An ideal transformer is rated to deliver 460 kVA at 380 Vrms to a customer. How

Question

An ideal transformer is rated to deliver 460 kVA at 380 Vrms to a customer. How much current can the transformer supply to the customer? If the customer's load is purely resistive (i.e.. pf= 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 the customer can receive? What is the maximum power if the power factor is 0.7? If the customer requires 300 kW to operate, what is the minimum power factor with the given size transformer?

Explanation / Answer

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P = 460 kVA and Vrms = 380 V

a) Current supplied to customer I = P/V = 1210.53 A

b) Maximum power recieved = 460 KVA = 460 KW.

c) If power factor = 0.8 then maximum usable power =

P(usable) = (460KW)(power factor) = (460KW)(0.8) = 368 KW.

d) If power factor = 0.7 then maximum usable power =

P(usable) = (460KW)(power factor) = (460KW)(0.7) = 322 KW.

e) The minimum power factor is p.f (min) = 300/460 = 0.652 .

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