戴维宁定理和应用
Davidin Theorem and Application
單元四 戴維寧定理
Unit four. Davinin's theorem.
一、重點整理
I. CONCLUSION OF THE CONTEXT
1. 對於任何複雜的線性網路系統，都可以用單一的等效電
1. For any complex linear network system, one-size-fits-all electrical equivalents are available
壓源 ETh 串聯一個等效電阻器 RTh 來表示即為戴維寧定理。
The pressure ETh connects to an equivalent electrical resister RTh to indicate that it is Davinin's theorem.
2. 戴維寧等效電路圖示
2. Davidin Equivalent Circuit Icons
a错误!
A mistake!
複
Repeat.
複
Repeat.
雜線雜性線網性路網
Multi-channel, inter-line, inter-line, inter-line, inter-networked, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line network, multi-line, multi-line, multi-line, multi-line, multi-line, online, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, online, multi-line, multi-line, multi-line, online, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line, multi-line,
R
RL
R
L
R
路
Lou.
3. 解題步驟
3. Steps to solve the problem
bb
Step1 選取戴維寧等效電路的範圍：欲求網路中任意二點
Step1 Selects the range of Daviden equivalent circuits: any two points in the web
間的戴維寧等效電路時，先移去此二點內的電路元件（並將此二端點標記為 a、b）
, move the two-point electrical component (and mark the two endpoints as a,b)
a
b
RLRL
Step2 計算戴維寧等效電阻 RTh：將原來網路中所有的
Step2 Calculating Davining Equivalence Resistment RTh: All that will be available on the Internet
電壓源短路、電流源斷路。戴維寧等效電阻 RTh即為 a、b二端點間的等效電阻值
Pressure short circuits, current circuits, etc. The Davidin Equivalent Resist RTh is an Equivalent Equivalence Retardation value between a and b-points
R
R
Step3 計算戴維寧等效電壓 ETh：戴維寧等效電壓 ETh即為
Step3 calculates Daviden's Equivalence.
RTh?Rab?(R1//R2)?R3
a、b二點間的開路電壓。對於較複雜的網路，我們
A two o'clock circuit voltage. For a more complicated network, we...
可以利用串並聯電路及重疊定理等方法來求 ETh
ETh can be found by connecting and connecting circuits and reglazing theorems.
（利用重疊定理求Vab）
(Vab using a regressed theory)
RRaETh
ETh
RTh
a
E
?b
b
ETh?
Vab
Step4 a、b二點間的複雜網路可用電壓 ETh串聯電阻 RTh
The complex network between Step4 a and b can use ETh interlocking RTh
來取代，並將移去之元件接回a、b二端點，然後計算負載電流 IL 及電壓 VL
To replace and return the removed widgets to the a, b, endpoint and then calculate the load IL and voltage VL
RL
ETh
IL?
VL
wWw.unjs.comL
ETh
RTh?RL
RL
?ETh
RTh?RL
4. 當RL=Rth時，RL可獲得最大功率為
4. When RL = Rth, RL achieves maximum power
5. 電功率公式 P=I R
5. Electrical power formula P=I R
甲、 計算出負載之最大功率
A. Calculating maximum capacity for load
PLmax
EThEE?()2RL?Th?Th
RTh?RL4RTh4RL
2
2
2
二、例題講解 Ex1：
Ex1:
Ex2：
Ex3：
2、
求總電容量？【解：20μF】
Total power capacity?
3、
【解：10H】
Solution: 10H
.圖中L1=3H，L2=5H，M=1H，則總電感量多少？
L1 = 3H, L2 = 5H, M = 1H in the graph, and what is the total electric sense?
4、兩電感10H及5H並聯，互感（M）為互助2H，則總電感量為
Four, two senses of 10H and 5H are connected.
多少【解：
How much?
4611
H】
5、兩電感3H及10H並聯，互感（M）為互消1H，則總電感量為多少【解：
Five, two senses of 3H and 10H are combined, and the senses (M) are equal to 1H.
6、某電感L＝20H，電感電流為10A，則此電感的儲存能量為多少焦耳【解：1000joul】
Six, a sense L = 20H, an electrosensive current of 10A, and how much of this energy is stored in J.
