# Transformer Question

Discussion in 'Homework Help' started by Zaraphrax, Oct 5, 2009.

1. ### Zaraphrax Thread Starter Active Member

Mar 21, 2009
47
3
Hi guys,

I have the following question and I'm a little confused.

Now, the no load current = 0.2 @ 70 degrees (0.2A magnitude, no load power factor angle is 70 degrees)

b) The primary current.

Secondary voltage = 2 * 240 = 480V

Total load impedence = Zl + Ze = 21 + J36.8 ohms

In polar form, this is 42.4 ohms @ 60.2 degrees.

Therefore, the secondary current (referred on the primary side) is:

I'2= 480 @ 0 degrees / 42.3 @ 60.2 degrees = 11.34 A @ -60.2 degrees

From here, I'm not sure which formula I should use. Should I use this:

I1 = I'2 + I'm

Or, I'2 = (N2/N1)I2 to find I2, then use the turns ratio to find I1.

(Hopefully I haven't completely stuffed this up)

Thanks.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Firstly, I think you should draw the transformer equivalent circuit to gain a better understanding of what you need to work out. How you draw the equivalent cct will depend to some extent on what you have been taught in class.

You should perhaps try to refer everything - including the secondary load impedance - back to the primary side.

Remember the open circuit test is intended to give you an estimate of the magnetizing impedance (Zm) which shunts the "virtual" primary winding. The short circuit test gives you the series impedance interposed between the actual primary input terminal and the "virtual" primary winding.

When you have all the values worked out you can then calculate the primary current. The impedance seen looking into the primary terminals is

Zp=Zs+Zm||Zs'

Where Zs=1+j0.8, Zm = the primary magnetizing impedance, Zs' = the secondary load impedance referred to the primary.