excitation branch in the equivalent circuit for transformers

I am reading a textbook about transformers. According to the book, the excitation branch in the equivalent circuit of a transformer are two elements (Rc and Xm) in parallel. It says the total excitation admittance is Y = 1/Rc - (1/Xm)j. Any idea why the indutance is minus instead of postive?

Thanks in advance.

Erica said:

It says the total excitation admittance is Y = 1/Rc - (1/Xm)j. Any idea why the indutance is minus instead of postive?

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Because this is talking about admittance (Y) - rather than impedance (Z).

t_n_k said:

Because this is talking about admittance (Y) - rather than impedance (Z).

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Ahhh! That makes much more sense!
So, \(Z = R_{c} + jX_{m}\)

\(Y = \frac{1}{Z} = \frac{1}{R_{c}} + \frac{1}{jX_{m}} = \frac{1}{R_{c}} - \frac{j}{X_{m}} \)

No quite right ....

Zm=Rc||jXm

\(Z_m=\frac{R_c*jX_m}{R_c+jX_m}\)

and so on ...

t_n_k said:

No quite right ....

Zm=Rc||jXm

\(Z_m=\frac{R_c*jX_m}{R_c+jX_m}\)

and so on ...

Click to expand...

jegues said:

Ahhh! That makes much more sense!
So, \(Z = R_{c} + jX_{m}\)

\(Y = \frac{1}{Z} = \frac{1}{R_{c}} + \frac{1}{jX_{m}} = \frac{1}{R_{c}} - \frac{j}{X_{m}} \)

Click to expand...

You are right. I just prove that (1/Rc)-(1/Xm)j=1/Z. Thanks so much!