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Using the nomogram of FIGURE 10, calculate the size of

capacitor bank to give a final power factor of 0.95 for a single phase,

230 V, 50 Hz lighting installation which has a load current of 17.5 A and

operates at 0.8 pf.

The nonogram gives a value k = 0.44

Relevant equations

Qc = Pk

Xc = 1/2pifC

P = VI

The attempt at a solution

I have come up with two ways but i think the first is wrong as the current though the capasitor (If in parrallel) will not be the same as the load current.

Attempt 1

P = 230 x 17.5 = 4025W

Qc = 4.03 x 0.44 = 1.77 KVAr

P = I^2 x Xc

Xc = P/I^2 = 1.77 x 1000 x 17.5^2 = 5.783Ω

so C would come out at 550μF (as I said I think this is wrong)

Attempt 2

Power still = 4025W and Qc = 1.77 KVAr

but P = V^2/Xc so Xc = V^2/P

Xc = 230^2 x 1.77 x 1000 = 18.38Ω

so C would work out at 173μF

Just a bit of guidance would be appreciated.