Transcript 作業解答

CHAPTER 5
Problems and Solutions
不要用看的，要動手！
Problem 1
a. Determine the currents I and IL for the network in Fig.5.77.
b. Find the power delivered to the resistor R2.
ZR1
ZC
ZR2
ZR2與ZL串聯成 Z’
ZL
ZC與Z’並聯成 Z”
Z'  ZR 2  ZL  3k  j3k  4.243k45
Z"  Z' // ZC  ...
 5.878k  11.31  5.764k  j1.152k  j3k
ZT  ZR1  Z"  7.764k  j1.152k  7.849k8.44
ZR1
ZC
ZR2
ZR2與ZL串聯成 Z’
ZL
ZC與Z’並聯成 Z”
I( j) 
E( j)
1200

 15.29mA8.44
ZT
7.849k  8.44
ZC
5k  90
I L ( j) 
 I( j) 
 15.29mA8.44
ZC  Z'
 j5k  3k  j3k
 21.18mA  47.87
PR 2  I L  R 2  ...  1.346W
2
延伸問題
a. E與I的相位差？
b.從電壓源E往右看，由R1…XL所組成電路，是那一種特徵
的電路？電感特徵？電容特徵？
Problem 2
a. Determine the vo1tage Vs for the network in Fig.5.78.
b. Find the power delivered to R2.
ZL
ZR1
ZR2
Z'  ZR 2 // ZC  ...  5k  j5k  7.701k  45
Z"  Z'ZL  5k  j5k  7.701k45
ZT  Z R1 // Z "  ...  4.472k25.565
Vs ( j)  I( j)  ZT  ...  89.44V26.565
ZC
Z"  Z'ZL
ZL
Z'  ZR 2 // ZC
ZR1
ZC
ZR2
Z'
VR 2  VS ( j) 
 ...  89.44V  63.44
Z' ZL
2
VR 2
PR 2 
 ...  0.8W
R2
Z R1
另一條路… I
 ...
R2  I 
Z R1  Z"
PR 2  I R 2 R 2  ...
2
延伸問題
求vs(t) ？
Problem 7
Determine the current I in Fig. 5.79 by first converting both
voltage sources to current sources and then combining the
parallel current sources.
I1 ( j) 
E1 ( j)
 10mA0
Z R1
I 2 ( j) 
E2 ( j)
40V0

 8mA90
ZC
5k  90
IT  I1 ( j)  I2 ( j)  10mA  j8mA  12.81mA38.66
將ZR1與ZC並聯起來Z’
Z'  ZR1 // ZC  ...  1.857k  21.8
Z'  ZR1 // ZC  ...  1.857k  21.8
利用Current divider rule
Z'
I( j)  IT ( j) 
 ...  6.37mA  45.63
Z' ZL
Problem 8
Convert the current source in Fig.5.80 to a voltage source and
find the current I by combining the series voltage sources and
using Ohm's law.
2kΩ
E1  I  ZC  2mA 0  2k  90
 4V  90
E1 ( j)  E( j)
I
 ...  2.41mA  48.37
ZC  Z R 2  Z L
Problem 9
a. Determine the current through the inductor of Fig. 5.79 using
mesh analysis.
b. Using the results for part (a) find the voltage across the
inductor.
Z1=2kΩ
Z3=5kΩ∠-90°
I1
I2
Z2=4kΩ∠90°
Mesh 1
E1  I1Z1  (I1  I2 )Z2  0
Mesh 2
 ( I 2  I1 )Z2  I 2Z3  E2  0
I1 ( Z1  Z2 )  I 2 Z2  E1
I1Z2  I 2 ( Z2  Z3 )  E 2
E1
 Z2
E 2  ( Z 2  Z3 )
I1 
 ...
Z1  Z2
 Z2
Z2
 ( Z 2  Z3 )
Z1  Z2 E1
Z2
E2
I2 
 ...
Z1  Z2
 Z2
Z2
 ( Z 2  Z3 )
I  I1  I2  ...  6.57mA  45.63
跨越電感的電壓降
VL  I  Z2  ...  25.48V44.37
Problem 10
Determine the current I in Fig.5.81 using mesh analysis.
I1
I2
Z1=10Ω∠0°
Z2=20Ω∠0°
Z3=30Ω∠90°
Z4=50Ω∠-90°
Mesh 1
Mesh 2
E  I1Z1  (I1  I2 )Z2  0
 (I 2  I1 )Z2  I 2 (Z3  Z4 )  0
I1 ( Z1  Z2 )  I 2 Z2  E
I1Z2  I 2 ( Z2  Z3  Z4 )  0
I1
I2
Z1=10Ω∠0°
Z2=20Ω∠0°
Z3=30Ω∠90°
Z4=50Ω∠-90°
I2 
Z1  Z2
Z2
Z1  Z2
Z2
E
0
 ...  0.632A71.565
 Z2
 ( Z 2  Z3  Z 4 )
Problem 11
Determine the current through resistor R2 in Fig 5.86.
Z1  1k0
I2
Z 2  3k0
Z 3  6 k0
I1
Z 4  6 k0
Z5  6k90
I3
Z 6  6k90
Mesh 1
E  I1Z1  (I1  I 2 )Z2  (I1  I3 )Z4  0
Mesh 2
 (I 2  I1 )Z2  I 2 Z5  (I 2  I3 )Z3  0
Mesh 3
 (I3  I1 )Z4  (I3  I 2 )Z3  I3 Z6  0
I1 ( Z1  Z 2  Z 4 )  I 2 Z 2  I3 Z 4  E
I1Z 2  I 2 ( Z 2  Z3  Z5 )  I3 Z5  0
I1Z 4  I 2 Z3  I3 ( Z3  Z 4  Z6 )  0
解出
I1  15.25mA   32.15
I 2  9mA   92.16
I R 2  I1  I 2  13.2mA 4.05
Problem 12
Determine the voltage across the resistor R2 in Fig 5.86.
Z3
Z1
Z2
Z6
Z5
Z4
Problem 13
a. Determine the nodal voltages for the network in Fig.5.82.
b. Calculate the current through each impedance using the results
for part (a).
V2
V1
Z1  2k0
Z2  10k0
NODE 1
V1 V1  V2
I1  I 2  
0
Z1
Z2
NODE 2
V1  V2
V2
 I2  I3 
0
Z2
Z3
Z3  4k  90
 1
 1 
1 


