Power Factors MCQ Quiz - Objective Question with Answer for Power Factors - Download Free PDF

Power Triangle

In the given figure, the given terms represent the type of power:

P = Active Power

Q = Reactive Power 

S = Apparent Power

The power factor is given by:

\(cosϕ={P\over S}\)

Calculation

Given, cos ϕ = 0.8

S = 10 kVA

The active power is given by:

\(0.8={P\over 10}\)

P = 8 kW

Concept

The apparent power in an AC circuit is given by:

\(S=V_{rms}I_{rms}\)

The active power in an AC circuit is given by:

\(P=V_{rms}I_{rms}\space cos(\theta_v-\theta_i)\)

The reactive power in an AC circuit is given by:

\(Q=V_{rms}I_{rms}\space sin(\theta_v-\theta_i)\)

Calculation

Given, v = 20 sin (ωt + 60°) V

i = 10 sin (ωt + 30°) A

\(S={20\over \sqrt{2}}\times{10\over \sqrt{2}}\)

S = 100 VA

\(Q={20\over \sqrt{2}}\times{10\over \sqrt{2}}\times sin(60-30)\)

Q = 50 VAr

Power Factor for Purely Inductive or Purely Capacitive AC Circuit

Definition: The power factor in an AC circuit is a dimensionless number that ranges from 0 to 1 and represents the ratio of real power flowing to the load to the apparent power in the circuit. It is a measure of how effectively the current is being converted into useful work output. A power factor of 1 indicates that all the power is being effectively converted into work, while a power factor of 0 indicates that no real work is being done, and all the power is reactive.

In the context of purely inductive or purely capacitive AC circuits, the power factor has specific characteristics:

• Purely Inductive Circuit: In a purely inductive circuit, the current lags the voltage by 90 degrees. This means that the current and voltage are out of phase, resulting in a power factor of 0.
• Purely Capacitive Circuit: In a purely capacitive circuit, the current leads the voltage by 90 degrees. Similar to the inductive circuit, the current and voltage are out of phase, resulting in a power factor of 0.

Detailed Solution:

To understand why the power factor is 0 for purely inductive or capacitive AC circuits, we need to delve into the relationship between voltage, current, and power in these circuits.

Inductive Circuit:

In a purely inductive circuit, the voltage (V) leads the current (I) by 90 degrees. This phase difference can be represented mathematically as:

V(t) = V_max sin(ωt)

I(t) = I_max sin(ωt - 90°) = I_max cos(ωt)

Where V_max and I_max are the maximum values of voltage and current, respectively, and ω is the angular frequency of the AC supply.

The instantaneous power (P) in the circuit is given by:

P(t) = V(t) x I(t)

Substituting the values of V(t) and I(t), we get:

P(t) = V_max sin(ωt) x I_max  cos(ωt)

Using the trigonometric identity for the product of sine and cosine:

P(t) = (V_max x I_max / 2) x sin(2ωt)

This shows that the instantaneous power oscillates at twice the frequency of the supply voltage and current, and the average power over a complete cycle is zero. This means that no real power is consumed in a purely inductive circuit, and all the power is reactive. Hence, the power factor is 0.

Capacitive Circuit:

In a purely capacitive circuit, the current (I) leads the voltage (V) by 90 degrees. This phase difference can be represented mathematically as:

V(t) = V_max sin(ωt)

I(t) = I_max sin(ωt + 90°) = I_max cos(ωt)

The instantaneous power (P) in the circuit is given by:

P(t) = V(t) x I(t)

Substituting the values of V(t) and I(t), we get:

P(t) = V_max sin(ωt) x I_max cos(ωt)

Using the trigonometric identity for the product of sine and cosine:

P(t) = (V_max  I_max / 2) x sin(2ωt)

This again shows that the instantaneous power oscillates at twice the frequency of the supply voltage and current, and the average power over a complete cycle is zero. This means that no real power is consumed in a purely capacitive circuit, and all the power is reactive. Hence, the power factor is 0.

Therefore, for both purely inductive and purely capacitive AC circuits, the power factor is 0.

Explanation:

Power Consumption in a Purely Capacitive AC Circuit

Definition: In an AC circuit containing only a capacitor, the current leads the voltage by 90 degrees. This phase difference plays a crucial role in determining the power consumption in such circuits. The power consumed in any AC circuit is given by the product of voltage, current, and the cosine of the phase angle between them, known as the power factor.

