Resultado:
La resistencia del circuito equivalente en alta frecuencia viene dada por:
![[tex]R=R_1\cdot+R_2\cdot (\frac{N_1}{N_2})^2=[/tex] [tex]=4.5\cdot+4.5\cdot (1)^2=9\,\Omega[/tex]](https://1.bp.blogspot.com/-ATSt3yq5YNg/XYehH9aY_TI/AAAAAAAAGf8/-GTgrUlndbEpuUwitHsAbd8H6PgATmAawCK4BGAYYCw/s400/formula9.JPG "[tex]R=R_1\cdot+R_2\cdot (\frac{N_1}{N_2})^2=[/tex] [tex]=4.5\cdot+4.5\cdot (1)^2=9\,\Omega[/tex]")
![[tex]R=R_1\cdot+R_2\cdot (\frac{N_1}{N_2})^2=[/tex] [tex]=4.5\cdot+4.5\cdot (1)^2=9\,\Omega[/tex]](http://1.bp.blogspot.com/-ATSt3yq5YNg/XYehH9aY_TI/AAAAAAAAGf8/-GTgrUlndbEpuUwitHsAbd8H6PgATmAawCK4BGAYYCw/s1600/formula9.JPG "[tex]R=R_1\cdot+R_2\cdot (\frac{N_1}{N_2})^2=[/tex] [tex]=4.5\cdot+4.5\cdot (1)^2=9\,\Omega[/tex]")
La inductancia del circuito equivalente en alta frecuencia viene dada por:
![[tex]L=L_{d1}\cdot+L_{d2}\cdot (\frac{N_1}{N_2})^2=[/tex]](http://3.bp.blogspot.com/-VT3W8JkYfZk/XYehQhNnEMI/AAAAAAAAGgE/Uaw2xEPr63s_fZTQWZ27Rw0smqrLZkrowCK4BGAYYCw/s1600/formula10.JPG "[tex]L=L_{d1}\cdot+L_{d2}\cdot (\frac{N_1}{N_2})^2=[/tex]")
![[tex]=5\cdot 10^{-6}\cdot+5\cdot 10^{-6}\cdot (1)^2=10\,\mu H[/tex]](http://4.bp.blogspot.com/-ciyEu3f4Y0Q/XYehZ5EDKZI/AAAAAAAAGgM/lSFQgEAoYVcAX5jLxBg_lu9-ECX33DcPACK4BGAYYCw/s1600/formula11.JPG "[tex]=5\cdot 10^{-6}\cdot+5\cdot 10^{-6}\cdot (1)^2=10\,\mu H[/tex]")
El condensador del circuito equivalente en alta frecuencia viene dada por:
![[tex]C=C_{1}\cdot+C_{2}\cdot (\frac{N_2}{N_1})^2+\frac{C_a}{3}\cdot (1+\frac{N_2}{N_1}+(\frac{N_2}{N_1})^2)=[/tex] [tex]=1.2\cdot 10^{-9}\cdot+1.2\cdot 10^{-9}\cdot (1)^2+\frac{0.1\cdot 10^{-9}}{3}\cdot (1+1+(1)^2)=2.5\cdot 10^{-9}\, F[/tex]](https://2.bp.blogspot.com/-8k5k9r-5TZM/XYehj7uB1RI/AAAAAAAAGgY/YeMU1OttFiogGWwMLttrtP2X_5lB8ZaiQCK4BGAYYCw/s640/formula12.JPG "[tex]C=C_{1}\cdot+C_{2}\cdot (\frac{N_2}{N_1})^2+\frac{C_a}{3}\cdot (1+\frac{N_2}{N_1}+(\frac{N_2}{N_1})^2)=[/tex] [tex]=1.2\cdot 10^{-9}\cdot+1.2\cdot 10^{-9}\cdot (1)^2+\frac{0.1\cdot 10^{-9}}{3}\cdot (1+1+(1)^2)=2.5\cdot 10^{-9}\, F[/tex]")
‹ Parte practica S.POTENCIA 2013 09 S1 (Tranformador de impulso)arribaParte practica S.POTENCIA 2013 09 S1 (Parametros circuito equivalente de baja frecuencia, Tranformador de impulso) ›
![[tex]C=C_{1}\cdot+C_{2}\cdot (\frac{N_2}{N_1})^2+\frac{C_a}{3}\cdot (1+\frac{N_2}{N_1}+(\frac{N_2}{N_1})^2)=[/tex] [tex]=1.2\cdot 10^{-9}\cdot+1.2\cdot 10^{-9}\cdot (1)^2+\frac{0.1\cdot 10^{-9}}{3}\cdot (1+1+(1)^2)=2.5\cdot 10^{-9}\, F[/tex]](http://2.bp.blogspot.com/-8k5k9r-5TZM/XYehj7uB1RI/AAAAAAAAGgY/YeMU1OttFiogGWwMLttrtP2X_5lB8ZaiQCK4BGAYYCw/s1600/formula12.JPG "[tex]C=C_{1}\cdot+C_{2}\cdot (\frac{N_2}{N_1})^2+\frac{C_a}{3}\cdot (1+\frac{N_2}{N_1}+(\frac{N_2}{N_1})^2)=[/tex] [tex]=1.2\cdot 10^{-9}\cdot+1.2\cdot 10^{-9}\cdot (1)^2+\frac{0.1\cdot 10^{-9}}{3}\cdot (1+1+(1)^2)=2.5\cdot 10^{-9}\, F[/tex]")
‹ Parte practica S.POTENCIA 2013 09 S1 (Tranformador de impulso)arribaParte practica S.POTENCIA 2013 09 S1 (Parametros circuito equivalente de baja frecuencia, Tranformador de impulso) ›
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