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

Concept:

In AC circuits, the power factor is defined as the ratio of the real power flowing to the load to the apparent power in the circuit.

Power factor = cos ϕ

Where ϕ is the angle between voltage and current.

• If the current lags the voltage, the power factor will be lagging.
• If the current leads the voltage, the power factor will be leading.

Power factor is calculated as:

\(cosϕ=\frac{R}{Z}\)

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

R = resistance

Z = impedance

X_L = Inductive reactance

XC = Capacitive reactance

Calculation:

Given R = 4 Ω, X_C = 3 Ω

\(Z=\sqrt{{{4}^{2}}+{({0-3)}^{2}}}\)

Z = 5 Ω

∴ The power factor will be:

\(cosϕ=\frac{4}{5}=0.8\)

Explanation:

If Current and Voltage are 90 degrees out of phase, then the power (P) will be zero. The reason is as follow:

We know that power in single-phase AC Circuits:

P= V I cos θ

Where;

P = Power in Watts

V = Voltage in Volts

I = Current in Amperes

cos θ = Power factor of the circuit i.e. phase difference between current and voltage waves.

If the angle between current and Voltage are 90° (θ = 90), then

Power = P = V I cos(90°) = 0

So if we put cos 90° = 0, then the overall power of the circuit will be zero (i.e. In a pure Inductive circuit where voltage is leading by 90° from respectively).

From the above phase diagram, we can say that the voltage Across and current through the inductor are in quadrature.

So the reason justifies the assertion,

Hence option (1) is correct.

In AC circuits, the power factor is defined as the ratio of the real power flowing to the load to the apparent power in the circuit.

Power factor = cos ϕ

Where ϕ is the angle between voltage and current.

If the current lags the voltage, the power factor will be lagging.

If the current leads the voltage, the power factor will be leading.

Power factor = cos ϕ = R/Z

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

Where,

R = resistance

Z = impedance

X = reactance

The power factor is unity when the circuit is purely resistive.

• The maximum value of the power factor is unity (1)
• The power factor for a purely resistive load is unity (1)
• The power factor for RL load is less than unity (1) and is lagging in nature
• The power factor for RC load is also less than unity (1) but leading in nature

Concept:

In AC circuits, the power factor is defined as the ratio of the real power flowing to the load to the apparent power in the circuit.

Power factor = cos ϕ

Where ϕ is the angle between voltage and current.

• If the current lags the voltage, the power factor will be lagging.
• If the current leads the voltage, the power factor will be leading.

Power factor is calculated as:

\(cosϕ=\frac{R}{Z}\)

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

R = resistance

Z = impedance

X_L = Inductive reactance

XC = Capacitive reactance

Calculation:

Given R = 4 Ω, X_C = 3 Ω

\(Z=\sqrt{{{4}^{2}}+{({0-3)}^{2}}}\)

Z = 5 Ω

∴ The power factor will be:

\(cosϕ=\frac{4}{5}=0.8\)

Explanation:

If Current and Voltage are 90 degrees out of phase, then the power (P) will be zero. The reason is as follow:

We know that power in single-phase AC Circuits:

P= V I cos θ

Where;

P = Power in Watts

V = Voltage in Volts

I = Current in Amperes

cos θ = Power factor of the circuit i.e. phase difference between current and voltage waves.

If the angle between current and Voltage are 90° (θ = 90), then

Power = P = V I cos(90°) = 0

So if we put cos 90° = 0, then the overall power of the circuit will be zero (i.e. In a pure Inductive circuit where voltage is leading by 90° from respectively).

From the above phase diagram, we can say that the voltage Across and current through the inductor are in quadrature.

So the reason justifies the assertion,

Hence option (1) is correct.

In AC circuits, the power factor is defined as the ratio of the real power flowing to the load to the apparent power in the circuit.

Power factor = cos ϕ

Where ϕ is the angle between voltage and current.

If the current lags the voltage, the power factor will be lagging.

If the current leads the voltage, the power factor will be leading.

Power factor = cos ϕ = R/Z

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

Where,

R = resistance

Z = impedance

X = reactance

The power factor is unity when the circuit is purely resistive.

