Power factor and frequency change

AI Thread Summary
The discussion focuses on calculating the power factor and supply voltage for a 240V, 50 Hz appliance rated at 2KW with a lagging power factor of 0.7 when operated at 60 Hz. The power factor at 60 Hz is determined to be 0.63 after calculating the impedance at the new frequency. Participants discuss the relationship between power (P), voltage (V), and impedance (Z) to find the necessary supply voltage to maintain the rated power. The confusion arises around the formulas used, particularly whether to apply P = V^2/R or another equation to find the supply voltage. Ultimately, the key calculations involve using the rated power and the new impedance to derive the required supply voltage.
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Homework Statement


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?


Homework 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



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.
 
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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
 
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.
 
should i be using P=V^2/Z
 
Note V^2/Z gives the apparent power S.
 
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.
 
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.
 
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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