7、當10μF的`電容器充電至100伏特時，其儲存的能量為多少焦耳？【解：0.05焦耳】
7. How much of the energy is saved when 10 mF's'capacitors'charge to 100 volts?
L1
L2
2915
H】
8
L1＝12H，L2＝18H，則總電感量多少？【解：30H】
L1 = 12H, L2 = 18H, and what's the total electric sense?
9、
L1＝10H，L2＝15H，則總電感量多少？【解：
L1 = 10H, L2 = 15H, and what's the total electric sense?
6H】
10、重點整理公式及電路符號、單位，再寫乙遍能背誦默寫。
10- Focus formulae and circuit symbols, units, and write " B" over and over and over again.
單元六 電阻與電容、電阻與電感的暫態
Unit six, resistance and contemplation of capacitation, resistance and electric sense.
一、重點整理
I. CONCLUSION OF THE CONTEXT
二、例題講解 1、
II. CURRENT PRESENTATIONS 1,
一RC串聯電路，R＝800k?，C＝0.5?F，試求其充電5個
A RC series of circuits, R=800k, C=0.5?F, trying to charge five.
時間常數，需耗時多久？ 【解：2秒】 2、
Time constant, how long does it take?
電阻R＝100k?與電感L＝500mH串聯之電路，其時間常數
Resist R = 100k? Connected circuits with electrosensive L = 500 mH. Time constant
?為多少？【解：5μ秒】 3、
♪ For what? ♪ 3 ♪ 3 ♪ 3 ♪ 3 ♪ 3 ♪ 3 ♪ 3
電阻R＝100?與電感L＝0.5H串聯之電路，其時間常數?
Resist R = 100? Connected circuits with electrosensive L = 0.5H. Time constant?
為多少？【解：5m秒】 4、
- How much? - Four.
一RC串聯電路，R＝5k?，C＝0.02?F，試求其充電時間常
A RC series of combined circuits, R=5k? C=0.02?F, trying to charge time.
數，需耗時多久？
How long will it take?
【解：100μ秒】 5、
Five, two, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, three, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four, four....
電阻R＝100?與電感L＝0.5H串聯之電路，其充電5個時
Resist R = 100? Connected circuits with electrosensive L = 0.5H, charged five times.
間常數5?為多少？【解：25m秒】 6、
Constant 5? What is it? 6?
達穩態時，電感視同 ，電流I為
When it's stable, the electric senses are the same.
多少？【解：4安培】
How much?
(t)C
7、 若E=90伏特，R=15KΩ，C=10μF，充
7. If E = 90 volts, R = 15K, C = 10 mF,
電瞬間，電容器視同 此時電流I為多少？【解：6mA】
In the instant, the capacitor looks the same. What's the current I?
8、同上題，達穩態時，電容器視同 ，此時電容器電壓為多少？【解：90V】
8: The same capacitator when it's stable. What's the capacitor's voltage at this time?
L(t)9、
電感器視同
The electrosensors look the same.
若E=100V，R=10Ω，L=50mH，充電瞬間，
If E=100V, R=10, L=50mH, charge instant,
此時電流I為多少？【解：0A】
What's the current?
10、同上題，達穩態時，電感器視同 ，此時電流I為多少？【解：10A】
10, op. cit. When you're stable, the sensor looks the same. What's the current I?
三、練習題
III. PRACTICAL ISSUES
(t)C
1、若E=100伏特，R=1KΩ，C=2μF，試求其充
If E = 100 volts, R = 1K, C = 2 mF, try to fill it.
電時間常數？【解：2mA】
Electricity Time Constant?
2、寫出RC、RL電路在充放電時的充電時間常數公式？【解：重點整理】
2 - Write RC, RL circuits for charging time constant?
3、寫出RC、RL電路在充放電時須經多久的時間常數才能達穩態？【解：重點整理】
Three, how long does it take for the RC and RL circuits to get steady when they charge?
4、RC、RL電路在充放電時，開關切入的瞬間（t=0），電容的狀態視同 ，電感的狀
4, RC, RL circuits when charged, the moment of switch cut (t = 0), the capacitor's condition is the same, the electrical senses are the same.
態視同 。【解：重點整理】
Same as always.
5
、
【解：重點整理】
[Setting: Collapse]
電容器充電的電壓曲線圖為何？
What's with the capacitor's voltage curve?
