EP IKASGELAKO PRAKTIKA 2009/02/24 (ebazpena)

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

Transcript of EP IKASGELAKO PRAKTIKA 2009/02/24 (ebazpena)

Page 1: EP IKASGELAKO PRAKTIKA 2009/02/24 (ebazpena)

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 IKASGELAKO PRAKTIKA 2009/02/24 (ebazpena)

(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 IKASGELAKO PRAKTIKA 2009/02/24 (ebazpena)

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 IKASGELAKO PRAKTIKA 2009/02/24 (ebazpena)

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 IKASGELAKO PRAKTIKA 2009/02/24 (ebazpena)

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 IKASGELAKO PRAKTIKA 2009/02/24 (ebazpena)

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 IKASGELAKO PRAKTIKA 2009/02/24 (ebazpena)

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 IKASGELAKO PRAKTIKA 2009/02/24 (ebazpena)

(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 IKASGELAKO PRAKTIKA 2009/02/24 (ebazpena)

(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