Help on AC Circuits: Power Factor = 1.0

AI Thread Summary
The discussion revolves around calculating the series capacitance needed to achieve a power factor of 1.0 for a motor connected to a 120 V/60 Hz power line drawing 8.50 A with an average energy dissipation of 800 W. The user has successfully determined the power factor to be 0.784, the RMS resistor voltage as 94.1 V, and the motor's resistance as 11.1 ohms. To increase the power factor to 1.0, it's necessary to add capacitance that cancels out the inductive component of the impedance. The approach involves calculating the inductive impedance from the voltage-to-current ratio and then determining the required capacitive impedance at 60 Hz. The discussion emphasizes the importance of ensuring that both voltage and current are RMS values to avoid inaccuracies in the power factor calculation.
RagincajunLA
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Hey guys I just need some help on something

The question is: A motor attached to a 120 V/60 Hz power line draws an 8.50 A current. Its average energy dissipation is 800 W.

I got the first 3 parts of this question correct...

1. What is the power factor
ans = .784

2. What is the rms resistor voltage?
ans = 94.1 V

3. What is the motor's resistance?
ans = 11.1 ohms

now all of these answers are right but i can't seem to get the last one and it is...
4. How much series capacitance needs to be added to increase the power factor to 1.0?
 
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I'm assuming 120V and 8.5A are also RMS. It's been a while, but it's the only thing that makes sense with 800W. If either one is peak-to-peak, that'd give you power factor that's greater than 1.0 and that's nonsense. Anyways...

You know the resistance. You know the total magnitude of the impedance, because you have voltage to current ratio. So you can compute inductive component of the impedance. Now all you have to do is find the capacitance such that capacitive impedance at 60Hz cancels your inductive impedance. Then the net impedance of the motor is just the resistance, and your power factor becomes 1.0.
 
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