6、同上題，電容器放電時電流曲線圖為何？【解：重點整理】
What's the current curve map when the capacitor discharges?
L(t)7、若E=35V，R=21Ω，L=420mH，試求其充電
L(t)7, if E=35V, R=21, L=420mH, tries to charge it
時間常數？【解：20mH】
Time constant?
8、同上題，再穩態時電感器可視為 ，此時穩態的時間常數為多少？【解：100mH】
8-Supremely, re-stabilization time sensors can be seen as, what is the steady time constant at this time?
9、同第五題的甲乙丙曲線圖，何者為電感器充電電流圖【解：甲】
9. The fifth question is the map of the meth-acetic curve, and what's the electrosensor charge?
10、同第五題的甲乙丙曲線圖，何者為電感器充電電壓圖【解：乙】
10. The fifth question is the map of the meth-acetic curve, and what is the electrosensor charger.
單元七 頻率與週期、相位與向量運算
Unit 7 frequency and periodicity, phase and vector operations
一、重點整理
I. CONCLUSION OF THE CONTEXT
二、例題講解
II. LESSONS ON THE QUESTION
Ex1：求v(t)=1002sin(314t+300)之頻率與週期大小？
Ex1: The frequency and frequency of v(t)=1002sin (314t+300)?
Ex2：將v(t)=1002sin(314t-450)以相量式表示？
Ex2: Express v(t)=1002sin (314t-450) in phase?
Ex3：若v(t)=100sin(377t-300)，i(t)=10sin(377t+600)；則兩波形之相位差為何？
Ex3: If v(t) = 100sin (377t-300), i(t) = 10sin (377t+600); what is the difference between the two waves?
Ex4：請將A=8-j6以極座標表示；B?10?600以直角座標表示。
Ex4: Indicate A=8-j6 at the polar coordinates; B?10?600 at the straight corner coordinates.
Ex5：若A?10?530，B?5370；則A?BA?B
Ex5: If A? 10.530, B? 5370; A?BA?B?
Ex6：若A=5?j53，B=3?
Ex6: If A=5? j53, B=3?
三、練習題
III. PRACTICAL ISSUES
1. v(t)=50sin(377t-370)之頻率為 ，週期為 。
The frequency of v(t) = 50sin (377t-370) is, and the weekly period is.
2. v(t)=20sin(1000t+530)之相量式為 。 3. 若v1(t)=2sin(314t+600)，v2(t)=10cos(377t-450)；
v(t)=20sin(1,000t+530) the phase is 3. If v1(t)=2sin(314t+600), v2(t)=10cos(377t-450);
則v1(t)超前v2(t) 度。
And v1(t) is overv2(t) degrees.
j3；則A?B
J3; A? B?
A
B
4. 將C=?6?j6以極座標表示 ；
4. Indicate C = 6?j6 as the polar coordinates;
D?52?1350以直角座標表示。
D?52? 1350 is indicated by the straight corner coordinates.
5. 若A?2450，B?6?j8；則A?B?A?B?。
If A-2450, B-6? j8; A?B?A?B?B?
6. 若C=4?j3，D=6?
If C=4? j3, D=6?
j8
；則C?D? ；
C?D? or C? or C?D? or C? or C? or C? or C? or C? or C? or C? or C? or C? or C? or C?
D
? 。 C
單元八 RLC串並聯電路
Unit 8 RLC connects and connects circuits
一、重點整理
I. CONCLUSION OF THE CONTEXT
二、例題講解
II. LESSONS ON THE QUESTION
Ex1：如圖所示，若v(t)=100sin(1000t+30)，R=20Ω，求
Ex1: If v(t) = 100sin (1,000t+30), R = 20 times, please
O
此電路之阻抗為何？
What's the resistance to this circuit?
O
Ex2：如圖所示，若v(t)=100sin(1000t+30)，C=250μF，求此電路之阻抗為何？
Ex2: If v(t) = 100sin (1,000t+30), C = 250 mF, as shown in the graph, why is this circuit blocked?
O
Ex3：如圖所示，若v(t)=100sin(1000t+30)，L=25mH，求此
Ex3: If v(t) = 100sin (1,000t+30), L = 25mH, please
電路之阻抗為何？
What's the resistance to the circuit?