• The maximum value of the power factor is unity (1)
• The power factor for a purely resistive load is unity (1)
• The power factor for RL load is less than unity (1) and is lagging in nature
• The power factor for RC load is also less than unity (1) but leading in nature

Concept:

In AC circuits, the power factor is defined as the ratio of the real power flowing to the load to the apparent power in the circuit.

Power factor = cos ϕ

Where ϕ is the angle between voltage and current.

• If the current lags the voltage, the power factor will be lagging.
• If the current leads the voltage, the power factor will be leading.

Power factor is calculated as:

\(cosϕ=\frac{R}{Z}\)

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

R = resistance

Z = impedance

X_L = Inductive reactance

XC = Capacitive reactance

Calculation:

Given R = 4 Ω, X_C = 3 Ω

\(Z=\sqrt{{{4}^{2}}+{({0-3)}^{2}}}\)

Z = 5 Ω

∴ The power factor will be:

\(cosϕ=\frac{4}{5}=0.8\)

Concept:

In AC circuits, the power factor is defined as the ratio of the real power flowing to the load to the apparent power in the circuit.

Power factor = cos ϕ

Where ϕ is the angle between voltage and current.

• If the current lags the voltage, the power factor will be lagging.
• If the current leads the voltage, the power factor will be leading.

Power factor is calculated as:

\(cosϕ=\frac{R}{Z}\)

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

R = resistance

Z = impedance

X_L = Inductive reactance

XC = Capacitive reactance

Calculation:

Given R = 4 Ω, X_C = 3 Ω

\(Z=\sqrt{{{4}^{2}}+{({0-3)}^{2}}}\)

Z = 5 Ω

∴ The power factor will be:

\(cosϕ=\frac{4}{5}=0.8\)

Concept:

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

Z = R + j (X_L - X_C)

When f > f₀, X_L > X_C, and the circuit will be inductive.

When f > f₀, X_L < X_C, and the circuit will be capacitive.

The current for a series RLC circuit will be:

\(I=\frac{V}{R+j{(X_L-X_C)}}\)

This can also be written as:

\(I=\frac{V}{Z\angleθ }\)

From the above relation, we can conclude that:

1) For f > f0, θ will be positive, i.e. the current will lag the applied voltage.

2) For f < f0, θ will be negative, i.e. the current will lead the applied voltage.

Analysis:

We know that V and I_R in phase to each other i.e.

So I leads V, hence circuit has a lead power factor. So circuit nature is capacitive.

For series RLC circuit at f < f₀ ⇒ capacitive

f = f₀ ⇒ resistive

Explanation:

Power Factor of Series R-L Circuit

Definition: In an R-L series circuit, the power factor is a measure of how effectively electrical power is converted into useful work. It represents the cosine of the angle (\(\theta\)) between the voltage and current vectors in the circuit. The power factor is crucial in AC circuits as it determines the efficiency of power usage.

The power factor of a series R-L circuit is given by:

\(\rm \cos \theta = \frac{R}{Z}\)

Where:

• \(R\): Resistance of the circuit
• \(Z\): Impedance of the circuit

Impedance (\(Z\)): In an R-L series circuit, the impedance is the vector sum of the resistance (\(R\)) and the reactive inductance (\(X_L\)). It is given by:

\(\rm Z = \sqrt{R^2 + X_L^2}\)

Here:

• \(X_L\): Reactive inductance, which is calculated as \(X_L = \omega L\), where \(\omega\) is the angular frequency and \(L\) is the inductance.

Detailed Explanation:

In an R-L series circuit, the total impedance (\(Z\)) is a combination of the resistive and inductive components. The resistance (\(R\)) represents the real part of the impedance and is associated with energy dissipation as heat. The inductive reactance (\(X_L\)) represents the imaginary part of the impedance and is associated with energy storage in the magnetic field of the inductor.

