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An electric load consumed 17.32 kW at a power factor of 0.707 (lagging). For changing the load power factor to 0.866 (lagging), the capacitor that is to be connected in parallel with the load should draw-
- a)7.32 kVAR
- b)10 kVAR
- c)27.32 kVAR
- d)10.32 kVAR
Correct answer is option 'A'. Can you explain this answer?
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An electric load consumed 17.32 kW at a power factor of 0.707 (laggin...
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An electric load consumed 17.32 kW at a power factor of 0.707 (laggin...
Calculation of Reactive Power Required
Given data:
Load power = 17.32 kW
Initial power factor (PF1) = 0.707 lagging
Final power factor (PF2) = 0.866 lagging
To calculate the reactive power required to improve the power factor, we can use the following formula:
Q = P x (tanθ1 – tanθ2)
Where Q is the reactive power required, P is the load power, θ1 is the initial phase angle, and θ2 is the final phase angle.
First, we need to calculate the initial and final phase angles using the power factors:
θ1 = cos⁻¹(PF1)
θ2 = cos⁻¹(PF2)
Substituting the given values:
θ1 = cos⁻¹(0.707) = 45.00°
θ2 = cos⁻¹(0.866) = 30.96°
Now, we can calculate the reactive power required:
Q = 17.32 x (tan45.00 – tan30.96) = 7.32 kVAR
Therefore, the capacitor required to improve the power factor should draw 7.32 kVAR of reactive power. Answer (a) is correct.
Given data:
Load power = 17.32 kW
Initial power factor (PF1) = 0.707 lagging
Final power factor (PF2) = 0.866 lagging
To calculate the reactive power required to improve the power factor, we can use the following formula:
Q = P x (tanθ1 – tanθ2)
Where Q is the reactive power required, P is the load power, θ1 is the initial phase angle, and θ2 is the final phase angle.
First, we need to calculate the initial and final phase angles using the power factors:
θ1 = cos⁻¹(PF1)
θ2 = cos⁻¹(PF2)
Substituting the given values:
θ1 = cos⁻¹(0.707) = 45.00°
θ2 = cos⁻¹(0.866) = 30.96°
Now, we can calculate the reactive power required:
Q = 17.32 x (tan45.00 – tan30.96) = 7.32 kVAR
Therefore, the capacitor required to improve the power factor should draw 7.32 kVAR of reactive power. Answer (a) is correct.
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