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

Construction of wattmeter

• By using a current and potential transformer in conjunction with a wattmeter, high power measurements can be done.
• The current and potential transformers are connected between the line and current and voltage coils of a wattmeter respectively.
• CT is connected in series with the wattmeter current coil, and PT is connected in parallel with the voltage coil.
• Both transformers step down the high voltage and current values to a value measurable by the voltage and current coils of the wattmeter.
• In order to extend the range of the wattmeter, many current transformers and potential transformers of varied ratios are to be used along with the current and voltage coils of the wattmeter respectively. Thereby, a single wattmeter can cover a wide range of power measurements.

Explanation:

Power Measurement in a 3-Phase Circuit Using Two Wattmeters

Definition: The two-wattmeter method is a commonly used technique for measuring power in a 3-phase circuit. This method requires two wattmeters to be connected to two of the three lines in a 3-phase system. The readings of these wattmeters give the total power in the circuit. The method is especially useful for both balanced and unbalanced loads.

Working Principle: In a 3-phase circuit, the total power can be calculated as the algebraic sum of the readings of the two wattmeters:

Total Power (P) = W₁ + W₂

Where W₁ and W₂ are the readings of the two wattmeters. Depending on the power factor of the circuit, the readings of the wattmeters can be positive, negative, or even zero. The phase angle (φ) between the line voltage and current determines the power factor, and this in turn affects the wattmeter readings.

Given Condition:

In the question, one of the wattmeter readings is positive, and the other is negative. The magnitudes of the readings are different. This condition indicates that the power factor of the circuit is less than 0.5 (lagging).

Analysis:

Let’s analyze the readings of the two wattmeters for various power factors:

• For a power factor of 1 (unity), both wattmeter readings will be positive and equal, as the angle between voltage and current is 0°.
• For a power factor between 0.5 and 1 (lagging), both wattmeter readings will still be positive but unequal. The difference in their magnitudes increases as the power factor decreases.
• For a power factor of 0.5 (lagging), one wattmeter reading becomes zero. This happens because the phase angle φ is 60°, leading to the wattmeter connected to a specific phase showing zero reading.
• For a power factor less than 0.5 (lagging), one wattmeter reading becomes negative. The negative reading occurs because the phase angle φ exceeds 60°, causing the current direction relative to the voltage to result in a negative product in one wattmeter.

In the given scenario, one wattmeter shows a positive reading, and the other shows a negative reading, with different magnitudes. This is a clear indication that the power factor of the circuit is less than 0.5 (lagging).

Conclusion:

The correct answer is:

Option 4: Less than 0.5 (lagging)

This conclusion is based on the characteristic behavior of wattmeters in the two-wattmeter method when the power factor of the circuit is less than 0.5. The negative reading of one wattmeter, combined with the positive reading of the other, confirms this condition.

Additional Information

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

Option 1: Unity

If the power factor were unity, both wattmeter readings would be positive and equal. This does not match the given condition where one wattmeter reading is negative, and the magnitudes are different.

Option 2: 2 (lagging)

A power factor of 2 (lagging) is not possible in any electrical circuit, as power factor is defined as the cosine of the phase angle (cos φ), which always lies between -1 and 1. Hence, this option is invalid.

Option 3: 5 (lagging)

Similar to Option 2, a power factor of 5 (lagging) is not physically possible, as power factor values cannot exceed 1. Therefore, this option is also invalid.

Option 4: Less than 0.5 (lagging)

This is the correct answer, as explained above. The negative reading of one wattmeter and the positive reading of the other, with different magnitudes, indicate a power factor less than 0.5 (lagging).

Conclusion:

Understanding the behavior of wattmeters in the two-wattmeter method is essential for determining the power factor of a 3-phase circuit. The key takeaway is that when one wattmeter reading is negative and the other is positive, with different magnitudes, the power factor is less than 0.5 (lagging). This knowledge is crucial for accurately analyzing and diagnosing power systems in electrical engineering applications.

