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《电路理论》试题 - 华中科技大学精品课程


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



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 。
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.
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,

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
)

L
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
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
=
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
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 .
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
* *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
+

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.
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
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:
=
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
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
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