V1     V2    I1  I 2
 Z1 Z2 
 Z2 
 1
 1 
1 
V1    V2     I 2  I 3
 Z2 
 Z2 Z3 
I1  I 2

1
Z2
1
1 
I 2  I 3    
 Z1 Z2 
V1 
1
1
1


Z2
Z1 Z2
1
1 
1
   
Z2
 Z1 Z2 
 13.133V72.58
1
1

I1  I 2
Z1 Z2
1
I2  I3
Z2
V2 
1
1
1


Z1 Z2
Z2
 1
1
1 
   
Z2
 Z1 Z2 
 36.72V  172.47
Problem 14
Find the Thévenin equivalent circuit for the network to the left of
the load ZL in Fig. 5.83.
ZR1  6k0
ZR 2  1k0
ZC  8k  90
ZL  6k90
Z'  ZR1  ZL  6k  j6k  8.485k45
Z"  Z' // ZC  ...  10.732k  26.565
ZTH、ETH
ZTH  Z " Z R 2  ...  10.599k  j 4.799k
 R  jX C
ETH
ZC
 E
 151.78V  71.565
Z R 1  Z L  ZC
Problem 15
a. Determine ZL for maximum power to ZL in Fig. 5.83.
b. Calculate the maximum power that can be delivered to ZL.
R + XL
ZTH  Z"ZR1  ...  10.599k  j4.799k
 R  jXC
For max. power
ZL  10.599k  j4.799k
2
E TH
PMAX 
 ...  544mW
4R TH
Problem 16
a. Find the Thévenin equivalent circuit for the network to the left
of XC= 50Ω in Fig. 5.81.
b. If XC in Fig. 5.81 is replaced by a load ZL, determine R and X
for the load so that the load will receive maximum power.
LOAD
ZTH  ZR1 // ZR 2   ZL  ...  6.667  j30
ETH
ZR 2
 E
 13.333V0
Z R1  Z R 2
For max. power
ZL  6.667  j30
由電阻與電容串聯而成，
電容的電抗（reactance）30Ω
Problem 17
a. Find the Thévenin equivalent circuit for the network to the left
of ZL in Fig. 5.84.
b. Determine ZL for maximum power to ZL.
c. Calculate the maximum power to ZL.
ZTH  ZR  ZL1  ZL2  // ZC  ...
 4.17  j9.17  10.07  65.56
ETH
ZC
 E
 17.7V  135
ZR  ZL1  ZL 2  ZC
For max. power
ZL  4.17  j9.17
由電阻與電感串聯而成，
電感的電抗（reactance）9.17Ω
Problem 23 For the network in Fig. 5.88, determine:
a. The current through each branch using Ohm's law.
b. The total power dissipated (real power).
c. The net reactive power.
d. The total apparent power
e. The network power factor.
E
1200

 30mA0
a. I R 
Z R 4k0
E
1200
IL 

 12mA  90
Z L 10k90
E
1200
IC 

 15mA90
ZC 8k  90
b. P  I 2 R  R  3.6W
c. Q L  I 2 L X L  1.44VAR
QC   I 2 C X C  1.8VAR
QT  Q L  QC  0.36VAR
d. Total apparent power ST  P 2  QT 2  3.62 VA
P
 0.994
e. Power factor FP 
ST
Leading
Problem 24 For the system in Fig. 5.89, determine:
a. PT b. QT c. ST d. I e. FP
a. Total power PT  300W  100W  600W  1,000W
b. Net reactive power
QT  200VAR  700VAR  400VAR  100VAR
Lagging 電感性較強
c. Total apparent power
ST  PT  QT  1,005 VA
2
2
d.
ST
I
 8.375 A
E
e. Power factor
PT
FP 
 0.995
ST
  cos1 0.995  5.732 Lagging