Tema IV: Transformadores

P
η
=
P
4.18 Rendimiento del
cedida =
absorbida
transformador
P
P
2
1
P 1 = P 1 P2
+ 2 + P
fe + fe + P
2
2
2
2
2
P cu = cu = R
1 ⋅ 1 ⋅I
I 1 + 1 + R
2'
2'⋅
⋅ I
2'
2'
≅
R
cc ⋅ cc ⋅I
I 1 = 1 = R
cc ⋅ cc ⋅I
I 1
n ⋅ n ⋅C
C =
P
cc
U
2I
2I2Cos
2Cosϕ
ϕ
η
=
U
I
Cos
ϕ
+
P
+
P
I
1 I 1 I2
C
2
C =
≅
I
1
n I n I2
2 n
C
⋅
U
2I
2I2
2 n
Cos
ϕ
=
2 =
C 2
C C
⋅
U
I
Cos
ϕ
+
P
+
P
2
C 2
C
2
2
U2
U − U2
− U
0
cc
2
2
n
n
2
C
ε C
εc
= c =
U (%)
U (%)
2 c = c = [ 1 −
ε
c ] c ] ⋅
U
2
n
U
2
n
η
=
C
⋅
C
⋅
[ 1
−
ε
c ] c ]
[ 1
−
ε
c ]
⋅
U
2nI
I 2
c
2n
⋅
U
2
nI
nI2
2 nCos
nCosϕ
ϕ
Cos
ϕ
+
P
+
P
2n
n
0
cc
C
2
=
C
⋅
cu
0
cc
P
η =
P
η =
⋅
C
2
2
P
P
+ P
P
+
2
fe
P + P +
C
⋅
[ 1
−
ε
c ]
⋅ c ⋅S
S nCos
nCosϕ
ϕ
2
[ 2
[ 1
−
ε
c ] ⋅
S
nCos
Cos ϕ
+
P
0 +
P
ccC
C
c
Ensayo de vacío
n
0
cc
cu
EL TRANSFORMADOR
TRABAJA CON UN
ÍNDICE DE CARGA C