Question

A 1,250 W electric motor is connected to a 220 Vrms, 60 Hz source. The power factor is lagging by 0.65. To correct the pf to 0.9 lagging, a capacitor is placed in parallel with the motor. Determine the value of the capacitor required to make the corretion in LaTeX: \mu Fμ F Note: if the calculated value of C is 0.000056, you should enter 56 in the answer box, as the canvas is asking for answer in microfarads

Answer 1

`C = 46.891 \mu F`

Step-by-step explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

`s = (P)/(P.F) < COS^(-1) 0.65`

`S = (1250)/(0.65) < 49.45`

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

`P.F new = cos [tan^(-1) (Q_(new))/(P)]`

solving for
`Q_(new)`

`Q_(new) = P tan [cos^(-1) P.F new]`

`Q_(new) = 1250 [tan[cos^(-1)0.9]]`

`Q_(new) = 605.40 VARS`

`Q_C = Q - Q_(new)`

`Q_C = 1461 - 605.4 = 855.6 vars`

`Q_C = \frac[v_(rms)^2}{xc} =v_(rms)^2 \omega C`

`C = (Q_C)/( v_(rms)^2 \omega)`

`C = (855.6)/(220^2 * 2\pi * 60)`

`C = 4..689 * 10^(-5)` Faraday

`C = 46.891 \mu F`