Single phase transformer question

[TheRedDevil18]

Homework Statement

A 100 kVA, 2300/230V, single phase transformer has the following parameters: Rp = 0.30Ω, Rs = 0.0030Ω, Rc = 4.5 kΩ, Xp = 0.65Ω, Xs = 0.0065Ω, Xm = 1.0 kΩ The transformer delivers 75 kW at 230 V at 0.85 power factor lagging, find:
a) The input current.
b) The input voltage.

Homework Equations

The Attempt at a Solution

[/B]
Referred to primary

I'm confused with the part where the transformer delivers 75kW. Then will the secondary current be 0.75*Israted ?

 

[NascentOxygen]

I'm confused with the part where the transformer delivers 75kW. Then will the secondary current be 0.75*Israted ?

You know the power and the PF so you can determine the secondary's VA output.

 

[TheRedDevil18]

You know the power and the PF so you can determine the secondary's VA output.

So,
cos(∅) = P/S
S = P/cos(∅)
= 75/0.85
= 88.24 kVA

So would the secondary current be
Is = 88.24/230 = 383.65A ?

 

[NascentOxygen]

So,
cos(∅) = P/S
S = P/cos(∅)
= 75/0.85
= 88.24 kVA

So would the secondary current be
Is = 88.24/230 = 383.65A ?

That should be right.

 

[TheRedDevil18]

Ok, so to find the input current, Ip, would this be right ?

From my schematic:

Ip = Io + Is/a.....1

Io = 2300(1/4.5k + 1/1k)

Is/a = 383.65∠-31.79 / 10
= 38.365∠-31.79

Therefore,

Ip = 2300(1/4.5k + 1/1kj) + 38.365∠-31.79
= 40.05∠-34.2

 

[NascentOxygen]

Ok, so to find the input current, Ip, would this be right ?

From my schematic:

Ip = Io + Is/a.....1

Io = 2300(1/4.5k + 1/1k)

Is/a = 383.65∠-31.79 / 10
= 38.365∠-31.79

Therefore,

Ip = 2300(1/4.5k + 1/1kj) + 38.365∠-31.79
= 40.05∠-34.2 ✔[/color][/size]

That looks right.

 

[TheRedDevil18]

That looks right.

Thanks