SINGLE PHASE INDUCTION MOTOR

Somebody correct my math, but I get PF=.757

XL= 2iifL = 9.3

Z = 14.25

I = 8.42 amps

theta = 40.73

cos theta = .757

Good luck,

James

You should know better than to temp the gods by saying "Someone correct my maths.."

2iifL = 2 * ii * 60 * 25.3 *10^-3 = 9.54 ohm

giving

PF = .75 ffej

cosΦ=pf= r/sqrroot(r²⁺w2^{l2)=10.3/√(10.3)2+(25.3*.001)2(60)2}