Working Principle: In a purely capacitive AC circuit, the voltage and current are out of phase by 90 degrees. When the voltage across the capacitor is at its maximum value, the current is zero, and when the current is at its maximum value, the voltage is zero. Because of this phase difference, the power consumed over a complete cycle is zero.

Concept:

The power triangle is shown below.

P = Active power (or) Real power in W = VrIr cos ϕ = (Ir)2R

Q = Reactive power in VAR = VrIr sin ϕ

S = Apparent power in VA = VrIr = (Ir)2Z

Where,

Vr and Ir is the RMS value of voltage and current.

R is resistance Ω.

Z is impedance in Ω.

S = P + jQ

\(S = \sqrt {{P^2} + {Q^2}} \)

ϕ is the phase difference between the voltage and current

Power factor \(\cos ϕ = \frac{P}{S}\)

Calculation:

Given,

P = 800 kW

cos ϕ = 0.8 (Leading)

Hence,

S = P/cos ϕ = 800/0.8 = 1000 kVA

Hence,

Reactive Power (Q) = \(\sqrt{S^2-P^2}=\sqrt{1000^2-800^2}=600\) kVAR

Power factor:

• The power factor is defined as the ratio of the active power to the apparent power.
• It is denoted by cos ϕ.
   

Formula:

Power factor = (Active power /Apparent power)

In series circuit,

Power factor = (Resistance / Impedance)cos⁡ϕ=R|Z|" tabindex="0">cosϕ=R|Z|

 \((cos\phi\ =\frac{R}{Z})\)

Total Equivalent impedance is given by,

Z = R ±  j X

Where,

R = Resistance of Circuit

+ j X = Inductive Reactance of Circuit

- j X = Capacitive Reactance of Circuit

The magnitude of Equivalent Impedance is given by

\((\left| Z \right| = \sqrt {{R^2} + {X^2}})\)

Calculation:

Equivalent Impedance Z_eq = j2 + ( - j2 ∥ j4 )

\( =j2+\dfrac{-j2\times j4}{-j2+j4}\)

|Z|=R2+X2" tabindex="0">|Z| = R₂ + X₂

= j2 - j4 Ω

= - j2 Ω

= - j X Ω

Negative Imaginary Term without Real Part Indicates a Pure Capacitive Circuit.

Therefore, a capacitive Reactance of 2 Ω with zero power factor leading.

The correct answer is option 3):(1)

Concept:

• kVAR is the measure of reactive power required by inductive loads such as induction motors, induction furnaces.
• For the purely resistive circuit, the power factor is 1 (perfect), because the reactive power equals zero.
• The power factor is the ratio between real and apparent power. That means the higher the percentage of the KVAR in your load, the lower the ratio of the kW to the kVA. The outcome gives you a poor power factor.
• A power factor of 0 indicates that the voltage and current are 90-degrees out of phase. In semiconductor circuits powered from the AC mains, a bridge rectifier converts the AC power into DC.

Concept:

The power triangle is as shown below.

P = Active power (or) Real power in W = V_rms I_rms cos ϕ

Q = Reactive power in VAR = V_rms I_rms sin ϕ

S = Apparent power in VA = V_rms I_rms

S = P + jQ

\(S = \sqrt {{P^2} + {Q^2}}\)

ϕ is the phase difference between the voltage and current

Power factor \(\cos ϕ = \frac{P}{S}\)

From power trinagle:

Q = P tan ϕ = S sin ϕ

The correct answer is option 4.

The ratio of voltage (V) and current (I) gives the impedance (Z) of an AC circuit.

Power factor:

The power factor for an AC circuit is given by:

\(cosϕ={True \space power\over Apparant\space power}\)

\(cosϕ={P\over VI}\)

\(cosϕ ={R\over Z}\)​

where, cos ϕ = Power factor

P = Active Power

R = Resistance

Z = Impedance

The correct answer is option 4): 1.11

Concept:

Peak factor

The ratio of the maximum value (peak value) to RMS value is known as the peak factor or crest factor.

Peak factor= \(maximum\:value\over rms\:value\)

Form factor.

The ratio of RMS value to the average value is known as the form factor.

