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damaskus [11]
3 years ago
9

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
Physics
1 answer:
Black_prince [1.1K]3 years ago
6 0

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

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