# How Do You Calculate Capacitance for Unity Power Factor in a Single-Phase Motor?

• Reefy
In summary, a single-phased motor connected across a 240-V source at 50Hz has a power factor of 1.0. To achieve a unity power factor, a capacitor must be connected in parallel with the load.
Reefy

## Homework Statement

Not sure if this is the correct place for Electrical Engineering Homework help but here goes

A single-phased motor connected across a 240-V source at 50Hz as shown in the figure has a power factor of 1.0, I = 20A, I1 = 25 A.

Find the capacitance required to give a unity power factor when connected in parallel with the load as shown.

## Homework Equations

Apparent Power S = VRMSIRMS
Angular Frequency ω = 2πf
cosθ = 1.0
Reactive Power Q = Ssinθ = VRMSIRMSsinθ
Active Power P = Scosθ = VRMSIRMScosθ
ZRL = R + ωLi
Zequiv = ZCZR/[ZC + ZR]

QC = 1/ωC

## The Attempt at a Solution

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I think I am misunderstanding the question because if it is already at unity, wouldn't that make θ = 0° meaning QL = 0 VAR?

If we want this at unity, QL = QC, correct? But I am getting 0 when the answer is allegedly non-zero.

The only thing I was able to find was P = (240 V)(20 A) = 4800 W and I think I2 = 5 A which is entering the node. Meaning I1 = I + I2. This I2 is an RMS value, correct?

Reefy said:
I think I am misunderstanding the question because if it is already at unity, wouldn't that make θ = 0° meaning QL = 0 VAR?
It can't be already at unity pf. Every ac motor has a lagging pf, which is the result of its magnetizing current. So, unity pf condition can be achieved only after putting a capacitor in parallel with the motor.
Reefy said:
This I2 is an RMS value, correct?
Yes and rms values aren't added algebraically.

Reefy
cnh1995 said:
It can't be already at unity pf. Every ac motor has a lagging pf, which is the result of its magnetizing current. So, unity pf condition can be achieved only after putting a capacitor in parallel with the motor.

Ok, so when the problem says that it has a power factor of 1.0, they are talking about after you correct it with the capacitor not before? The wording made it seem as if it is already at unity which made no sense to me. So basically, I have to find the power factor before the capacitor is added.

cnh1995 said:
Yes and rms values aren't added algebraically.

Ok so I have multiply the rms values by √2 to find the peak values and then add the currents using phasors in rectangular form aka complex numbers?

Reefy said:
Ok so I have multiply the rms values by √2 to find the peak values and then add the currents using phasors in rectangular form aka complex numbers?
That would complicate the problem. Start with the phasor diagram of the circuit. Using elementary geometry, you can find I2. Once you know I2, you can find capacitive reactance Xc and capacitance C.

Reefy
cnh1995 said:
That would complicate the problem. Start with the phasor diagram of the circuit. Using elementary geometry, you can find I2. Once you know I2, you can find capacitive reactance Xc and capacitance C.

Hmm ok so I drew the diagram on a graph and got that IC = √[IL2 - I2] which gave me IC = 15 A

Using the voltage source, which is in parallel with the capacitor, I found reactance Xc which gave me C = 1.19mF. Did I make a mistake somewhere?

Reefy said:
Hmm ok so I drew the diagram on a graph and got that IC = √[IL2 - I2] which gave me IC = 15 A
Right.
Reefy said:
which gave me C = 1.19mF
I am getting a different value for C. How much is the capacitive reactance?

Reefy
cnh1995 said:
Right.

I am getting a different value for C. How much is the capacitive reactance?

Zc = Vc/Ic = (240 < 0°)/(15 < 90°) = 2.67 < -90° which means Xc = -2.67i = -i/(ωC)

Edit: Nvm I see that I must've mistyped in my calculator. Zc = 16 < -90°

Now I got C = 199 μF

Reefy said:
Edit: Nvm I see that I must've mistyped in my calculator. Zc = 16 < -90°
Right.

Reefy
Thanks, I really appreciate it.

cnh1995
Reefy said:
Thanks, I really appreciate it.
You're welcome!

Reefy

## 1. What is "Power Correction Unity"?

"Power Correction Unity" is a concept in physics that refers to the ability of a system to maintain its power output at a constant level, despite changes in the input power. It is also referred to as "unity power factor".

## 2. Why is "Power Correction Unity" important?

"Power Correction Unity" is important because it allows for efficient energy usage and reduces power losses in electrical systems. It also helps to improve the overall stability and reliability of the system.

## 3. How is "Power Correction Unity" achieved?

"Power Correction Unity" is achieved through the use of power correction devices, such as capacitors, inductors, and transformers. These devices are designed to adjust the reactive power in the system and bring it closer to unity power factor.

## 4. What are the benefits of "Power Correction Unity"?

There are several benefits of achieving "Power Correction Unity" in a system. These include reduced energy costs, improved voltage regulation, increased system capacity, and extended equipment lifespan.

## 5. How can I determine if my system has "Power Correction Unity"?

To determine if your system has "Power Correction Unity", you can measure the power factor using a power meter. If the power factor is close to 1, then your system has achieved "Power Correction Unity". You can also consult with a qualified electrician or engineer to evaluate your system's power factor.

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