\(Form\;factor = \frac{{rms\;value}}{{average\;value}}\)

Calculation:

Given

 v = 250sin(2π × 50t)

V_m = 250 V 

RMS value = 0.707 × peak value = 176.6 V

Average value =  \(2V_m\over π\)

\(2V_m\over π\) = 159.15 V

Form factor = 1.11

In the AC circuit, the load power factor is given by:

Load power factor (cos φ) \(= \frac{{Total\;Active\;power\;consumed\;by\;load\left( P \right)}}{{Total\;Apparent\;Power\left( S \right)}}\)

Total apparent power is given by

Total apparent power (S) = \(\sqrt {{P^2} + {Q^2}}\)

Where,

P = Total active power consumed

Q = Total reactive power consumed

Load power factor (cos φ) = \(\frac{P}{{\sqrt {{P^2} + {Q^2}} }}=\frac{P}{S}\)

Application:

Given that, P = Q

From above concept,

\(S=\sqrt{P^2+Q^2}=\sqrt{P^2+P^2}=P\sqrt{2}\)

Load power factor (cos φ) = \(\frac{P}{P\sqrt{2}}=\frac{1}{\sqrt{2}}\)

cos ϕ = 0.707 

Since the load current is inductive in nature, therefore power factor will be lagging.

Power factor:

• The overall power factor is defined as the cosine of the angle between the phase voltage and phase current. 
• In AC circuits, the power factor is also defined as the ratio of the real power flowing to the load to the apparent power in the circuit. 
  Hence power factor can be defined as watts to volt-amperes.
• It is also defined as the ratio of resistance to the impedance of the circuit.
• Power factor is a unitless quantity.

Power factor \(= \cos \phi = \frac{R}{Z}= \frac{kW}{kVA}\)

ϕ is the angle between the voltage and the current.

In a series RLC circuit, the impedance is given by

\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}\)

∴ If Xc increases, then Z decreases. So that the power factor is increases.

Where R = resistance

Z = impedance

XL = inductive reactance

XC = capacitive reactance

• For a purely resistive circuit, the angle between the voltage and current is 0°

          So power factor for a purely resistive circuit is:

          P.F. = cos 0° 

          P.F. = 1 (unity)

• In a purely inductive circuit, the current lags the voltage by 90° and the power factor is zero lagging
• In a purely capacitive circuit, the current leads the voltage by 90° and the power factor is zero leading
• In a circuit consists of a resistor and a capacitor, the current lead the voltage by an angle less than 90° and it is leading.

Concept:

For a series RLC circuit, the net impedance is given by:

Z = R + j (XL - XC)

XL = Inductive Reactance given by:

XL = ωL

XC = Capacitive Reactance given by:

XL = 1/ωC

The magnitude of the impedance is given by:

\(|Z|=\sqrt{R^2+(X_L-X_C)^2}\)

\(\tan \phi = \frac{{{X_L} - {X_C}}}{R}\)

\(\therefore \cos \phi = \frac{R}{{\sqrt {{{\left( {{X_L} - {X_C}} \right)}^2} + {R^2}} }} = \frac{R}{Z}\)

 Power factor \( = \frac{R}{Z}\)

Calculation:

R = 8 Ω

X_C = 10 Ω

X_L = 4 Ω

\(|Z|=\sqrt{R^2+(X_L-X_C)^2}\)

\(|Z|=\sqrt{8^2+(10-4)^2}\)

|Z| = 10 Ω

Power factor \( = \frac{R}{Z}\)\( = \frac{8}{10}\)

= 0.8 lead (Because X_C > X_L)

Apparent Power (PA): The combination of reactive power and true power is called apparent power. It is measured in the unit of Volt-Amps (VA) and is symbolized by the capital letter S.

Active Power or true power (PT): The actual amount of power being dissipate or used, in a circuit, is called true power. It is measured in watts and symbolized by the capital letter P

Reactive Power (PR): It is measured in Volt-Amps-Reactive (VAR). The mathematical symbol for reactive power is the capital letter Q.

The relation between all the three powers is as shown:

\({P_A} = \sqrt {P_T^2 + P_R^2}\)

Calculation:

With PT = 100 W, and PR = 100 VAR, the apparent power will be:

\({P_A} = \sqrt {100^2 + 100^2}\)

\({P_A} = \sqrt {2}\times 100\)

PA = 141.4 VA