what value of resistance should be placed.in parallel with 50microfarad capacitor in order to have a total power factor of 0.8 on a 6p-cycle ac system?
3 answers
It should be 60-cycle
power factor= resistive power/apparent power
resistive power (parallel ckt)=V^2/R
apparent power= V^2/sqrt(R^2+(Xc)^2)
.8= sqrt(R^2+Xc^2)/R= sqrt(1-(2PI*f*C/R)^2) where Xc= j*2PI*fC
squaraing both sides...
.64=1-4PI^2*60^2*(50e-6)^2/R^2)
and from then, just solve for R.
resistive power (parallel ckt)=V^2/R
apparent power= V^2/sqrt(R^2+(Xc)^2)
.8= sqrt(R^2+Xc^2)/R= sqrt(1-(2PI*f*C/R)^2) where Xc= j*2PI*fC
squaraing both sides...
.64=1-4PI^2*60^2*(50e-6)^2/R^2)
and from then, just solve for R.
Xc = -j1/WC = -j1/377*50*10^-6 = -j53.1 Ohms. = 53.1 Ohms[-90o].
Pf = Cos A = 0.8, A = 36.9o.
Tan36.9 = R/Xc = R/53.1,
R = 39.9 Ohms.
Check: The impedance (Z) should have an angle of (-36.9o).
Z = R*Xc/(R - jXc) = 39.9*(53.1[-90o]/(39.9 - j53.1) = 2117[-90o] / 66.4[-53.1o] = 31.9 Ohms[-36.9o].
Pf = Cos A = 0.8, A = 36.9o.
Tan36.9 = R/Xc = R/53.1,
R = 39.9 Ohms.
Check: The impedance (Z) should have an angle of (-36.9o).
Z = R*Xc/(R - jXc) = 39.9*(53.1[-90o]/(39.9 - j53.1) = 2117[-90o] / 66.4[-53.1o] = 31.9 Ohms[-36.9o].
1 answer