《电路理论》试题 - 华中科技大学精品课程

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8. (1) 设 UA = 220∠0o V ， 则 UB = 220∠ −120o V ， UC = 220∠120o V ，
•
UAC = 380∠ − 30o V ，
•
•
•
UBC = 380∠ − 90o V ；功率因素角为θ ，那么 IA = I∠(−θ )A ， IB = I∠(−120o −θ )A 。
解得，uo = − 2 ui 11
3．解：
⎧ ⎪
I
l1
=
2
⎨⎪⎪(I2l2
⎪ ⎪
I
l4
= 0.5U + 4 + 2) = −1
I
l3
− (2 + 4)Il1
+
2Il2
− 2Il4
=8
⎪⎩U = 4(Il3 − Il1 )
解得
⎧ ⎪
I
l1
=
2A
⎪ ⎪
I
l2
=
2 3
A
⎨
⎪ ⎪
I
l3
= 13 6
A
⎪ ⎩
I
2Ω 1A
+
uC (s) 1
−
S
解得,
(1
+
s)
I
L
(
s)
=
(1
−
I
L
(
s))(2
+
1 s
)
IL
(s)
=
2s +1 s2 + 3s +1
1.（6）设
N
的参数矩阵为：
⎡ ⎢ ⎢
Z11 Z 21
Z12 ⎤
Z
22
⎥ ⎥
由于 N 为对称松弛网络，所以 Z11 = Z22 ， Z12 = Z21
当右边端口开路时：
2A
3Ω
4Ω
N1
2Ω
+
U
6Ω
6V N2
（4）Find the admittance parameter matrix Y(s) of the two-port.
R n:1
C
*
*
L
-2-
（5）Find the Laplace transform of the impulse response iL(t) in the circuit.
6Ω
2A
2Ω 8V
4Ω U 0.5U
2Ω 1A
-4-
得 分 评卷人 4．（10 Mark）The circuit is in steady state before the switch is opened. The switch is opened at t = 0 ，Find the voltage u
1Ω
ab
6Ω
+
3
S−
2
2Ω
S−S
9
3Ω
S
+
il (0− )
ua (s)
=
3 s
⋅
1
2 +
// 2/
2 s /
2
=
6 s(s + 3)
=
2 s
−
2 s+3
s
ub
(s)
=
9 s
⋅
s
+
3 6+
3
⋅
s
+
6
⋅
s
+
s 6
+
3
−
6
=
−
27 s+9
ua (t) = (2 − 2e−3t )V
ub (t) = −27e−9tV
maximum power absorbed by it.
10Ω
10Ω
R
6A
+
12V
20Ω
-5-
得分
is
评卷人
6．（10 Mark）In the steady state circuit, is = (3cost + cos3t)A ，
the reading of the watt meter is 5W and the reading of the voltage meter is 8V （rms）.Find the values of R and L .
=
U
2
⎪
⎪ ⎪⎩
I3
=
I4 n
⇒
⎧ ⎪⎪
I1
=
(1 R
−
jω L)U1
−
n R
U
2
⎨
⎪ ⎪⎩
I2
=
−
n R
U1
− (n2 R
+
j
1 ωC
)U
2
于是
Y
(s)
=
⎡ ⎢ ⎢
1 R
−
jω L
⎢ ⎢⎣
−n R
−n R
⎤ ⎥ ⎥
n2 R
−
j
1 ωC
⎥ ⎥⎦
1．（5）运算电路为：
1Ω IL (s) S
+
•
U1 −
1
2+
•
U2
− 2'
-3-
得分
评卷人
2．（10 Mark）Find the voltage gain
uo ui
ideal operational amplifiers.
4kΩ
in the circuit with
2kΩ
+
2kΩ ui
4kΩ uo
1kΩ
得 分 评卷人 3．（10 Mark）Find the supplied power by each independent source in the circuit.
⎧⎪⎨⎪⎩UU12
= =
Z11I1 Z21I1
⇒
Z11
=
2Z21
当右边端口短路时：
⎧⎪⎨⎪⎩UU12
= =
Z11I1 + Z12I2 Z21I1 + Z22I2
⇒
⎧⎪15∠0° =
⎨ ⎪⎩0
=
Z21I1
2Z21I1 +1∠180°Z21 + 2∠180°Z21
-9-
解得，
Z21 = 5Ω，所以Z11 = Z22 = 10Ω，Z12 = 5Ω
voltage U 2 in the Fig.(b). the voltages and currents are effective value phasors in
this problem.
+ 15∠0o V
−
•
1' I1
+
•
U1 −
1
•
I2 2 +
•
U2
− 2'
+ 15∠0o V
−
5Ω
•
1' I1
对于图（b）有：
⎧⎪⎨UU12
= 10I1 = 5I1
⎪⎩U1 = 15∠0° − 5I1
解得，U2 = 5∠0°V
⎧ui − u1
⎪ ⎪
2000
=
u1 − u2 2000
+
u1 − uo 4000
2．解：
⎪ ⎨ ⎪
u2 − u3 4000
⎪u1 = 0
= u3 1000
⎪⎩u3 = uo
（2）The voltage u and the current i are in phase in the sinusoidal steady circuit.
