Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW at a 0.707 pf lagging. Determine:

(a) the pf of the second load,

(b) the parallel element required to correct the pf to 0.9 lagging for the two loads.

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V = 120V, 60 Hz

P = 2.4 kw, cos θ = 80

P = S sin θ - (p/cos θ) sin θ

= P tan θ (cos^-1 (0.8)

=2.4 tan (36.87) = 1.8KVAR

S = 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P = 1.5 kW, cos θ = 0.707 and θ=45 degree

Q = Ptan θ = tan 45°

Q=P=1.5kw

S1 = 1.5 + 1.5j KVA

S1 + S2 = S

2.4+j1.8 = 1.5+1.5j + S2

S2 = 0.9 + 0.3j KVA

S2 = 0.949 = 18.43 °

Pf = cos (18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ = 0.9

θ = 25.84 °

(WC) V^2 = P (tan θ1 - tan θ2)

C = 2400 (tan (36. 87°) - tan (25.84°)) / 2 πf * 120^2

f=60, π=22/7

C = 117.5µf