24. 2. T = Tu + Ts =
T
u s
W
W W+
=
5400
1.5 19m 10 60 1.5 200 50 60× × × × µ× ×+
=
300hr
3.
R '
R
= 2
N
9
1
⇒ = 2
N ⇒ N=3⇒ 'l = Nl =100× 3 = 300m
4. I =
1
1
P
R
=
18
2
= 3A
E = IRT =3× (2+ 3 +4)= 27V
5. I2 =
1
2
V
R
=
10
2
= 5A
由 Node A 利用 K.C.L.可得 I1 + I2 = 2 + 4
I1 = 2 + 4 - I2 =(2+4)- 5 = 1A
再利用 K.C.L.可得 7 + I1 + 3 + I = 6 + 4 + 1 + 5
∴I = 5A
6. ∵I2 = 0A 代表電橋平衡⇒ R× 6 =4× 9⇒ R = 6Ω
∴I1 =
T
E
R
=
30
(4 6) // (6 9) 4+++
= 3A
7. 利用密爾門定律可得:
Vc= ΣIin(c)Rcb = ETH
=
1
o 1 o
1
E
( I )(R // R )
R
+
=
18
( 3) (6 //3)
6
×+ = 12V
RN = RTH = Rab(oc)| 移去電源
25. =R2+ 1 o(R // R ) = 3 +(6 // 3) = 5Ω
∴IN =
TH
TH
E
R
=
12
5
=2.4A
8. Vc=ETH
2 L
2 L
R R
R R R
+
++
⇒ 分壓律
9 =12
L
L
3 R
2 3 R
×
+
++
⇒ RL = 3Ω
9. Vo =E1=+ 60V ,利用節點電壓法
1 o
1
V V
R
-
+
1
3
V 0
R
-
+
1 2
2
V V
R
-
= 0
⇒ 1V 60
4
-
+
1V
5
+
1 2V V
4
-
= 0
⇒ 0.7V1 -0.25V2 =15…………
2 o
5
V V
R
-
+
2 1
2
V V
R
-
+
2 2
4
V E
R
-
= 0
⇒ 2V 60
5
-
+
2 1V V
4
-
+
2V 10
2
-
= 0
-0.25V1 +0.95V2 =17…………
10. CT =CX // CY ⇒ 20 Fµ = Y Y60 // C C 30 Fµ ⇒ µ=
QS = CTVT =20 300µ× = 6000 Cµ
VY =
S
Y
Q
C
=
6000
30
µ
µ
= 200V
11. WL =
21
LI
2
=
21
0.5 4
2
× × =4J
∵L =
N
I
φ
⇒ 0.5 =
N 0.01
4
×
∴ N= 200 匝
12. LX =
X X
X
N
I
φ
=
4
500 4 10
5
× × -
= 0.04H = 40mH
26. M=
Y XY
X
N
I
φ
=
4
1000 4 10 90%
5
× × ×-
= 72mH
13. 利用戴維寧定理化成下圖等效電路
Eo= E
2
1 2
R
R R+
=
6k
60
3k 6k
×
+
= 40V
Ro=R1 // R2=3k // 6k= 2kΩ
ic(0) =
o c
o 3
E v (0)
R R
-
+
=
40 12
2k 2k
-
+
=7mA
RTH =R3+(R1 // R2) =2k+(3k // 6k) = 4kΩ
τ = RTHC =4k 50× µ = 0.2sec
14. RTH =(6 // 6)+ 2 = 5Ω ETH = E
2
1 2
R
R R+
=
6
12
6 6
×
+
= 6V
τ =
TH
L
R
=
5m
5
= 1ms iL(tm) =
TH
TH
E
R
=
6
5
=1.2A
t =1sec=1000τ 故電感已達儲能穩態
iL(1)= iL(tm) =1.2A
15. f =
2
ω
π
=
157
2π
= 25Hz
Vrms =
mV
CF
=
100 2
2
= 100V
16. I =
mI
90 45
CF
∠ ° °- =
10
45
2
∠ ° =5 2 45 A∠ °+
17. E =
m
e
E
CF
∠θ =
4
0
2
∠ ° =2 2 0 V∠ °
I =
m
i
I
CF
∠θ =
2
45
2
∠ °+ =1 45 A∠ °+
27. Z =
E
I
=
2 2 0
1 45
∠ °
∠ °+
=2 2 45∠ °Ω-
R = Zcosθ =2 2 cos( 45 )°- = 2Ω
18. iR =
L
L
X
i(t)
R X+
=
j3
8sin377t
3 j3
×
+
=
j1
8sin377t
2 45
×
∠ °
=4 2 sin(377t 45 )A°+
19. abZ = L CR // X // X =10 // j20 // ( j4)-
=
20 4
10 //( j )
20 4
×
-
-
=10 // ( j5)- =
10 ( j5) 10 j5
10 j5 10 j5
×
×
-+
-+
= 2 2
250 j500
10 5
-
+
=2 j4Ω-
∴ 1v (t) =
ab
s
o ab
Z
v (t)
R Z+
=
(2 j4)
100 2 sin377t
10 (2 j4)
×
-
+-
=
(2 j4)(12 j4)
100 2 sin377t
(12 j4)(12 j4)
×
-+
-+
= 2 2
(24 16) j(8 48)
100 2 sin377t
12 4
×
++-
+
=
40 j40
100 2 sin377t
160
×
-
=50sin(377t 45 )V°-
20. S =EI=110× 5 =550VA
21. IR =
sV
R
=
40
4
= 10A
I =
S
E
=
500
40
= 12.5A
IL = 2 2
RI I- = 2 2
12.5 10- = 56.25A
XL =
sV
I
=
40
56.25
=
40
7.5
≒5.33Ω