电路分析第二章习题答案
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2-1
K
Ω
解:)
(6A
闭合时: 总电阻Ω
=
+
?
+
=4
6
3
6
3
2
R
)
(5.7
4
30
30
1
A
R
I=
=
=
此时电流表的读数为:)
(5
5.7
3
2
6
3
6
1
A
I
I=
?
=
+
=
2-2 题2-2图示电路,当电阻R2=∞时,电压表12V;当R2=10Ω时,解:当∞
=
2
R时可知电压表读数即是电源电压
S
U.
.
12V
U
S
=
∴
当Ω
=10
2
R时,电压表读数:4
12
10
10
1
2
1
2=
?
+
=
+
=
R
U
R
R
R
u
S
(V)
Ω
=
∴20
1
R
2-3 题2-3图示电路。求开关K打开和闭合情况下的输入电阻R i。
解:K )(18.60//(10Ω+=∴i R
K
)(8//30//(10Ω==∴i R
2-4 求题2-3图示电路的等效电阻R ab 、R cd 。
解:电路图可变为:
)
(1548
82.2148
82.2148//82.21)4040//10//(80//30)
(08.1782.294082
.294082.29//40)80//3040//10//(40)(4020
800)(8010800)
(4020
800
20201020202010123123Ω=+?==+=Ω=+?==+=Ω==Ω==Ω==?+?+?=cd
ab R R R R R 2-5 求题2-5图示电路的等效电阻 R ab 。
题2-59Ω
Ω
Ω
解:(a)图等效为:
)(73.3
56
87)25//(8Ω==?=+=∴ab R (b)
)(96325
150
Ω=+=+
=∴ab R
(c)图等效为:Ω
Ω
注意到10电阻可断去
)(67.127
147
148)25//()410(8Ω=+?+=+++=∴ab R
(d)图等效为:
18
1818912+?=
R
)
(2272//)36//1436//54()
(722)(3612311223Ω=+=Ω==Ω==ab R R R R R
2-6 题2-6图示电路中各电阻的阻值相等,均为R ,求等效R ab .
(b)
(a)
解:e 、f 、g 电位点,所以 (a)图等效为:
)]//()(//[)(R R R R R R R R R R R ab +++++++=
R R R R R R R 4
5310//
2]
4//22//[2==+=