science question?

Question

A 120v appliance draws 30A of current. If electrcity cost 0.065/kW*h, how much does it cost to operate the appliance for 3.5 hours?

Answer 1

Power = Current x Voltage
P = 20(120)
P = 2400 Watts
P = 2.4 kW

Cost = 2.4(3.5)(0.065)
Cost = $0.55

Answer 2

Since the utility company's "kw" meter doesn't measure true KW, but "amps", thus KVA (no power factor considerations), your answer becomes a simple:

{ [(E x I)/1000] kw x 3.5 h } x .065 $/kw-h

or

{[(120 x 30) / 1000] x 3.5} x .065 = $0.82

the units are figured as:

E (volts) x I (amps) = watts
watts/1000 = kw

kw-h x $
------------ = $
kw-h

If the device measured "true" power, or kw, you'd have to know the Power Factor of your applicance, thus its efficiency (or lack thereof) and the formula to calculate true kw would be:

E x I x pF = kw

where pF is the cosine of the phase angle, which happens to be the effective efficiency, with a pF of "1" being 100 percent, or meaning that KVA = KW. Most inductive loads have a pF that varies between .6 and .85

Answer 3

consider power factor if it's an ac
formula for power is P= EI cos(theta)
where cos(theta)=power factor=80% mostly