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Power factor and frequency change 
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#1
Nov2510, 10:39 PM

P: 9

1. The problem statement, all variables and given/known data
240V,50 Hz electrical appliance is rated at 2KW and has a lagging power factor of0.7. a)Determine the appliance's power factor when it is used on a 60Hz supply? b)Calculate the supply voltage required to maintain the appliance at its rated power when operated off a 60Hz supply? 2. Relevant equations p.f.=P/S S^2=P^2+Q^2 to obtain Q S=VsI to obtain I Z=Vs/I to obtain Z p.f. = R/Z to obtain R Z^2=R^2+Xl^2 to obtain Xl Xl=2pifL to obtain L at 50Hz 3. The attempt at a solution I have done the above then used the value for L in Xl=2pifL at 60 Hz but not sure were to go from here. 


#2
Nov2610, 02:31 AM

P: 241

As you have already found Xl at 60 Hz you can find power factor at 60 Hz from value of new Z and R (Note R is unchanged)
Since you know Z and P (rated power) you can find V 


#3
Nov2610, 03:28 AM

P: 9

Thanks, I get the first section now but I don't see how I can work out the supply voltage using the new Z and the rated power.



#4
Nov2610, 03:32 AM

P: 9

Power factor and frequency change
should i be using P=V^2/Z



#5
Nov2610, 03:54 AM

P: 241

Note V^2/Z gives the apparent power S.



#6
Nov2610, 03:59 AM

P: 9

I'm confused now, how do I calculate the supply voltage required to maintain the appliance at its rated power at 60Hz.
I have the rated power and the impedence at 60Hz. 


#7
Nov2610, 04:57 AM

P: 241

Hi, you know P and Z, from Z you can find p.f.
From above values you can find S. Once you know S you can find V. 


#8
Nov2610, 05:24 AM

P: 9

I have worked out the pf to be 0.63
P(rated power) = 2000 w Z=22.31 ohms Would it be correct to use P(rated power)= V^2/R as R remains the same to obtain V(supply voltage) 


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