AC electricity question?

Question

I have this electricity question in my exam:
"A manufacturing plant uses 2.1 MW of 220-V single phase power at a power factor of 0.8. Determine the apparent power and the effective, or rms, value of the current."
Getting the apparent power was simple, you just have to divide the true power by the power factor (which, in this case, gave you 2.625 MW), but I'm a bit confused as to the rms value.
To get the rms, you need the peak wave value (or the peak-to-peak value). I'm not sure how to get it. Could it be the amperage itself? Or maybe the voltage?

Yes, sorry about that. I put it in VA in as the answer. But, is the current the peak value? Or is it the peak-to-peak value? Or is it none of those?

All that they gave me was the wattage, voltage, and the power factor. Is the peak value of the voltage simply the voltage itself? Because that would seem logical.

Answer 1

Edit:
P(avg) = V(rms)A(rms)cosΦ = V(rms)A(rms)*p.f.

Dividing 2.625 MVA by 220 V will give rms current:
(2,625,000 VA) / (220 V(rms)) ≈ 11,900 A(rms)

BTW, no real utility would provide 2.625 MVA at 220 V1Φ, nor is there a 1Φ transformer that big. Also, cabling for that amperage is near impossible.

Answer 2

Given:
2.1 MW
220 V 1Ø AC (assumed to be an RMS value)
PF = 0.8 (assumed to be lagging)

Real power = 2.1 MW or 2100 kW
Apparent power = kW ÷ 0.80 = 2626 kVA

The values furnished appear to be RMS values. 220 V AC is a common RMS voltage value.

RMS current = 2626 kVA ÷ 0.220 kV = 11,932 amps RMS 1Ø

Answer 3

Your apparent power is in MVA not Mega watts. Current is then VA divided by volts (this will be in rms because the voltage should be stated rms)

Answer 4

You need the peak to peak value from voltage