A single phase transformer operates from a 230V supply. The primary and secondary resistances are 0.03 Ω and 1.12Ω respectively while the corresponding leakage reactances are 0.1Ω and 6.4Ω; the magnetising branch can be neglected. The secondary winding has four times as many turns as the primary. Calculate :i) The equivalent impedance referred to the primary circuitii) The secondary terminal voltage when the load has a resistance of 200Ω and an inductive reactance of 100Ω.iii) The secondary terminal voltage when the load has a resistance of 200Ω and a capacitive reactance of 100Ω.Can someone show me how to tackle this question please?
i) the secondary transformer impedance is 1.12 + j6.4Ω. Since Ns/Np 4, this impedance reflected to the input is (1.12 +j6.4Ω)/(Np/Ns)? 007 + j0.4Ω ii) Reflect the primary transformer voltage and impedance to the secondary side. 230(Ns/Np) (1.12Ω + j6.4Ω + (Ns/Np)?(0.03Ω + j0.1Ω ) +200Ω + j100Ω )I 920 (1.6 +200 + j8 +j100)I (201.6 + j108)I |Z| 228.7Ω √(201.6? + 108?) at angle of inverse tan (108/201.6) 28.18° I 4.02 amps at -28.18° 3.54 - j1.90 amps The terminal voltage is 920 - (1.6Ω +j8Ω)(3.54 - j1.90 amps) iii) solve 920 (1.6Ω +200Ω + j8Ω - j100Ω)I (201.6Ω - j92Ω)I The terminal voltage is 920 - (1.6Ω + j8Ω)I
The is something wrong with your leakage reactances or the turn ratio, because N^2Lls/Llp. I'll neglect this issue. i) The total impedance of the transformer referred to the primary is Zt .03+j.1+(1.12+j6.)/16. ii) VloadVin*ZLoad/16/(Zt+ZLoad/16) ii) Same as above, except ZLoad now becomes the parallel combination of 200 and -j100.
