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《电路》(英文版)1-5 Network reduction by A-Y transformation

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Any given arm of the Y network is found by taking the product of the two adjacent arms of the network and dividing by the sum of the network arms.
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§1-5 Network reduction by△- YY transformation R R R R RI b R R R R R+ (R3+R2)(R2+Ra) 1 R3+R+ 3 R+R 2 =R3∥(Rn∥R1+Rn∥R2)

§1-5 Network reduction by Δ-Y transformation e d e d ad R R R R R R R R R R + + + + + = + 3 2 3 2 1 ( )( ) //( // // ) ' 2 ' 1 ' Ra d = R3 Ra R + Rd R Ra Rb Rc Re Rd a c b d R1 Re Rd a c d R3 R2 b Ra Rd a b d ' R3 ' R2 ' R1

△→Y R R Ro b C Raly=[r R1+R2 R, (R,+R) ab△ +Rb+ [rbe=trbj .R2+s r,(R+ra) Ra +r,+R (2) +r R(R+Rb R Ra+r,+r R,,+rbR+rr (1)+(2)+(3):R1+R2+R3=R+B+R (4)-(2):R1 RR R+B+e(1) RR (4)-(:0Rn+R+R(2) △相邻电阻之积 R= (4)-(1):R3 R,R △电阻之和 (3) R,+ R+r

(1) (2) (3): (4) 1 2 3 a b c a b b c c a R R R R R R R R R R R R + + + + + + + + = ' 1 (4) (2): (1) a b c c a R R R R R R + + − = ' 2 (4) (3): (2) a b c a b R R R R R R + + − = ' 3 (4) (1): (3) a b c b c R R R R R R + + − = 电阻之和 相邻电阻之积   RY =     Rab Y = Rab  (1) ( ) 1 2 a b c a b c R R R R R R R R + + +  + =     Rbc Y = Rbc  (2) ( ) 2 3 a b c b c a R R R R R R R R + + +  + =     Rca Y = Rca  (3) ( ) 3 1 a b c c a b R R R R R R R R + + +  + =  Y a c b Ra Rb Rc a c b R1 R2 R3

R1=△相邻电阻之积 R R R R3 △电阻之和 REb Any given arm of the Y network is found by taking the product of the two adjacent arms of the A network and dividing by the sum of the A network arms K'=k+k,+K' If ra=rb=r=r(balanced load) I then r=r=r2=R/3 +g+(5 y→△:(1)(2)+(2)(3)+(3)(1): Ka=k+k+\(3). R,R2+R2R3+r3R RRbR(R+R,+R) RRoR (Ra+r+r (Ra+Rb+r) (4)÷(3):Rn=(R1R2+R2R3+R3R1)/R3=R1+R2+R1R2/R3 (4)÷(1):R=(RR2+R2R3+R3R1)R1=R2+R3+R2R3/R1 (4)÷(2):R2=(R1R2+R2R3+R3R1)/R2=R3+R1+R3R1/R2

Any given arm of the Y network is found by taking the product of the two adjacent arms of the Δ network and dividing by the sum of the Δ network arms. 电阻之和 相邻电阻之积   RY = If Ra=Rb=Rc=R (balanced load) then R1=R2=R3=R/3. Y   : (1) (2) (2) (3) (3) (1) : ' ' ' ' ' ' + + ' 1 2 2 3 3 1 2 (4) ( ) ( ) ( ) a b c a b c a b c a b c a b c R R R R R R R R R R R R R R R R R R R R R + + = + + + + + + = 1 2 2 3 3 1 3 1 2 1 2 3 ' ' (4)  (3) : Ra = (R R + R R + R R )/ R = R + R + R R / R 1 2 2 3 3 1 1 2 3 2 3 1 ' ' (4)  (1) : Rb = (R R + R R + R R )/ R = R + R + R R / R 1 2 2 3 3 1 2 3 1 3 1 2 ' ' (4)  (2) : Rc = (R R + R R + R R )/ R = R + R + R R / R ' 1 (1) a b c c a R R R R R R + + = ' 2 (2) a b c a b R R R R R R + + = ' 3 (3) a b c b c R R R R R R + + = a c b Ra Rb Rc a c b R1 R2 R3

(4)÷(3):Rn=(RR2+R2R3+R3R)R3=R1+R2+RR2/R (4)÷(1):Rb=(R1R2+R2R3+R3R1)R1=R2+R3+R2B3/R1 (4)÷(2):R2=(R1R2+R2R3+R3R1)/R2=R3+R1+R3R1R2 Y两电阻之积 R3 g、g △ Y相对电阻 Rob Any resistance of A network is equal to the sum of the products of all possible pairs of the Y resistance divided by the opposite resistance of the Y network. If R2R3=R(balanced load), then Ra= 3R

相对电阻 两电阻之积 Y Y R   = If R1=R2=R3=R (balanced load), then Ra=Rb=Rc =3R. Any resistance of Δ network is equal to the sum of the products of all possible pairs of the Y resistance divided by the opposite resistance of the Y network. 1 2 2 3 3 1 3 1 2 1 2 3 ' ' (4)  (3) : Ra = (R R + R R + R R )/ R = R + R + R R / R 1 2 2 3 3 1 1 2 3 2 3 1 ' ' (4)  (1) : Rb = (R R + R R + R R )/ R = R + R + R R / R 1 2 2 3 3 1 2 3 1 3 1 2 ' ' (4)  (2) : Rc = (R R + R R + R R )/ R = R + R + R R / R a c b Ra Rb Rc a c b R1 R2 R3

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