《电路》课程教学资源（课件讲稿）L11-4 戴氏2+最大功率传输

$4-3戴维宁定理与诺顿定理（续）(Thevenin -Norton Theorem)

§4-3 戴维宁定理与诺顿定理 （续） (Thevenin-Norton Theorem)

单选题设置UocF10Uoc-1010V+-22D提交

Uoc = V 10 -10 -2 2 A B C D 提交 单选题 a b + – 10V – + U2 + – U1 + – Uoc

戴维宁等效电路（法2开路短路法ReqaaUocRLusbboc/R=P开路电压等效电阻(短路电流；保留所有独立源)WocUocReq= Uocli,S

戴维宁等效电路(法 1 ) a b a b us RL + - Req u RL oc + - + - us uoc + - Req 开路电压 等效电阻 (所有独立源置零) i sc i sc= uoc/Req Req= uoc/i sc i u sc oc + - 等效电阻 (短路电流; 保留所有独立源) 2 开路短路法

Ex Find the Thevenin equivalent of the circuitoa+4252Solution:O+?25V2023AWab1.Open circuit voltagebetween a, b:一bWocWaEquivalent resistance (short2.5242十circuit current; keep all theW+Oindependent sources)20225V3Au?ocMa4252b上+M20225V3AiBy node-voltage analysis method:b25?号+325+3u35204Uoc = 32 V= 16/4 = 4Auni = 16VR. = uo. / i. = 82

Ex Find the Thévenin equivalent of the circuit. 25V + - 5Ω 20Ω 3A 4Ω + - uab a b Solution: 1 1 1 25 ( ) 3 5 20 4 5 oc     u 2. Equivalent resistance when all independent sources killed: 5Ω 20Ω 4Ω a b Req Equivalent resistance (short circuit current; keep all the independent sources) i 25V sc + - 5Ω 20Ω 3A 4Ω a b 0 1 1 1 1 1 25 ( ) 3 5 20 4 5 n     u 1 un 16V i sc   16 4 4A eq / 8 R u i    oc sc 1. Open circuit voltage between a, b: 25V + - 5Ω 20Ω 3A 4Ω + - uab a b uoc uoc By node-voltage analysis method: uoc  32V

3. Thévenin equivalent circuit:=-8QReqMaMa42+52=32VW420225V3AUabLObuoc = 32VRe = uoc / isc = 32/4 = 82

3. Thévenin equivalent circuit: 25V + - 5Ω 20Ω 3A 4Ω + - uab a b uoc +- Req a b =32V =8 eq 32V 32 4 8 oc oc sc u R u i        

Ex. Find the Thevenin equivalent with respect to terminals a, b20ig（网络内含受控源）5k210kQ2kΩ②①?aMM开路短路法mio50k220kg40k250Vb?Solution:1. Open circuit voltage between a, b:v,-50Vnl50VnlVn2=0×10-Y02×10320×1035×1032x10320Vn2Vn2Vn2 - Vnl-Vn320ig =(×10-y +(-)×10-31<10-320i5×10350×10310 ×10355010Vn3Yn3 -1220ig =0<10-3 v,)×10-3yn3 = +20ig40×1010×103104010m2Vip10V; =100V = Voc50x10350×103

Ex. Find the Thévenin equivalent with respect to terminals a, b. 50V + - 2kΩ 20kΩ a b 5kΩ 50kΩ 40kΩ 10kΩ 20ib ib (网络内含受控源) Solution: 1. Open circuit voltage between a, b: 0 1 开路短路法 2 3 1 1 1 2 3 3 3 2 1 2 2 3 3 3 3 3 3 2 3 3 2 3 50 0 2 10 20 10 5 10 20 0 5 10 50 10 10 10 20 0 40 10 10 10 50 10 n n n n n n n n n n n n n v v v v v v v v v i v v v i v i b b b                                     3 100V oc v v   - + + 3 3 1 2 3 3 3 3 1 2 3 3 3 2 3 2 3 1 1 1 1 50 ( ) 10 10 2 20 5 5 2 10 1 1 1 1 1 10 ( ) 10 10 20 5 5 50 10 10 1 1 1 10 ( ) 10 20 10 40 10 50 10 n n n n n n n n v v v v v i v v i v i b b b                                            - +

20ig2. Short circuit current between a, b10kD5k?2kQ?2aMMWWi0三50k220k2>40k2lsc50VbOVnl - 50VnlVnl12=0V12i.=20ig2×1035×10320×10310×10320vm2Vn2Vn2V2-Vn+ 20is = 0Vnl = 36V10×1035×1010×1050x10350×103(Vn2 =10VVn2= 5mAip50x10320 k2M3.Thevenin equivalent circuit:voc = 100V100V100Voc120k2Rbec5mA1SC

2. Short circuit current between a, b: 50V + - 2kΩ 20kΩ a b 5kΩ 50kΩ 40kΩ 10kΩ 20ib ib isc 0 1 2 0 1 1 1 2 3 3 3 2 1 2 2 3 3 3 2 3 50 0 2 10 20 10 5 10 20 0 5 10 50 10 10 10 50 10 n n n n n n n n n v v v v v v v v i v i b b                            1 2 36V 10V n n v v      2 3 20 10 10 n sc v i i  b  3. Thévenin equivalent circuit: eq 100V 100V 20k 5mA oc oc sc v v R i           100V + - 20 kΩ a b 2 2 3 3 20 50 10 10 10 n n v v      5mA

戴维宁等效电路(法3外施电源法Reqaa不下十口一RLDUocRLusbb等效电阻开路电压（所有独立源置零）外加电源iWocUThReq= ur/in

戴维宁等效电路(法 1 ) a b a b us RL + - Req u RL oc + - + - us uoc + - Req 开路电压 等效电阻 (所有独立源置零) 外加电源 iT a b uT + - Req= uT /iT 3 外施电源法

Ex Find the Thevenin equivalent with respect to a, b602602Wi000a?40i40i800800=bbVnt = 40i.Solution:Vn2Vn22MVn21. Open circuit voltage:Vn3=0802016Voc = OVVn3= VTVTVr - Vnl- Vn22. Short circuit current :it2060-Vn22. Apply a test source:80Voltage source → find currentVTT=18.75QR.ediT

Ex Find the Thévenin equivalent with respect to a, b. + - 80Ω 16Ω 20Ω 60Ω iD 40iD a b Solution: 1. Open circuit voltage: voc  0V 2. Short circuit current : 2. Apply a test source: + - iT + - 80Ω 16Ω 20Ω 60Ω iD 40iD a b vT 0 1 2 3 1 2 1 2 2 3 3 2 40 0 16 80 20 80 n n n n n n n T n v i v v v v v v v v i D D                   2 1 20 60 T n T n T v v v v i     Voltage source  find current eq 18.75 T T v R i   

3. Thevenin equivalent circuit:vo= 000VT =18.7592R..eqirMα18.75 Q

3. Thévenin equivalent circuit: oc eq 0 18.75 T T v v R i          18.75 Ω a b