In a Capacitive Circuit, Why the Power Factor (Cos θ) Decrease, When Capacitive Reactance (XC) Increases or Capacitance (C) Decreases?
Another question from electrical and electronics engineering interviews question and answers series.
Explain the statement that ” In a capacitive circuit, when Capacitance (C) decreases, or capacitive reactance (XC) increases, the Circuit power factor (Cos θ) decreases“.
Related Questions: Why Power Factor Decreases When Inductance or Inductive Reactance Increases?
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 + (XL – XC2)”
In case of capacitive circuit:
- Z = √ (R2 + XC2)
- I = V / XC or I = V / Z
- Why Current Increases When Capacitance Increases or Capacitive Reactance Decreases?
- Why Current Decreases When Inductance or Inductive Reactance Increases?
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;
XC = 1 / 2πfC
XC = 1 / (2 x 3.1415 x 60 x 200 µF)
XC = 13.33 Ω
Now circuit impedance:
Z = √ (R2 + XC2)
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
- Related Question: Why Power in Pure Inductive and Pure Capacitive Circuit is Zero?
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.
XC = 1 / 2πfC = 1 / (2 x 3.1415 x 60 x 1400 µF) = 1.88 Ω
Z = √ (R2 + XC2) = √ (152 + 1.882) = 15.12 Ω
Power Factor = Cos θ = R / Z = 15 Ω / 15.12 Ω
Power Factor = Cos θ = 0.99
- In a Capacitive Circuit, Why the Current Increases When Frequency Increases?
- In an Inductive Circuit, Why the Current Increases When Frequency Decreases?
We can see that, when capacitance (C) was 200 µF, then circuit power factor were 0.75,
But when the circuit capacitance increased from 200 µF to 1400 µF, then the power factor increased from 0.75 to 0.99.
Similarly, when capacitive reactance was higher i.e. 13.33 Ω in case of 200 micro farads capacitor, the circuit power factor was low i.e. 0.75. When capacitive reactance decreased from 13.33 Ω to 1.88 Ω in case of 1400 micro farad capacitace, the overall power factor improved from 0.75 to 0.99.
In a capacitive circuit, when capacitive reactance XC increases, capacitance and the circuit power factor decreases and vice versa.
In oral or verbal,
- Capacitive reactance is a kind of resistance. When capacitive resistance increases, the circuit power factor decreases and vice versa.
- Capacitance is inversely proportional to the capacitive reactance
C ∝ 1 / XC
- Capacitance is directly proportional to the current and inversely proportional to the capacitive reactance.
C ∝ I and C ∝ 1 / XC
- Power factor is inversely proportional to the capacitive reactance and directly proportional to the capacitance.
Cos θ ∝ 1 / XC and Cos θ ∝ XC
- Why Flux in Primary and Secondary Winding is Always Equal?
- Why the reactance of a system under fault condition is low and faults currents may raise to the dangerously high value?
- Why Inductive Reactance (XL In DC Supply Is Zero (0)?