Find the angular frequency of the source.
i
1mH
1µF * *
u
3mH 2mH
20 Ω
-1-
（3）N1 and N2 are active resistive one-port networks in the circuit. Find the voltage U of the 3Ω resistor.
=
576 4 ×10
= 14.4W
- 11 -
6.解： 解得：
⎧⎪⎪(
3⋅ 2
R 2 + L2 )2 + ( 1 ⋅ 2
R 2 + (3L)2 )2 = 64
⎨ ⎪(
3
)2 ⋅R + (
1
)2 ⋅ R
=5
⎪⎩ 2
2
⎧ ⎪L =
59 H
⎨3
⎪⎩R = 1Ω
7.位形图如下：
d
c
a
b
UV3 = Ubc = 50V
U 2 are both 50 V . Find the reading(rms) of the voltage meter V3.
++
U1 R1
C
U
V3
+
L
R2 U 2
-
-
-7-
1.（1）解：
试题答案
⎧i = ⎨⎩(iin
uin −i)
+
4(iin
−
i)
+
4(iin
−
i
+
2i)
=
uin
⇒ Req
⎧⎪ ⎨ ⎪⎩
U U
AC I A BC I B
cos(θ − 30o ) = 866 cos(−90o +120o +θ
)
=
433
⇒ θ = 30o
- 12 -
电路的功率因数为 cos 30o ，即 3 。 2
(2) 3UAIA cos 30o = W1 + W2
⇒
IA
=
433 110 3
A
+
•
IA
=
uin iin
= −2.5Ω
1．（2）解：原电路可以等效为：
I + U
0.002 jω 0.001 jω
1µ F
−
20Ω 0.001 jω
进一步化简得：
I + U
−
1µ F 20Ω
0.0025 jω
由于电源电压和端口电流同相位，所以电路发生并联谐振，于是
ω = ω0 =
1
= 2×104 rad / s
所以
u = ua (t) − ub (t) = (2 − 2e−3t + 27e−9t )V (t>0)
5.解：先求 ab 端的开路电压 UOC
UOC = −24V
再求从 ab 端看进去的等效电阻 Req
R eq = 10Ω
即 R=Req = 10Ω 时，获得最大功率，
且
Pmax =
UOC2 4R eq
0.0025 ×10−6
1．（3）解：
⎧⎨⎩3I1I
+I 2+
2= 6−
2 6
I1
=
0
U = 3I2 = 2V
解得
⎧ ⎪⎪ ⎨ ⎪ ⎪⎩
I1 I2
= =
4 3 2 3
A A
-8-
1.（4）解：
⎧nU ⎪
2
+
I3
=
U1
⎪⎪⎪(I3
− I1)
1 jωC
= U1
⎨ ⎪(
I4
−
I2
)
jω
L
核对人
2007－2008 学年第一学期
《电路理论》试题
（88 学时、闭卷）
专业与班级
学号
姓名
题号 题分
得分
1
2
3
4
5
6
7
8 总分
30 10 10 10 10 10 10 10 100
得分
评卷人
1．（30Mark）There are 6 problems and 5marks for each
problem. （1）Find the equivalent resistance of the circuit.
1Z 3
•
UA
−
1
•
Z IA
Βιβλιοθήκη Baidu
=
•
UA
3
•
⇒
Z=
3UA
•
=
IA
660∠0o 433 ∠ − 30o
=
290.4∠30o Ω
110 3
- 13 -
for t > 0 .
1Ω
3V
2Ω 0.5F
6Ω
u
1H
3Ω 9A
得 分 评卷人 5．（10 Mark）The resistor R absorbs the maximum power in the circuit by adjusting its value，Find the value of R and the
l4
=
−1A
8V 电压源提供的功率为： P8V
= 8(Il2
+ Il3 ) =
68 W 3
2A
电流源提供的功率为：
P2 A
=
2(12Il1
−
6Il3
−
2Il2
)
=
58 W 3
1A
电流源提供的功率为：
P1A
=
2( I l3
−
Il4
)
=
19 W 3
- 10 -
4.解： 做出运算电路如下：
uC (0− ) = 0V ， iL (0− ) = 6A
* *W
R
L
得 分 评卷人
7．（10 Mark）In the balanced three-phase circuit, the line voltage of the source is 380V(rms), the readings of the watt meter W1
and W2 are 866W and 433W, respectively. Find: (a) the power
1Ω iL
1H
δ (t)A
2Ω
uC 1F
（5）Find the symmetric dead two-port. In the Fig.(a), the open circuit voltage of 2-2’ U 2 = 7.5∠0°V ,the short circuit of 2-2’ I2 = 1∠180°A ,Find the open circuit
factor of the circuit; (b) the value of the load impedance Z .
* A * W1
Z
*
B
* W2
Z
Z
C
-6-
得 分 评卷人 8．（10 Mark）In the sinusoidal steady-state circuit, the rms of the voltage U is 100 V ，the rms of the voltage U1 and