An ideal transformer is rated to deliver 460 KVA at 380 V to

An ideal transformer is rated to deliver 460 KVA at 380 V to a customer. How much current can the transformer supply to the customer? If the customer\'s load is purely resistive (power factor, cosO = 1), what is the maximum power that the customer can receive? If the customer\'s power factor is 0.8 (lagging) what is the maximum usable power that the customer can receive? What is the maximum power if the power factor (pf) is 0.7 (lagging)? If the customer requires 300kW to operate, what is the minimum power factor with the given size transformer?

Solution

3 ans) Ideal Transormer No losses S=460 KVA and V=380V

a) Current supplied by transformer to Customer I=S/V=460 KVA/380=1210.5 A

so I=1210.5 Amperes

b)Maximum Power that customer can recieve= S*powerfactor as powerfactor=1

P=460*1=460 kW

c)powerfactor=0.8

P=460*0.8=368kW

d)powrfactor=0.7

P=460*0.7=322 kW

e)

P=300 kW given

powerfactor=P/S=300/460=0.6522