Ex4：如圖所示，若v(t)=100sin(1000t+30)，R=6Ω，C=125μ
Ex4: If v(t) = 100sin (1,000t+30), R = 6 times, C = 125 m, as shown in the figure
O
F，
求此電路之總阻抗
I'm asking you to stop this circuit all the time.
Z
=?
Ex5：如圖所示，若v(t)=100sin(500t+30)，R=4Ω，L=6mH，
Ex5: If v(t) = 100sin (500t+30), R = 4 times, L = 6mH,
O
求此電路之總阻抗Z
I'm going to ask for this circuit to stop Z.
O
Ex6：如圖所示，若v(t)=100sin(1000t+30)，R=8Ω，L=16mH，
Ex6: If v(t) = 100sin (1,000t+30), R = 8 times, L = 16mH,
C=100μF，求此電路之總阻抗Z=?
C=100mF, the total resistance to this circuit?
O
Ex7：如圖所示，若v(t)=100sin(1000t-45)，R=5Ω，C=200
Ex7: If v(t) = 100sin (1,000t-45), R = 5 times, C = 200, as shown in the figure
μF，求此電路之總阻抗ZO
MIF, call this circuit to resist ZO.
Ex8：如圖所示，若v(t)=100sin(1000t-53)，R=4Ω，L=3mH，
Ex8: If v(t) = 100sin (1,000t-53), R = 4 times, L = 3mH,
求此電路之總阻抗Z
I'm going to ask for this circuit to stop Z.
O
Ex9：如圖，v(t)=1202sin(1000t+30)，R=3Ω，L=2mH，C=250
Ex9: Figure, v(t) = 1202sin (1,000t+30), R = 3 times, L = 2mH, C = 250
μF，求電路之總阻抗Z=?
MIF, the circuits are always against Z =?
三、練習題
III. PRACTICAL ISSUES
1. 如圖所示，若v(t)=141.2sin(377t-37)，R=5Ω，則此電
1. As shown in the figure, if v(t)=141.2sin (377t-37) R=5 times
O
路之阻抗Z=。
Road blocker Z=.
2. 如圖所示，若v(t)=30sin(250t-53)，C=400μF，則此電
2. As shown in the figure, if v(t) = 30sin (250t-53), C = 400 mF, then the electricity
O
路之阻抗Z=。O
Road blocker Z=.
3. 如圖所示，若v(t)=100sin(400t+45)
If v(t) = 100sin (400t+45) as shown in the graph
，L=30mH，則此電
L = 30 mH, this power.
路之阻抗Z=。4. 如圖所示，若v(t)=500sin(1000t+37)，R=2Ω，C=500
Road resistance Z = 4. If v(t) = 500sin (1,000t+37), R = 2 times, C = 500 as shown in the graph
O
μ
F，此電路之總阻抗
F, there's always resistance to this circuit.
Z
O
5. 如圖所示，若v(t)=100sin(1000t+30)，R=9Ω，L=12mH，此電路之總阻抗
As shown in the graph, if v(t) = 100sin (1,000t+30), R = 9, L = 12mH, total resistance of this circuit
Z
=。
O
6. 如圖，若v(t)=250sin(400t+30)，R=30Ω，L=25mH，C=50
6. If v(t) = 250sin (400t+30), R = 30, L = 25mH, C = 50
μF，此電路之總阻抗Z。
MIF, this circuit is always against Z.
7. 如圖，若v(t)=1202sin(1000t)，R=30Ω，C=25μF，此
If v(t) = 1202 sin(1,000t), R = 30, C = 25 mF,
電路之總阻抗Z=
The circuits are always against Z.
。
8. 如圖，若v(t)=200sin(2000t+53)，R=10Ω，L=5mH，此
If v(t) = 200sin (2000t+53), R = 10, L = 5mH,
O
電路之總阻抗Z=。
The overall circuit resistance is Z =.
O
9. 如圖，若v(t)=102sin(1000t-30)，R=5Ω，L=5mH，C=50μF，此電路之總阻抗Z。
9. As can be seen, if v(t) =102sin (1,000t-30), R = 5 times, L = 5mH, C = 50 mF, this circuit is in total resistance to Z.
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