Question

A motor attached to a 120 V/60 Hz power line draws an 8.40 A current. Its average energy dissipation is 850 W. How much series capacitance needs to be added to increase the power factor to 1.0?

Answer 1

power factor is .843. rms resisttor voltage is 101 V and resistance is 12 ohms. how do i do the last part?

Answer 2

@radar

Answer 3

Did you work out the power factor, or was it provided as part of the problem?

Answer 4

i worked it out. 850/(120*8.4)

Answer 5

The apparent power (VARS) is 1,008 watts and the real power is kind of given as "850 Watts of average energy dissapation" we probably can assume that it power that is actually consumed. Sure enough 850/1008= 0.8433. That is probably good. Now that value of 12 Ohms, was that given or did you work that out and is that suppose to be the inductive reactance offered by the motor?

Answer 6

i worked it out. the only thing i didn't work out was the main part of the problem.

Answer 7

120^2/(tan(arccos.084)*850)

Answer 8

typo. .084 was .843. sorry

Answer 9

What is the 12 Ohms represent?
I have calculated a much lower value needed for the reactance needed for the capacitor.. You see the value of the capacitor reactance will equal the value of the inductive reactance to adjust the power factor to equal 1.

Answer 10

What is the motor's resistance?
12.0
Ω
SubmitMy AnswersGive Up
Correct

Answer 11

ok, open study is having issues and keeps kicking me off.....

Answer 12

We need to construct an impedance triangle. First there is the resistance of the motor which is 850/(8.4)^2 = 12.046 Ohm.
Second there is the total impedance (the hypotenuse) of 120/8.4 = 14.288 Ohms
Now we must find the Inductance reactance: