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## Homework Statement

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

_{1}.

Transformation ratio, a = N1/N2 = 0.1

R

_{1}= 0.12 Ω; R

_{2}= 12 Ω

X

_{1}= 0.4 Ω ; X

_{2}= 40 Ω

R

_{C}= 560 Ω

V

_{2}= 2300 V

R

_{L}= 1 kΩ

*'X*

_{m}= 800 Ω'## Homework Equations

R

_{1eq}= R

_{1}+ a

^{2}R

_{2}

X

_{1eq}= j(X

_{1}+ a

^{2}X

_{2})

V

_{2}' = aV

_{2}

Current divider equation.

## The Attempt at a Solution

[/B]

Approximate equivalent circuit:

R

_{1eq}= 0.12 + (0.1)

^{2}(12)

R

_{1eq}= 0.24 Ω

X

_{1eq}= j(0.4 + (0.1)

^{2}(40))

X

_{1eq}=j0.8 Ω

V

_{2}' = 0.1(2300)

V

_{2}' = 230 V

a

^{2}Z

_{L}= (0.1)

^{2}(1000)

a

^{2}Z

_{L}= 10 Ω

I

_{p}= V

_{2}'/a

^{2}Z

_{L}

I

_{p}= 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 X

_{m}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:

I

_{p}= I

_{1}*Z

_{2}/(Z

_{1}+Z

_{2})

⇒I

_{1}= I

_{p}*(Z

_{1}+Z

_{2})/Z

_{2}

⇒I

_{1}= 23*(R

_{c}// X

_{m}+ X

_{1eq}+ R

_{1eq}+ a

^{2}Z

_{L})/ (R

_{c}// X

_{m})

I

_{0}= I

_{1}- I

_{p}

∴ V

_{1}= I

_{0}(R

_{c}// X

_{m})

I'd be thankful for any input.