习题答案
时间：2025-05-15
清华大学出版社
Problem 4.19/20/99
1
i
5
io
1V
+
8
3
8 (5 3) 4 io 1 2 i 1 10
,
i
1 1 4
1 5
0.1A
Problem 4.26 ( 4 2 ) 3 ,io 1 2 i1 1 4 ,
i1 i 2
1 2
A
v o 2i o
0.5V 5 i2 i1 4 io
1A
8
6
2
If is = 1 A, then vo = 0.5 V Problem 4.3 R 3R io 3R Vs + 3R R + vo (a) (b) 1V + 3R 1.5R
清华大学出版社
We transform the Y sub-circuit to the equivalentR 3R vo vs 2
.
3R
2
3 4
R,
3 4
R
3 4
R
3 2
R
4R
independent of R
io = vo/(R) When vs = 1V, vo = 0.5V, io = 0.5A (b) (c) When vs = 10V, vo = 5V, io = 5A When vs = 10V and R = 10 , vo = 5V, io = 10/(10) = 500mA
Problem 4.4 If Io = 1, the voltage across the 6 resistor is 6V so that the current through the 3 resistor is 2A. 2A 1A 3 3A 6 4 Is 2 3A + 2 v1 (a)3 6 2
2 i1 4 Is
(b)i1 vo 4 3A .
, vo = 3(4) = 12V,
Hence Is = 3 + 3 = 6A If Is = 6A Is = 9A Io = 1 Io = 6/(9) = 0.6667A 2 v1 3 vo
Problem 4.5
Vs
+
6
6
6
清华大学出版社
If vo = 1V,
1 V1 1 2 V 3 10 2 Vs 2 v 1 3 3
If vs =
10 3
vo = 1 vo =3 10 x 15
Then vs = 15 Problem 4.6 Let i = i1 + i2,
4.5V
where i1 and iL are due to current and voltage sources respectively. 6 i1 6 4 5A 20V + (b)20 6 4 2A
i2 4
(a) i1 =6 6 4 (5 ) 3 A , i 2
Thus i = i1 + i2 = 3 + 2 = 5A Problem 4.7 Let wherei x1 i x i x1 i x 2 i x2
is due to 15V source and
is due to 4A source,
12 +
i ix1 ix2 12 10 40 -4A 40
15V
10
(a)
(b)
清华大学出版社
For ix1, consider Fig. (a). 10||40 = 400/50 = 8 ohms, i = 15/(12 + 8) = 0.75 ix1 = [40/(40 + 10)]i = (4/5)0.75 = 0.6 For ix2, consider Fig. (b). 12||40 = 480/52 = 120/13 ix2 = [(120/13)/((120/13) + 10)](-4) = -1.92 ix = 0.6 – 1.92 = -1.32 A p = vix = ix2R = (-1.32)210 = 17.43 watts
Problem 4.8 Let vab = vab1 + vab2 where vab1 and vab2 are due to the 4-V and the 2-A sources respectively.
10
3vab1 + +
10
3vab2 + + 2A vab2 (b)
4V
+
vab1 (a)
For vab1, consider Fig. (a). Applying KVL gives, - vab1 – 3 vab1 + 10x0 + 4 = 0, which leads to vab1 = 1 V For vab2, consider Fig. (b). Applying KVL gives, vab2 – 3vab2 + 10x2 = 0, which leads to vab2 = 5 vab = 1 + 5 = 6 V
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Problem 4.9 the 4-A source.
