Why Power Factor Decreases When Capacitive Reactance Increases or Capacitance Decreases?

Why Power Factor Decreases When Capacitive Reactance Increases or Capacitance Decreases?

Explain the statement that ” In a capacitive circuit, when Capacitance (C) decreases, or capacitive reactance (X_C) increases, the Circuit power factor (Cos θ) decreases“.

Explanation:

We know that in DC circuits:

I = V / R,

But in case of AC circuits:

I = V / Z

Where “total resistance of AC circuits = Impedance = Z = √ (R² + (X_L – X_C²)”

In case of capacitive circuit:

• Z = √ (R² + X_C²)
• I = V / X_C or I = V / Z

It means, if capacitance increases, capacitive reactance decreases and the circuit current increases which leads to improve the power factor. Similarly, when the circuit current and capacitance decreases due to increase in capacitive reactance, the overall power factor will be decreased as power factor is directly proportional to the capacitance and inversely proportional to the capacitive reactance. That is the reason we use capacitors banks to correct the power factor.

Let’s check with a solved example to see how power factor decreased by increase in capacitive reactance.

When Capacitance = 200 µF

Suppose a capacitive circuit where:

• Capacitance = C = 200 µ Farads
• Resistance = R = 15 Ω
• Frequency = f = 60 Hz

To find the inductive reactance;

X_C = 1 / 2πfC

X_C = 1 / (2 x 3.1415 x 60 x 200 µF)

X_C = 13.33 Ω

Now circuit impedance:

Z = √ (R² + X_C²)

Z = √ (15² + 13.33²)

Z = 20.06 Ω

Finally, Power factor in capacitive circuit:

Cos θ = R / Z

Cos θ = 15 Ω / 20.06 Ω

 Power Factor = Cos θ = 0.75 

When Capacitance = 1400 µF

Now we increased the capacitance (C) of capacitor form 200 µF  to 1400 µF.

R = 15 Ω, C = 1400 µF, f = 60 Hz.

X_C = 1 / 2πfC = 1 / (2 x 3.1415 x 60 x 1400 µF) = 1.88 Ω

Z = √ (R² + X_C²) = √ (15² + 1.88²) = 15.12 Ω

Power Factor = Cos θ = R / Z = 15 Ω / 15.12 Ω

 Power Factor = Cos θ = 0.99 

FAQ

What happens to capacitive reactance when capacitance increases?

But when circuit capacitance increased from 10 µF to 60 µF, then the current increased from 0.72 A to 4.34 A. Hence proved, In a capacitive circuit, when capacitance increases, the capacitive reactance X_C decreases which leads to increase the circuit current and vise versa

How does capacitance affect power factor?

A capacitor corrects the power factor by providing a leading current to compensate the lagging current. Power factor correction capacitors are designed to ensure that the power factor is as close to unity as possibe

What happens when capacitive reactance decreases?

From this summary, it is apparent that as frequency of source voltage increases, capacitive reactance decreases and current increases

How does reactance affect power factor?

Similarly, when the circuit current increases due to decrease in inductance or inductive reactance, the overall power factor will be improved as power factor is directly proportional to the inductance and inductive reactances. Related Question: According to Ohm’s Law, I ∝ V, But I ∝ 1/V in Power Equation

What is the effect of increasing capacitance?

Also, the more capacitance the capacitor possesses, the more charge will be forced in by a given voltage. This relation is described by the formula Q = C V or C = Q V where is the charge stored, is the capacitance, and is the voltage applied

How to connect capacitor to improve power factor?

Power factor can be improved by connecting the static capacitor in parallel with the equipment operating at lagging power factor. The capacitor draws leading currents from the supply voltage by 90° and compensates for the lagging reactive components of the load current

What happens if capacitance is decreased?

V = q / C, where V is voltage, q is charge, and C is capacitance. So if you charge a capacitor, then reduce its capacitance, the voltage will go up. The energy on a capacitor is 1/2 C V^2. If you reduce the capacitance to half the original, then the voltage will be double the original

What is the relationship between reactance and capacitance?

The formula for calculating the Capacitive Reactance, or impedance of a capacitor is: Capacitive reactance, denoted as x sub c (X_C), is equal to the constant one million (or 106) divided by the product of 2^p ( or 6.28) times frequency times the capacitance . where: X_C = Capacitive reactance measured in ohms