# Find V2 in Single Phase Transformer

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1. Dec 7, 2016

### jsammut

1. The problem statement, all variables and given/known data
Find V2. Everything else is available in the picture

2. Relevant equations
r1=r1+a^2*r2
x1=x1+a^2*r2
z2=a^2*r2
I1=vp/z2+r1+x1

3. The attempt at a solution
So I have used the first three formulas but it's the last one I am getting stuck on.
Since my x1 is an imaginary number how do I do this?
I'm currently stuck at I1=1200/192+4+j16
Can someone help me?

EDIT: VP=1200 NOT 1000 LIKE THE PICTURE STATES. EVERYTHING ELSE IS THE SAME

2. Dec 7, 2016

### cnh1995

The impedance 192+4+j16 can be written as 196+j16.
Convert this into polar form. For multiplication and division of two phasors, you need their polar forms.

3. Dec 8, 2016

### jaus tail

You can also solve as:
I1 = 1200/(196 + j6). Multiply numerator and denominator by 196 - j6.
This will remove 'j' from denominator and then you can proceed.
j * j = -1.

4. Dec 8, 2016

### jsammut

Okay I got.
But then wouldn't I still have "j" in my numerator?

5. Dec 8, 2016

### cnh1995

Yes. If you want to keep the answer in complex form, you should keep it in the form
Ireal+jIreactive. Or you can simply convert it into polar form.

6. Dec 8, 2016

### jsammut

Okay. It's my first time working with Complex numbers. Just to confirm.
In order to do 1200/196+j16 I have to convert the numerator into polar form first. Reason for this is because 1200 is basically 1200∠0 which is polar form. You can not divide polar by rectangular.

Can you divide rectangular by polar?

7. Dec 8, 2016

Right.
Yes and no.

8. Dec 8, 2016

### jsammut

T
Thank you, been a big help?

9. Dec 8, 2016

### jaus tail

It's easier to convert into polar form when dividing or multiplying, and to convert to rectangular form when adding or subtracting. But if the denominator has simple terms like 1 + i or 1 - i then multiplying num and den by 1- i or 1 + i respectively is quicker than using calculator.