# Calculating Transformer Input Voltage

1. Apr 28, 2015

### MechEngJordan

1. The problem statement, all variables and given/known data

Calculate the primary current, and hence the voltage at the transformer input winding, V1.

Transformation ratio, a = N1/N2 = 0.1
R1 = 0.12 Ω; R2 = 12 Ω
X1 = 0.4 Ω ; X2 = 40 Ω
RC = 560 Ω
V2 = 2300 V
RL = 1 kΩ
'Xm = 800 Ω'

2. Relevant equations

R1eq = R1 + a2R2
X1eq = j(X1 + a2X2)
V2' = aV2

Current divider equation.

3. The attempt at a solution

Approximate equivalent circuit:

R1eq = 0.12 + (0.1)2(12)
R1eq = 0.24 Ω

X1eq = j(0.4 + (0.1)2(40))
X1eq =j0.8 Ω

V2' = 0.1(2300)
V2' = 230 V

a2ZL = (0.1)2(1000)
a2ZL = 10 Ω

Ip = V2'/a2ZL
Ip = 23 A

From here is where I believed that there is likely a more efficient way to solve the problem -- particularly because the value of Xm was not actually given in the paper, but told to us during the tutorial, more-or-less made up on the spot.

Here is the outline of the given solution:

Ip= I1*Z2/(Z1+Z2)

⇒I1 = Ip*(Z1+Z2)/Z2

⇒I1 = 23*(Rc // Xm + X1eq + R1eq + a2ZL)/ (Rc // Xm)

I0 = I1 - Ip

∴ V1 = I0(Rc // Xm)

I'd be thankful for any input.

2. Apr 28, 2015

### Hesch

It seems that in the given solution another equivalent is used (T-equivalent) than the one in #1. It's hard to explain what happens in that other equivalent because I cannot see what is meant by Z1 and Z2.

Don't you like to calculate with complex numbers? I don't know if you find complex calculations "more efficient", but they will be more structured. Example:

Z0 = (Rc || jXm) = (560 || j800) = (375.8 + j263). That's it.

In the same way you calculate other impedances, that can be used in Kirchhoffs laws, etc.

It becomes simple as in the given solution: Ip= I1*Z2/(Z1+Z2). As for the numerical calculation I assume a calculator can do the job for you.

3. Apr 28, 2015

### MechEngJordan

Hi,

I should have probably clarified that in this case,

Z1 = Rc // Xm

and

Z2 = X1eq + R1eq + a2ZL

I have no issues with complex numbers or the numerical computations; I suppose what I was referring to by 'efficient' was that in the original question, there was no value of Xm given (this was only given during the tutorial and seemed to be done off the cuff). This gave me the impression that the solution to this problem should have been different than the one given. Also, this was a 5-mark question and relative to other 5 mark questions, the given approach seemed like a lot of work.