The phase angle (\(\theta\)) between the voltage and current in the circuit is determined by the relationship between the resistance (\(R\)) and the reactive inductance (\(X_L\)):

\(\rm \tan \theta = \frac{X_L}{R}\)

From this, the cosine of the phase angle (\(\cos \theta\)) is calculated as:

\(\rm \cos \theta = \frac{R}{Z}\)

This formula indicates that the power factor depends on the ratio of the resistance to the total impedance of the circuit. A higher resistance relative to the impedance results in a better power factor, whereas a higher reactive inductance results in a lower power factor.

Correct Option Analysis:

The correct option is:

Option 3: \(\rm \cos \theta = \frac{R}{Z}\)

This option correctly defines the power factor of a series R-L circuit. The power factor is the ratio of the resistance (\(R\)) to the total impedance (\(Z\)) of the circuit. It is a measure of how efficiently the electrical power is being converted into useful work.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: \(\rm \cos \theta = \frac{X_L}{Z}\)

This option is incorrect. The ratio of the reactive inductance (\(X_L\)) to the impedance (\(Z\)) does not represent the power factor. Instead, it represents the sine of the phase angle (\(\sin \theta\)), which is associated with the reactive component of the circuit.

Option 2: \(\rm \cos \theta = \frac{Z}{R}\)

This option is incorrect. The ratio of impedance (\(Z\)) to resistance (\(R\)) does not define the power factor. Instead, it is inversely related to the power factor and does not have a direct physical interpretation in this context.

Option 4: \(\rm \cos \theta = \frac{R}{X_L}\)

This option is incorrect. The ratio of resistance (\(R\)) to reactive inductance (\(X_L\)) is related to the tangent of the phase angle (\(\tan \theta\)), not the cosine of the phase angle (\(\cos \theta\)). It is not a measure of the power factor.

Conclusion:

In summary, the power factor of a series R-L circuit is given by \(\rm \cos \theta = \frac{R}{Z}\), which accurately represents the ratio of the resistance to the total impedance. This formula is essential for understanding the efficiency of power usage in AC circuits. Correctly identifying the power factor formula is crucial for analyzing and optimizing the performance of R-L circuits, especially in applications where power efficiency is critical.

Explanation:

If Current and Voltage are 90 degrees out of phase, then the power (P) will be zero. The reason is as follow:

We know that power in single-phase AC Circuits:

P= V I cos θ

Where;

P = Power in Watts

V = Voltage in Volts

I = Current in Amperes

cos θ = Power factor of the circuit i.e. phase difference between current and voltage waves.

If the angle between current and Voltage are 90° (θ = 90), then

Power = P = V I cos(90°) = 0

So if we put cos 90° = 0, then the overall power of the circuit will be zero (i.e. In a pure Inductive circuit where voltage is leading by 90° from respectively).

From the above phase diagram, we can say that the voltage Across and current through the inductor are in quadrature.

So the reason justifies the assertion,

Hence option (1) is correct.

Power factor = 45° lead

\(\begin{array}{l} \tan 45^\circ = \frac{{{X_C} - {X_L}}}{R}\\ 1 = \frac{{\frac{1}{{\omega c}} - \omega L}}{{10}} \end{array}\)  

1 - ω²LC = 10ωc

ω = 2π × 25

= 50 π

1 – (50 π)² × 60 × 10^-3 C = 500 πc

\(\begin{array}{l} C = \frac{1}{{500\pi + {{\left( {50\pi } \right)}^2} \times 60 \times {{10}^{ - 3}}}}\\ \simeq 327.73\;\mu F \end{array}\)

∴ Resonance Frequency \(= \frac{1}{{2\pi \sqrt {LC} }}\)

\(= \frac{1}{{2\pi \sqrt {60 \times 327.73 \times {{10}^{ - 9}}} }}\)

At \(\rm {f = {f_L} \to I = {{{I_m}} \over {\sqrt 2 }}}\)

\(\rm { \to {V \over Z} = {V \over {\sqrt 2 R}} \to Z = \sqrt 2 R}\)

Power factor \(\rm {cos\phi = {R \over Z} = {1 \over {\sqrt 2 }} = 0.707\;\left( {lead} \right)}\)

Because in series RLC circuit at \(\rm f={f_L}\)the circuit is having capacitive nature so it is having leading power factor.

At \(\rm {\rm {f = f_0 → R = Z → cosϕ = 1}}\)