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

Answer:

$C = 46.891 \mu F$

Explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

$s = \frac{P}{P.F} < COS^{-1} 0.65$

$S = \frac{1250}{0.65} < 49.45$

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

$P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]$

solving for $Q_{new}$

$Q_{new} = P tan [cos^{-1} P.F new]$

$Q_{new} = 1250 [tan[cos^{-1}0.9]]$

$Q_{new} = 605.40 VARS$

$Q_C = Q - Q_{new}$

$Q_C = 1461 - 605.4 = 855.6 vars$

$Q_C = \frac[v_{rms}^2}{xc} =v_{rms}^2 \omega C$

$C = \frac{Q_C}{ v_{rms}^2 \omega}$

$C = \frac{855.6}{220^2 \times 2\pi \times 60}$

$C = 4..689 \times 10^{-5}$ Faraday

$C = 46.891 \mu F$