$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