Let i = i1 + i2, where i1 is due to the 12-V source and i2 is due to 6 + io i1 12V 2 3
(a)
i2 6 2 3
4A 2
ix2 2
4A
(b) For i1, consider Fig. (a). 2||3 = 2x3/5 = 6/5, io = 12/(6 + 6/5) = 10/6 i1 = [3/(2 + 3)]io = (3/5)x(10/6) = 1 A For i2, consider Fig. (b), 6||3 = 2 ohm, i2 = 4/2 = 2 A i = 1+2 = 3A
Problem 4.10 Let vo = vo1 + vo2 + vo3, where vo1, vo2, and vo3 are due to the 2-A, 12-V, and 19-V sources respectively. For vo1, consider the circuit below. 2A 2A 4
5 + vo1 6 3
io 5 + vo1 5
12
清华大学出版社
6||3 = 2 ohms, 4||12 = 3 ohms. He
nce, io = 2/2 = 1, vo1 = 5io = 5 V For vo2, consider the circuit below. 6 + 5 + vo2 12V 3 12 12V + 4 6 + v1 5 + vo2 3
3
3||8 = 24/11, v1 = [(24/11)/(6 + 24/11)]12 = 16/5 vo2 = (5/8)v1 = (5/8)(16/5) = 2 V For vo3, consider the circuit shown below. 5 + vo3 6 3 12 4 + 19V 5 + vo3 2 12 4 + v2 7||12 = (84/19) ohms, v2 = [(84/19)/(4 + 84/19)]19 = 9.975 v = (-5/7)v2 = -7.125 vo = 5 + 2 – 7.125 = -125 mV Problem 4.11 Let vo = vo1 + vo2 + vo3, where vo1, vo2 , and vo3, are due to the 20V, 1-A, and 2-A sources respectively. For vo1, consider the circuit below. 6 4 + 20V 2 + vo1 3 + 19V
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6||(4 + 2) = 3 ohms, vo1 = (½)20 = 10 V For vo2, consider the circuit below. 6 4 2 + 1A vo2 3||6 = 2 ohms, vo2 = [2/(4 + 2 + 2)]4 = 1 V For vo3, consider the circuit below. 6 2A 4 2 + vo3 3 3 vo3 + 3 2A 3 4V + 4 6 2 + vo2 3
6||(4 + 2) = 3, vo3 = (-1)3 = -3 vo = 10 + 1 – 3 = 8 V
清华大学出版社
Problem 4.12 Let i = i1 + i2 + i3, where i1 , i2 , and i3 are due to the 20-V, 2-A, and 16-V sources. For i1, consider the circuit below. io + i1 2 3 1 4
20V
4||(3 + 1) = 2 ohms, Then io = [20/(2 + 2)] = 5 A, i1 = io/2 = 2.5 A For i3, consider the circuit below. + 2 vo’ 2||(1 + 3) = 4/3, vo’ = [(4/3)/((4/3) + 4)](-16) = -4 i3 = vo’/4 = -1 For i2, consider the circuit below. i3 3
1
4 +
16V
2
1 i2 3
2A (4/3) 4
1 i2 3
2A
清华大学出版社
2||4 = 4/3, 3 + 4/3 = 13/3 Using the current division principle. i2 = [1/(1 + 13/2)]2 = 3/8 = 0.375 i = 2.5 + 0.375 - 1 = 1.875 A p = i2R = (1.875)23 = 10.55 watts
Problem 4.13 Let io = io1 + io2 + io3, where io1, io2, and io3 are due to the 12-V, 4-A, and 2-A sources. For io1, consider the circuit below. io1 + 4 3 2
12V
10
5
10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A For io2, consider the circuit below. 4A
io2 4 10
3
2
5 i1
2 + 5 + 4||10 = 7 + 40/14 = 69/7 i1 = [3/(3 + 69/7)]4 = 84/90, io2 =[-10/(4 + 10)]i1 = -6/9 For io3, consider the circuit below. io3 i2 4 10 5 2A 3 2
清华大学出版社
3 + 2 + 4||10 = 5 + 20/7 = 55/7 i2 = [5/(5 + 55/7)]2 = 7/9, io3 = [-10/(10 + 4)]i2 = -5/9 io = (12/9) – (6/9) – (5/9) = 1/9 = 111.11 mA
Problem 4.14 Let vx = vx1 + vx2 + vx3, where vx1,vx2, and vx3 are due to the 90V, 6-A, and 40-V sources. For vx1, co …… 此处隐藏：5826字，全部文档内容请下载后查看。喜欢就下载吧 ……
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