hnc question

Thread Starter

redshaw

Joined Jul 15, 2008
12
could anyone help

a single phase motor takes 50 A at a power factor of 0.6 lagging from a 240V, 50 Hz supply. Determin a.) the current taken by the capacitor in parrallel with the motor to correct the power factor to unity, and b.) the value of the supply current after power factor correction.

i worked out the Ic by

pf=cosO
0=cos -1 (0.6)
0=53.13

true power=Vcoc0
=240 cos 53.13
144W

Q=true power * tan 0
Q=144 * tan 53.13
Q=191.99 VA

Ic=Q/V
Ic=191.99/240
Ic=0.799A

but am struggling with the value of the supply current question b

i have calculated Xc as 300.38 ohms and C as 0.106uF but dont know where to go from here

can anyone help

cheers
 

Thread Starter

redshaw

Joined Jul 15, 2008
12
is the answer to b just simply

supply corrected to power factor of 0.6 = 144W/(240V*0.6)

=1A ?????????
 

Ratch

Joined Mar 20, 2007
1,068
redshaw,

could anyone help
I'll try.

The apparent power is 50*240 = 12000 volt-amps

The average power is 12000*.6 = 7200 watts

The reactive capacitor power is sqrt(12000^2-7200^2) = 9600 vars

The reactance of the capacitor is (240^2)/9600 = 6 ohms

The reactive current is sqrt(9600/6) = 40 amps

The current after correction is 7200/240 = 30 amps

So correcting the power factor reduced the current load of the power supply by 20 amps. Ask if you have any questions.

Ratch
 
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