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A single phase motor connected to 400V, 50Hz supply takes 31.7 A at a power factor of 0.7 lagging. Calculate the capacitance (in μF) required in parallel with the motor to raise the power factor to 0.9 lagging.
Correct answer is between '93,97'. Can you explain this answer?
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A single phase motor connected to 400V, 50Hz supply takes 31.7 A at a ...
Active component of IM = IM cosϕM = 31.7 × 0.7 = 22.19 A
Active component of I = I cosϕ = I × 0.9 = 0.9I A
These components are represented by OA
0.9I = 22.19
Reactive component of IM = IM sinϕM =
Reactive component of I = I sinϕ =
Now, IC = Reactive component of IM – Reactive component of I
Reactive component of IM = IM sinϕM =
Reactive component of I = I sinϕ =
Now, IC = Reactive component of IM – Reactive component of I
= 22.63 – 10.74 = 11.88 A
⇒ 11.88 = 400 × 2π × 50 × c
⇒ c = 94.58 μF
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A single phase motor connected to 400V, 50Hz supply takes 31.7 A at a ...
Microfarads) required to improve the power factor to 0.9 leading.
First, we need to calculate the apparent power (S), active power (P), and reactive power (Q) of the motor:
S = V * I = 400V * 31.7A = 12,668 VA
P = S * cos(θ) = 12,668 VA * 0.7 = 8,867.6 W
Q = S * sin(θ) = 12,668 VA * sin(arccos(0.7)) = 8,087.6 VAR
To improve the power factor to 0.9 leading, we need to reduce the reactive power to -8,867.6 VAR (since the motor was originally operating at a power factor of 0.7 lagging). We can do this by adding a capacitor in parallel with the motor.
The formula for capacitance (C) required to adjust the reactive power is:
C = |Q| / (ω * V^2)
where ω is the angular frequency (2πf) and V is the supply voltage.
Plugging in the values, we get:
C = 8,867.6 VAR / (2π*50Hz * (400V)^2) = 1.10 μF
Therefore, a capacitance of 1.10 microfarads is required to improve the power factor of the motor to 0.9 leading.
First, we need to calculate the apparent power (S), active power (P), and reactive power (Q) of the motor:
S = V * I = 400V * 31.7A = 12,668 VA
P = S * cos(θ) = 12,668 VA * 0.7 = 8,867.6 W
Q = S * sin(θ) = 12,668 VA * sin(arccos(0.7)) = 8,087.6 VAR
To improve the power factor to 0.9 leading, we need to reduce the reactive power to -8,867.6 VAR (since the motor was originally operating at a power factor of 0.7 lagging). We can do this by adding a capacitor in parallel with the motor.
The formula for capacitance (C) required to adjust the reactive power is:
C = |Q| / (ω * V^2)
where ω is the angular frequency (2πf) and V is the supply voltage.
Plugging in the values, we get:
C = 8,867.6 VAR / (2π*50Hz * (400V)^2) = 1.10 μF
Therefore, a capacitance of 1.10 microfarads is required to improve the power factor of the motor to 0.9 leading.
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A single phase motor connected to 400V, 50Hz supply takes 31.7 A at a power factor of 0.7 lagging. Calculate the capacitance (in μF) required in parallel with the motor to raise the power factor to 0.9 lagging.Correct answer is between '93,97'. Can you explain this answer?
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A single phase motor connected to 400V, 50Hz supply takes 31.7 A at a power factor of 0.7 lagging. Calculate the capacitance (in μF) required in parallel with the motor to raise the power factor to 0.9 lagging.Correct answer is between '93,97'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A single phase motor connected to 400V, 50Hz supply takes 31.7 A at a power factor of 0.7 lagging. Calculate the capacitance (in μF) required in parallel with the motor to raise the power factor to 0.9 lagging.Correct answer is between '93,97'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single phase motor connected to 400V, 50Hz supply takes 31.7 A at a power factor of 0.7 lagging. Calculate the capacitance (in μF) required in parallel with the motor to raise the power factor to 0.9 lagging.Correct answer is between '93,97'. Can you explain this answer?.
A single phase motor connected to 400V, 50Hz supply takes 31.7 A at a power factor of 0.7 lagging. Calculate the capacitance (in μF) required in parallel with the motor to raise the power factor to 0.9 lagging.Correct answer is between '93,97'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A single phase motor connected to 400V, 50Hz supply takes 31.7 A at a power factor of 0.7 lagging. Calculate the capacitance (in μF) required in parallel with the motor to raise the power factor to 0.9 lagging.Correct answer is between '93,97'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single phase motor connected to 400V, 50Hz supply takes 31.7 A at a power factor of 0.7 lagging. Calculate the capacitance (in μF) required in parallel with the motor to raise the power factor to 0.9 lagging.Correct answer is between '93,97'. Can you explain this answer?.
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