A 50 Hz single-phase transformer has a turns ratio of 6. The resistanc

1.	A 50 Hz single-phase transformer has a turns ratio of 6. The resistances and reactances are 0.90 Ω and 1.5 Ω for HV winding and 0.13 Ω and 0.1855 Ω for low voltage winding respectively.Calculate the voltage to be applied to the high voltage side to obtain full load current of 200 A in the low voltage winding on short circuit
Answer» Answer: Voltage applied to obtain full LOAD current is 330V . Explanation: Suppose, *PRIMARY side is High Voltage & secondary side is Low Voltage* ; Turn Ratio $(k)=\frac{N_{1}}{N_{2}}=6$ Resistance $(R_{eff1})=0.90+k^{2}(0.13)\\ =0.9+6^{2}0.13\\ =5.58$ Ω Reactance $(X_{eff1})=1.5+k^{2}(0.1855)\\ =1.5+6^{2}0.1855\\ =8.178$ Ω Impedence $(Z_{eff1})=\sqrt{5.58^{2}+8.178^{2} }\\=9.9$ Ω Formula: *Turn Ratio= ratio of current/voltage on both sides* $\frac{V_{1}}{V_{2} }=\frac{I_{1}}{I_{2}} =k$ Voltage on Low Voltage Side (I_{2}) (secondary side): $k=6=\frac{I_{2}}{I_{1}}\\ 6=\frac{200}{I_{1}} \\I_{1}=\frac{100}{3}A$ Voltage onHigh Voltage Side (primary side): formula: =I_{1} * (Z_{eff1}) $=(\frac{100}{3})9.9\\ =330V$ To obtain the full load current* of 200A on LV WIRING on short circuit , 330 voltage should be applied on the HV side. Hopw u find it helpful :)