Question

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 1

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.

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