The result is as expected for the given transformer parameters. You provided a very low frequency for the transformer; instead of 50 kHz, 50 Hz was given. As a result, the leakage inductance became too high for a 50 kHz input, causing most of the voltage to drop across the secondary leakage inductance.
The R and L values of the transformer are given in per-unit. For the secondary winding:
- Rbase2 = V2^2/P = 0.2916 ohm
- ,Lbase2=Rbase2/(2*pi*f)=> 0.2916/(2*pi*50)=9.2819e-04
- The actual secondary leakage inductance L2 = 0.08*Lbase=74e-6
This value is too large for a 50 kHz resonant input, so most of the voltage drops across the secondary leakage inductance. The transformer output terminal voltage is given by: Vo = Vs-w*L2*Isec-R2*Isec, due to high L2, w*L2*Isec is also very high, (because the base frequency given in the transformer input parameter is 50Hz), For a 10 A secondary current, the voltage drop across the secondary leakage inductance is: 2*pi*50e3*74*1e-6*10A = 232V.
Since 50 kHz is the resonant frequency, please enter 50,000 (50e3) in the transformer parameters for nominal power and frequency.
