# Capacitance power factor question

• oxon88
In summary, the capacitor needed to be placed in parallel with the load to achieve a 90% lagging power factor has a value of 1.126 μF.
oxon88

## Homework Statement

A 50 KW load operates from a 60 Hz 10KV rms line with a power factor of 60% lagging. Determine the capacitance that must be placed in parallel with the load to achieve a 90% lagging power factor.

## The Attempt at a Solution

for 60% p.f:-

θ1 = Cos-1(0.6) = 53.13 degrees

Q1 = 50,000*tan(53.13) = 66.667 kVAR

for 90% p.f:-

θ2 = Cos-1(0.9) = 25.84 degrees

Q2 = 50,000*tan(25.84) = 24.22 kVAR
Qcapacitor = Q2 - Q1 = -42.45 kVAR
Xc = VRMS /Qc = (104)2 / -42450 = -2356 Ωω = 2.∏.f = 2.∏.60 = 377

C = 1 / ω.Xc = 1 / 377*2356 = 1.126 μF
can anyone check if I'm on the right path here please?

Last edited:
anyone?

I posted a response to this about 6 hours ago, saying that's the answer I got.

Not sure where my reply went. https://www.physicsforums.com/images/icons/icon9.gif

Last edited by a moderator:
Ok that's good. thank you

yes it is correct.

message deleted

Last edited:
oxon88 said:

## Homework Statement

A 50 KW load operates from a 60 Hz 10KV rms line with a power factor of 60% lagging. Determine the capacitance that must be placed in parallel with the load to achieve a 90% lagging power factor.

## The Attempt at a Solution

for 60% p.f:-

θ1 = Cos-1(0.6) = 53.13 degrees

Q1 = 50,000*tan(53.13) = 66.667 kVAR

for 90% p.f:-

θ2 = Cos-1(0.9) = 25.84 degrees

Q2 = 50,000*tan(25.84) = 24.22 kVAR
Qcapacitor = Q2 - Q1 = -42.45 kVAR
Xc = VRMS /Qc = (104)2 / -42450 = -2356 Ωω = 2.∏.f = 2.∏.60 = 377

C = 1 / ω.Xc = 1 / 377*2356 = 1.126 μF
can anyone check if I'm on the right path here please?
why isn't it - 2356 in the final formula ?

grinder76 said:
why isn't it - 2356 in the final formula ?
oxon88 calculated the signed reactance of the capacitor to be -2356 Ω. The minus sign reflects the fact that it is capacitive reactance. He then used the formula for the magnitude of the reactance, which is unsigned, to find the value of the capacitor.

An alternative approach might have been to use complex impedance which retains the sign throughout, but that would achieve the same end result.

grinder76
Thanks for the reply i now understand

## 1. What is capacitance power factor?

Capacitance power factor is a measure of how efficiently an electrical device uses energy. It is the ratio of the device's true power (the power actually consumed) to its apparent power (the power supplied by the source).

## 2. How is capacitance power factor calculated?

Capacitance power factor is calculated by dividing the device's true power by its apparent power. It can also be calculated by finding the cosine of the angle between the voltage and current in an AC circuit.

## 3. Why is capacitance power factor important?

A low capacitance power factor can result in wasted energy and higher electricity costs. It can also cause electrical devices to operate less efficiently and potentially lead to overheating and damage.

## 4. How can capacitance power factor be improved?

Capacitance power factor can be improved by adding power factor correction devices, such as capacitors, to the electrical circuit. These devices help to balance the reactive power and improve the overall power factor.

## 5. What are the consequences of a low capacitance power factor?

A low capacitance power factor can result in higher electricity bills, decreased efficiency of electrical devices, and potential damage to the devices. It can also cause power quality issues and lead to power outages.

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