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# EP MINTEGIA 2009/02/25 (ebazpena)

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### EP MINTEGIA 2009/02/25 (ebazpena)

1. 1. ISARE  (J1)2  (J1)3  (J1)1  λ  Δ  Δ   n1   n1   n1    = 15   = 30   = 30  n 2 1  n2 2  n2 3 λ  λ  λ  € € € (IC)1= 30A  (IC)2= 90A  (IC)3= 60A
2. 2.  n2  1 =   n1 1 15  n2   n2  1  n2  1  = ⇒   = ⋅   n1  2 30  n1  2 2  n1 1 € (V1)2  n2   n2  1  n2  1  = ⇒   = ⋅  30o (V13)3 n1  3 30  n1  3 2  n1 1  € 30o (V13€ )1 (IC)1= 30A (IC)2= 90A  (IC)2 = 3(IC)1 (IC)3= 60A  (IC)3 = 2(IC)1
3. 3. (J1)1 = (Ip1)1  n2  (J1)1   ⋅ ( IC )1 = x  n1 1 € λ n  −  2  ⋅ ( IC )1 = −x  n1 1  n1    = 15 (Is1)1  n 2 1 € λ (IC)1 -(IC)1 € (VL)1 (V32)1 (V12)1 (V13)1 (V23)1 (V21)1 (V31)1 (V32)1 (IC)1 = 30A 60 90 120 150 180 210 240 270 300 330 360 30
4. 4. (Ip1)2  2  1  n    n    2   n2    ⋅   ⋅ ( IC ) 2 =   ⋅   2   ⋅ [ 3⋅ ( IC )1 ] =  2  ⋅ ( IC )1 = x (J1)2  3   n1  2  3  2  n1 1  n1 1  € Δ   1  1  n   1  n  1 n  1 −   ⋅ 2  ⋅ ( IC ) 2 = −   ⋅  2   ⋅ [ 3⋅ ( IC )1 ] = − ⋅  2  ⋅ ( IC )1 = − x  3   n1  2  3  2  n1 1 2  n1 1 2  n1    = 30 (Is1)2  n2 2 € λ (IC)2 € (VL)2 (V1)2 (V2)2 (V3)2 (IC)2 = 90A 30 60 90 120 150 180 210 240 270 300 330 360
5. 5. (Ip3)2 (J1)2 x € Δ  1 −x 2  n1    = 30 € (Is3)2  n2 2 λ (IC)2 € (VL)2 (V1)2 (V2)2 (V3)2 (IC)2 = 90A 30 60 90 120 150 180 210 240 270 300 330 360
6. 6. (J1)2 = (Ip1)2 - (Ip3)2 3 x 2 (J1)2 € Δ 3 −x 2  n1    = 30  n2 2 (Ip1)2 € λ x € 1 −x € 2 € (Ip3)2 x (IC)2 = 90A € 1 −x 2 180 € 210 30 60 90 120 150 240 270 300 330 360
7. 7. (Ip1)3 1  n    n2   n2    ⋅ ( IC ) 3 =  ⋅    ⋅ [2 ⋅ ( IC )1 ] =   ⋅ ( IC )1 = x 2 (J1)3  n1  3 2  n1 1  n1 1 € n  Δ −  2  ⋅ ( IC ) 3 = −x  n1  3  n1  (Is1)3   = 30 € (IC)3  n2 3 λ -(IC)3 € (VL)3 (V12)3 (V13)3 (V23)3 (V21)3 (V31)3 (V32)3 (IC)3 = 60A 30 60 90 120 150 180 210 240 270 300 330 360
8. 8. (Ip3)3  n2    ⋅ ( IC ) 3 = x (J1)3  n1  3 € n  Δ −  2  ⋅ ( IC ) 3 = −x  n1  3  n1  (Is3)3   = 30 € (IC)3  n2 3 λ -(IC)3 € (VL)3 (V12)3 (V13)3 (V23)3 (V21)3 (V31)3 (V32)3 (IC)3 = 60A 30 60 90 120 150 180 210 240 270 300 330 360
9. 9. (J1)3 = (Ip1)3 - (Ip3)3 2x (J1) x € 3 € Δ −x  n1  −2x   = 30 €  n2 3 (Ip1)3 λ € x € € −x (Ip3)3 € x (IC)3 = 60A € −x 30 60 90 120 150 180 210 240 270 300 330 360 €
10. 10. 5x (J1)1 x 4x ISARE € 3x € € 2x € x −x € € € 3 −x x (J1)2 2 −2x € −3x € € −4 x € −5x 3 −x 2 30 60 90 120 150 180 210 240 270 300 330 360 € (J1)3 €  π  5 2  2 2 π 7  π 9  1 π π 2 2 ⋅  2⋅ ⋅  x  + 2⋅ ⋅  x  + 2⋅ ⋅  x  + 2⋅ ⋅ ( 2x) + 2⋅ ⋅ ( x)  I SARE = € 2x 2π  6 2  6 2  3 2  6 6    x  π 25 π 1 π 49 π 81 π € = x⋅ ⋅  2⋅ ⋅ + 2⋅ ⋅ + 2⋅ ⋅ + 2⋅ ⋅ 4 + 2⋅  2π 64 64 34 6 6 € 1  50 98 162 8 2 1  512 32 = x⋅ + +  = x⋅  = x⋅ ⋅ + + ⋅ 2  24 24 12 6 6 2  24  3 −x 32  n2   32  1  35 −2x € =   ⋅ ( I C )1 ⋅ I SARE = x ⋅ =  ⋅ 30 ⋅ = 6.53A 3  n1 1 3 15  3    30 60 90 120 150 180 210 240 270 300 330 360 €