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://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj2exjR-rR-CCEl3AIkE-oNWa2g7t-7v2c5lDthIH-g5XQZr53HndrykfZUiRCpmTK-Uaw8m23rvSj4u2wToVakE7nwuUPHhKnt6tAOFjAw_dg3DcRTaZ_rHM7cwyLim56PRQ8j-vB3GSQ8/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]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj2exjR-rR-CCEl3AIkE-oNWa2g7t-7v2c5lDthIH-g5XQZr53HndrykfZUiRCpmTK-Uaw8m23rvSj4u2wToVakE7nwuUPHhKnt6tAOFjAw_dg3DcRTaZ_rHM7cwyLim56PRQ8j-vB3GSQ8/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]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjciVUzbP0L4xGeVjXEw3qOijoECKFsrwAf1zWW38RUPzNsqRHR8kGXw1e9Qq5kD5QBBwK6-uciLE26mwQhZnpklqpmFVLX5zp_VayxHl8TRO0t8BVuPks_be74b1-BLJQfAvaE1ql2eaiw/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]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiPSgw6CZsF6xwLdmx2sKRfHOUEkjT7_Jk3EWKGoL7RzdfmEEK1TexgTHAvLmcSohbhFq02SWuKqN4PZUITN85_WB3IrGSw7-xiHIiCFDboxFe4q8NOflrUh4chYLWb_ybmI5hmNMsKwdzX/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://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3WoHSIrFYtRTb9RMi4JjfCAo3DU3I9mT9419zevGx-i7wQ2viXdZjIx426gp8DBQlq3F_3yeCbhT4nK6gxNnr888Cg05C5Ci5Yeje4MHfSw2gM-50SyXjINXiMrvXgAER8Q0ec2XD66SJ/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]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3WoHSIrFYtRTb9RMi4JjfCAo3DU3I9mT9419zevGx-i7wQ2viXdZjIx426gp8DBQlq3F_3yeCbhT4nK6gxNnr888Cg05C5Ci5Yeje4MHfSw2gM-50SyXjINXiMrvXgAER8Q0ec2XD66SJ/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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