电阻电路的等效变换课件.ppt
等效由KCL:i=i1+i2+……+in=u/R1+u/R2+……+u/Rn=u(1/R1+1/R2+……+1/Rn)R1R2i+ui1i2_inRnb).電阻並聯(ParallelConnection)+u_iReq=u/Req結構:頭與頭、尾與尾相連。定義:承受相同的電壓。並聯電阻的分流公式對於兩電阻並聯*Req=R1R2R1+R2R1R2i1i2ioo熟記即1/Req=1/R1+1/R2+……+1/RnGeq=G1+G2+……+Gn=?Gk=?1/Rk等效電導等於並聯的各電導之和R=4∥(2+3∥6)=2?R=(40∥40+30∥30∥30)=30?c)電阻的混聯:串、並聯的組合30?40?40?30?30?oR°40?30?30?40?30?ooR例2例14?2?3?6?ooR解:①用分流方法做②用分壓方法做求:I1,I4,U4..AAA例312V+_2R2R2R2RRRI1I2I3I4U4U2U1+-+-+-+-U32、求含受控源電阻電路的等效電阻Ri外加電源法U=6I1+3I1=9I1I1=I?6/(6+3)=(2/3)IRi=U/I=6?3?6?I1+–6I1U+–IU=9?(2/3)I=6I*二、等效互換?當滿足R1R4=R2R3時,電橋平衡:R5短路、開路或任意值,(自然平衡)Uab=0,I=0,混聯電路。?當不滿足R1R4=R2R3時,如果人為使a、b點短路,(強迫平衡)Uab=0,I?0混聯電路。?當不滿足R1R4=R2R3時,求Req=?電橋°°ReqR5R4R3R2R1°°Req?問題的提出Iba星形聯接與三角形聯接的電阻的等效變換(Y—?變換)Y型T型oooo+R1R2R3i1Yi2Yi3Y123++–––u12Yu23Yu31Yi1?=i1Yi2?=i2Yi3?=i3Yu12?=u12Yu23?=u23Yu31?=u31Y等效的條件:?型oooo?型R12R31R23i3?i2?i1?123++–––u12?u23?u31?Y接:u12Y=R1i1Y–R2i2Yu31Y=R3i3Y-R1i1Yu23Y=R2i2Y–R3i3Y(1)KVL::KCL:i1Yi2Yi3Y++=0?接:i3?=u31?/R31–u23?/R23i2?=u23?/R23–u12?/R12i1?=u12?/R12–u31?/R31(2)KCL:KVL:u12?u23?u31?++=0證明方法一:由式(2)解得;並滿足等效條件,i3?=u31?/R31–u23?/R23i2?=u23?/R23–u12?/R12i1?=u12?/R12–u31?/R311332212Y313Y12Y1RRRRRRRuRui++-====已知??求Y已知Y?求?聯立方程求解,可以得到:特例:若三個電阻相等(對稱),則有R??=3RY證明方法二、?滿足等效條件?R12Y=R12AR23Y=R23AR31Y=R31AR12Y=R1+R2R23YR2R3R3R1R31Y==++R12AR23+R12R31=//()R23AR31+R23R12=//()R31AR12+R31R23=//()聯立方程求解,可以得到同樣的上述結果。?型R1,R2,R5R3,R4,R5Y型R1,R3,R5R2,R4,R51、任意變換一個Y型或?型,都可以變成混聯電路。2、變換時,先變結構,然後根據公式特點,計算參數。Y變??變Y°°R31R12R23123°