Question

The motor inside a blender can be modeled as a resistance in series with an inductance, as shown in the given figure. The wall socket source is modeled as an ideal 120 Vrms voltage source in series with a 2-ohm output resistance. Assume the source frequency is w = 377 rad/s. Given: R = 10.5 ohm 202 w w 120 V 20 m Wall socket Blender motor References eBook & Resources Section Break Problem 07021 10. Required information value 5.00 points You did not receive full credit for this question in a previous attempt Problem 07.021.d What value of capacitor when placed in parallel with the motor will change the power factor seen by the voltage source to 0.9 lagging? (Round the final answer to two decimal places) The value of the capacitor that will change the power factor to 09 lagging is 235 F

Answer 1

2 10.5 EXC jx 3 IZ -3 XL = WL = 377 X 20x10 = 7.54 Xc = toc 3770 Z=2+ Z=2+ (10:5+9754) |-jxc (10.5+j7:54) (x2) (10 +39.54) + six 10+ $7.54 110Xcape 7.54*c+I c ( z = 2+ jioxc Z = z = jQoXc -1508*c + a +10+j7.54 (1-7.54x) +j Voxc (12 - 15.08Xc) + 3 (7.54 +2010) (1-754Xc) + j 10Xc [(12 - 15-08x) +j 47-su +20x]{1_7544 (-7-544)-110x] (1-7.5uxc ) + (lovc) 3 2- CS Scanned with CamScanner

z = (12 - 19.9442) + j (7.54 –156.84%C*— 68x2) ((Asuxe) ²+ (108) @so) Re {z} Izl -0.9 Loso 12-19.90x² (12-19.943.2)+(7:54-156-8486 46.5*3)2 2 2 (12-19.9442) = 0.81) >$){(12-19.4ur.)+(9.54–156.54–6842)? 0.19(12-19.94x2) -19.9482) = 0.81 (7-54 - 15684 – 6.8XC) Applying square root on both sides. 0.43 (12-19.94x2) 0.9 (7.54-15.848c 6.542) 1-7942xŹ - 14.256XC +1:626 = 0 Xc 0 7.916 abone olan 2.916 COC cs 377 x 7.96 Bir 3.3508x104 335.08 ef CS Scanned wila CamScal-ne-