9
λ λ λ λ Δ λ (J 1 ) 1 (J 1 ) 2 (J 1 ) 3 (I C ) 1 = 30A n 1 n 2 1 = 15 n 1 n 2 2 = 15 n 1 n 2 3 = 30 I SARE (I C ) 2 = 90A (I C ) 3 = 60A

[EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

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Page 1: [EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

λ  

λ  

λ  

λ  

Δ  

λ  

(J1)1   (J1)2   (J1)3  

(IC)1  =  30A  

n1n2

⎝ ⎜

⎠ ⎟ 1

=15

n1n2

⎝ ⎜

⎠ ⎟ 2

=15

n1n2

⎝ ⎜

⎠ ⎟ 3

= 30

ISARE  

(IC)2  =  90A   (IC)3  =  60A  

Page 2: [EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

(IC)1=  30A  

n1n2

⎝ ⎜

⎠ ⎟ 1

=15

n1n2

⎝ ⎜

⎠ ⎟ 2

=15⇒ n1n2

⎝ ⎜

⎠ ⎟ 2

=n1n2

⎝ ⎜

⎠ ⎟ 1

(IC)2=  90A    (IC)2  =  3(IC)1    

(IC)3=  60A    (IC)3  =  2(IC)1    

n1n2

⎝ ⎜

⎠ ⎟ 3

= 30⇒ n1n2

⎝ ⎜

⎠ ⎟ 3

= 2 ⋅ n1n2

⎝ ⎜

⎠ ⎟ 1

(V13)1  

(V1)2  =  (V13)3    

30o  

Page 3: [EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

λ  

λ  

(J1)1  

(IC)1  =  30A  

n1n2

⎝ ⎜

⎠ ⎟ 1

=15

30   60   90   120   150   180   210   240   270   300   330   360  

(VL)1  

(Is1)1  

(V13)1   (V23)1   (V21)1   (V31)1   (V32)1  (V12)1  

(J1)1  =  (Ip1)1  

(IC)1  

-­‐(IC)1  

n2n1

⎝ ⎜

⎠ ⎟ 1

⋅ IC( )1 = x

−n2n1

⎝ ⎜

⎠ ⎟ 1

⋅ IC( )1 = −x

(V32)1  

Page 4: [EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

30   60   90   120   150   180   210   240   270   300   330   360  

(VL)2  

(Is1)2  

(V2)2   (V3)2  (V1)2  

(J1)2  =  (Ip1)2  

(IC)2  

23⎛

⎝ ⎜ ⎞

⎠ ⎟ ⋅

n2n1

⎝ ⎜

⎠ ⎟ 2

⋅ IC( )2 =23⎛

⎝ ⎜ ⎞

⎠ ⎟ ⋅

n2n1

⎝ ⎜

⎠ ⎟ 1

⎣ ⎢

⎦ ⎥ ⋅ 3 ⋅ IC( )1[ ] = 2 ⋅ n2

n1

⎝ ⎜

⎠ ⎟ 1

⋅ IC( )1⎡

⎣ ⎢

⎦ ⎥ = 2x

−13⎛

⎝ ⎜ ⎞

⎠ ⎟ ⋅

n2n1

⎝ ⎜

⎠ ⎟ 2

⋅ IC( )2 = −13⎛

⎝ ⎜ ⎞

⎠ ⎟ ⋅

n2n1

⎝ ⎜

⎠ ⎟ 1

⎣ ⎢

⎦ ⎥ ⋅ 3 ⋅ IC( )1[ ] = −

n2n1

⎝ ⎜

⎠ ⎟ 1

⋅ IC( )1 = −xλ  

λ  

(J1)2  

n1n2

⎝ ⎜

⎠ ⎟ 2

=15

(IC)2  =  90A  

Page 5: [EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

30   60   90   120   150   180   210   240   270   300   330   360  

(VL)3  

(Is1)3  

(V13)3   (V23)3   (V21)3   (V31)3   (V32)3  (V12)3  

(IC)3  

-­‐(IC)3  

n2n1

⎝ ⎜

⎠ ⎟ 3

⋅ IC( )3 =12⋅n2n1

⎝ ⎜

⎠ ⎟ 1

⎣ ⎢

⎦ ⎥ ⋅ 2 ⋅ IC( )1[ ] =

n2n1

⎝ ⎜

⎠ ⎟ 1

⋅ IC( )1 = x

−n2n1

⎝ ⎜

⎠ ⎟ 3

⋅ IC( )3 = −xΔ  

