Show your work.
(okay, I admit, I'm just trying to trick you into showing me an alternate method to solving this problem.? I don't like the method used in the study guide I lifted it from, nor an alternate graphical method I saw used elsewhere and wonder if there's an easier way.)
Close? here goes...
draw the power triangle for the original situation...
arc cos 10/15, ~48.2 deg = pf 10/15 or .666
base = real power = 10W
hypot = complex power; real + reactive = 15KVA
rise = reactive power = sin 48.2 deg x 15 ~ 11.2 kvar
then superimpose the new 0.95 scenario over it...
angle =arc cos 0.95 ~ 18.2 deg
real power of 10 KW stays the same
complex equals 10kw/cos 18.2 = 10.5 kva
reactive = sine 18.2 x 10.5 ~ 3.3 kvar
subtract the reactive components, that's the vars you need to add in order to lower the pf: 11.2 - 3.3 = 7.9 kvar to be added
|Complex P| = V^2/Xc
Xc = 1/(2 Pi f C)
|Complex P| = 7.9 KVA
V = 440
f = 60
substitute and solve for C
C = 7900/(440^2 2 Pi 60) ~ 108 x 10^-6 or 108 uF