多路输出反激变压器如何漏感,请pi帮忙

应该是把所有输出的绕组都短接起来,因为任何一个输出绕组都可以传递通过原边耦合的励磁能量,而反激电源原边漏感的物理意义是无法通过励磁耦合到输出部分的电感,所以测原边漏感要把所有输出绕组都短路.

非常感谢您的解答,可不可以帮我做一个题目:

   To measure leakage inductance, short circuit tests are performed on the 3-winding transformer below. The results are as follows:
   measure A with B shorted: 1uH
   measure A with C shorted: 1uH
   measure B with C shorted: 1uH
   What will you measure at A if both B and C are shorted?
   Give your reasons?

图为:

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谢谢您的解答,我这里有答案,就是看不懂:

  L1+L2=1uH   (1)
  L1+L3=1uH   (2)
  L2+L3=1uH   (3)
由式(1)(2)(3)得
L2=L3  (4)
  L1=L2  (5)
由式(3)(4)得
  L2=L3=0.5uH (6)
由式(5)(6)得
  L1=L2=L3=0.5uH (7)
如果B和C短接,则
L2//L3＝0.25uH
进而A端的感值
L1+L2//L3=0.75uH

我主要想知道他为什么这算?

设1,2,3绕组漏感(单独)的分别是L1,L2,L3,绕组匝数分别是N1,N2,N3
则有:
L1+L2(N1*N1/N2/N2)=1UH
L1+L3(N1*N1/N3/N3)=1UH
L2+L3(N2*N2/N3/N3)=1UH
应该不能简单地那么列方程计算

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