Electronics question 10 point

So It says that I have to solve for the following transformer for all the unknowns:

Vp=240V
Ns=10,000
Efficiency=90%
Pp=400W
ls=30A
Np=
Vs=
Vs actual=
Ps=
Ps actual=
lp=
Turns ratio=
impedance ratio=

Thanks, Best answer gets 10 points!! :)

The primary input power is Pp =400 watt
Since the efficiency is 90%, the secondary power output is 400 watt*90% = 360 watt
The input current Ip = 400 watts/240 volt = 1.666 amps
Since the output current is 30 amps, the ideal secondary voltage Vs= 400 watts/30 amps = 13.333 volts
Turns ratio = Np/Ns = Vp/Vs = 240/13.333= 18
Np = 18*10,000 = 180,000 turns
Impedance ratio = 18² = 324
Vs actual = 360 watts/30 amps = 12 volts
Ps = 400 watts assuming a perfectly loss less transformer
Ps actual = 400 watts*90% = 360 watts

It would help if you defined your symbols.

If Pp is primary power, then secondary power is 0.9x400 = 360 watts
primary current is 400/240

Is, if that is secondary current, then secondary voltage is 360/30 = 12 volts
turns ratio is 240/12 = 20:1
impedance ratio is √20

that should answer them all.

I'm going to assume the magnetizing current is zero and the transformer can be represnted by a resistor (for all the losses) in series with an ideal transformer.
Determine Ip=Pp/Vp=400/240=1.667 amps
Rp=24/1.667=14.4 ohms. Note Ip^2*Rp=40 watts which is the loss.
Ps=440-40=400 watts
Vs_actual=240-24=216 volts
Vs=Ps/Is=13.333 volts
Turns ratio=Vp_actual/Vs=16.2
Np=16200 (This is huge number of turns.)
Impedance ratio=16.2^2=262.44
In my model Vs=Vs_actual, but you could split the losses between the primary and secondary and add a seconday series resistance. You don't need to do this for an accurate model.