λ  

(J1)3  

n1n2

⎝ ⎜

⎠ ⎟ 3

= 30

(IC)3  =  60A  

(Ip1)3  

Page 6: [EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

30   60   90   120   150   180   210   240   270   300   330   360  

(VL)3  

(Is3)3  

(V13)3   (V23)3   (V21)3   (V31)3   (V32)3  (V12)3  

(IC)3  

-­‐(IC)3  

n2n1

⎝ ⎜

⎠ ⎟ 3

⋅ IC( )3 = x

−n2n1

⎝ ⎜

⎠ ⎟ 3

⋅ IC( )3 = −xΔ  

λ  

(J1)3  

n1n2

⎝ ⎜

⎠ ⎟ 3

= 30

(IC)3  =  60A  

(Ip3)3  

Page 7: [EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

x

−x

Δ  

λ  

(J1)3  

n1n2

⎝ ⎜

⎠ ⎟ 3

= 30

(IC)3  =  60A  

(Ip3)3  

x

−x

(Ip1)3  

(J1)3  =  (Ip1)3  -­‐  (Ip3)3  

x€

2x

−x

−2x

30   60   90   120   150   180   210   240   270   300   330   360  

Page 8: [EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

(J1)3    

x€

2x

−x

−2x30   60   90   120   150   180   210   240   270   300   330   360  

(J1)2    

2x

−x

(J1)1    

x

−x

J1( )1 =12π

⋅ 2 ⋅ 2π3⋅ x 2

⎣ ⎢ ⎤

⎦ ⎥ = x ⋅ 2

3

J1( )1 = x ⋅ 23

=n2n1

⎝ ⎜

⎠ ⎟ 1

⋅ IC( )1⎡

⎣ ⎢

⎦ ⎥ ⋅

23

=115

⋅ 30⎡

⎣ ⎢ ⎤

⎦ ⎥ ⋅23

=1.63A

J1( )2 =12π

⋅ 1⋅ 2π3⋅ 2x( )2 +1⋅ 4π

3⋅ x( )2

⎣ ⎢ ⎤

⎦ ⎥ = x ⋅ 1

2⋅23⋅ 4 +

43

⎣ ⎢ ⎤

⎦ ⎥ = x ⋅ 1

2⋅123

⎣ ⎢ ⎤

⎦ ⎥

J1( )2 = x ⋅ 2 =n2n1

⎝ ⎜

⎠ ⎟ 1

⋅ IC( )1⎡

⎣ ⎢

⎦ ⎥ ⋅ 2 =

115

⋅ 30⎡

⎣ ⎢ ⎤

⎦ ⎥ ⋅ 2 = 2.83A

J1( )3 =12π

⋅ 2 ⋅ π3⋅ 2x( )2 + 4 ⋅ π

3⋅ x( )2

⎣ ⎢ ⎤

⎦ ⎥ = x ⋅ 1

2⋅ 2 ⋅ 1

3⋅ 4 + 4 ⋅ 1

3⎡

⎣ ⎢ ⎤

⎦ ⎥ = x ⋅ 1

2⋅123

⎣ ⎢ ⎤

⎦ ⎥

J1( )3 = x ⋅ 2 =n2n1

⎝ ⎜

⎠ ⎟ 1

⋅ IC( )1⎡

⎣ ⎢

⎦ ⎥ ⋅ 2 =

115

⋅ 30⎡

⎣ ⎢ ⎤

⎦ ⎥ ⋅ 2 = 2.83A

Page 9: [EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

(J1)3    

x€

2x

−x

−2x30   60   90   120   150   180   210   240   270   300   330   360  

(J1)2    

2x

−x

(J1)1    

x

−x

ISARE    

30   60   90   120   150   180   210   240   270   300   330   360  

−2x

−3x

5x

−4x

4x

ISARE =12π

⋅ 4 ⋅ π6⋅ 4x( )2 + 2 ⋅ π

6⋅ 5x( )2 + 2 ⋅ π

6⋅ 2x( )2 + 2 ⋅ π

6⋅ 3x( )2

⎣ ⎢ ⎤

⎦ ⎥

= x ⋅ 12π

⋅ 4 ⋅ π6⋅16 + 2 ⋅ π

6⋅ 25 + 2 ⋅ π

6⋅ 4 + 2 ⋅ π

6⋅ 9

⎣ ⎢ ⎤

⎦ ⎥

= x ⋅ 12⋅646

+506

+86

+186

⎣ ⎢ ⎤

⎦ ⎥ = x ⋅ 1

2⋅1406

⎣ ⎢ ⎤

⎦ ⎥ = x ⋅ 70

6

ISARE = x ⋅ 353

=n2n1

⎝ ⎜

⎠ ⎟ 1

⋅ IC( )1⎡

⎣ ⎢

⎦ ⎥ ⋅

353

=115

⋅ 30⎡

⎣ ⎢ ⎤

⎦ ⎥ ⋅353

= 6.83A