Power Factor Question

[ElectroChem]

OK, you've had a go so I can in good conscience step in and help, always a good idea when asking for help in here to show us what you've attempted first.

The power dissipated in a circuit or part thereof is equal to:

Volts x Amps x Power Factor

Which in this case is

230 V x 10 A x 0.8 = 1.84kW

You're answer of 12.5A would be correct if the question was "What current will a 2300W, 230V machine draw if its Power Factor is 0.8?"

NB: This applies to single phase circuits, once you reach three phase power you'll learn a slightly different formula.

 

[shnabz]

OK, you've had a go so I can in good conscience step in and help, always a good idea when asking for help in here to show us what you've attempted first.

The power dissipated in a circuit or part thereof is equal to:

Volts x Amps x Power Factor

Which in this case is

230 V x 10 A x 0.8 = 1.84kW

You're answer of 12.5A would be correct if the question was "What current will a 2300W, 230V machine draw if its Power Factor is 0.8?"

NB: This applies to single phase circuits, once you reach three phase power you'll learn a slightly different formula.

I don't understand why though, because I thought that a power-factor of anything less that 1 meant that the circuit was inefficient. So would the machine not dissipate more energy? In theory, the motor would draw 2.3KW of efficient power if PF was at unity.

 

[shnabz]

Oh I get it now, it's the way I read the question.
"A 230V, single phase machine, draws a current of 10A"
I thought this meant that it WAS a 2.3KW machine. But it means that all together, including the 'reactive' power, it draws 10A.

Thanks for clarifying.

 

[happysteve]

Good answer and explanation, ElectroChem! Power dissipated is due to the resistive part of the load, and this part is the total current consumed multiplied by the power factor, as you say.

shnabz, do you understand why a power factor less than 1 is considered "a bad thing"? (If your answer is "no" then we can help with your understanding, but it might be that you've "got it" and don't need any more help ).

 

[shnabz]

Good answer and explanation, ElectroChem! Power dissipated is due to the resistive part of the load, and this part is the total current consumed multiplied by the power factor, as you say.

shnabz, do you understand why a power factor less than 1 is considered "a bad thing"? (If your answer is "no" then we can help with your understanding, but it might be that you've "got it" and don't need any more help ).

Yes I understand the principles behind it all, with inductive loads, the current ends up lagging the voltage, which means the circuit is out of phase. This creates reactance which is a waste of energy.

I Just didn't comprehend the question properly.

Thanks for offering additional help.

 

[Simonslimline]

Has anyone been shown the fluorescent light example at college, with the PF correction cap being switched in and out of the circuit?

What will happen when it is switched out of circuit therefore giving it a power factor lower than unity?

Sorry just realised what i have wrote is not relevant to the question anyway. :90::seeya:

 

[happysteve]

Sorry just realised what i have wrote is not relevant to the question anyway. :90::seeya:

Relevant or not, reminded me I had this photo on my phone. Our tutor, Alan, scored this beautiful power factor meter off ebay for 99p. Used it in the fluorescent / power factor practical.