Explanation:

Measurement of Three-Phase Power using the Two Wattmeter Method

Introduction: The two wattmeter method is a widely used technique for measuring the total power in a three-phase system. It is particularly suitable for balanced and unbalanced loads. The method involves the use of two wattmeters connected to the system, and the readings of these wattmeters help determine the total power and the power factor of the system.

Condition: When the two wattmeter readings are equal, the power factor of the circuit is unity (1). This occurs because the phase angle between the line voltage and current in the system is zero, indicating a purely resistive load. Let us delve deeper into the working principle and analysis to understand why this happens.

Working Principle:

In the two wattmeter method:

• Wattmeter 1 measures the power in one phase (P1).
• Wattmeter 2 measures the power in another phase (P2).
• The total power in the system is given by the sum of the readings of the two wattmeters: P = P1 + P2.

For a three-phase system with a balanced load, the power factor can be determined using the formula:

Power Factor (cos φ) = (P1 - P2) / (P1 + P2)

Analysis for Unity Power Factor:

When the power factor is unity (1), it implies that the load is purely resistive. In this case, the current and voltage are in phase, and the phase angle φ is zero. Substituting this condition into the formula:

• P1 = P2 (since both wattmeter readings are equal for a purely resistive load).
• cos φ = (P1 - P2) / (P1 + P2).
• cos φ = (P1 - P1) / (P1 + P1) = 0 / 2P1 = 0.

Thus, the power factor of the circuit is unity (cos φ = 1) when the two wattmeter readings are equal. This is a unique condition that arises due to the absence of any phase difference between the voltage and current.

Advantages of Two Wattmeter Method:

• Applicable for both balanced and unbalanced loads.
• Simple and effective for three-phase power measurement.
• Provides a direct way to measure power factor.

Correct Option Analysis:

The correct option is:

Option 2: Unity

This option is correct because the power factor of the circuit is unity when the two wattmeter readings are equal. The condition of equal readings arises when the load is purely resistive, and the current and voltage are in phase.

Additional Information

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

Option 1: 0.8 lag

This option is incorrect. When the power factor is 0.8 lag, the phase angle is not zero, and the wattmeter readings will not be equal. Instead, one wattmeter will record a higher value than the other, depending on the phase angle of the load.

Option 3: 0.8 lead

Similar to option 1, this option is also incorrect. A power factor of 0.8 lead indicates a capacitive load, where the current leads the voltage. In this case, the wattmeter readings will again be unequal, and the condition of equal readings will not be satisfied.

Option 4: Zero

This option is incorrect because a power factor of zero implies a completely reactive load (either inductive or capacitive). In such a scenario, the current and voltage are 90° out of phase, and the two wattmeter readings will not be equal. Instead, one wattmeter may show a negative reading depending on the connection.

Conclusion:

The two wattmeter method is a reliable technique for measuring three-phase power and determining the power factor of a circuit. When the two wattmeter readings are equal, it indicates that the power factor of the circuit is unity, corresponding to a purely resistive load. This condition is unique and helps distinguish purely resistive loads from other types of loads in three-phase systems.

Two Wattmeter Method of Power Measurement

The wattmeter meter readings are:

\(W_1=V_LI_Lcos(30-ϕ)\)

\(W_2=V_LI_Lcos(30+ϕ)\)

where, \(ϕ=tan^{-1}({√{3}(W_1-W_2)\over W_1+W_2})\)

The power factor is given by cosϕ.

Calculation:

Given, cosϕ = 0 ⇒ ϕ = 90°

\(W_1=V_LI_Lcos(30-ϕ)\)

\(W_1=V_LI_Lcos(30-90)\)

\(W_1={1\over 2}\times V_L\times I_L\)

\(W_2=V_LI_Lcos(30+ϕ)\)

\(W_2=V_LI_Lcos(30+90)\)

\(W_2=-{1\over 2}\times V_L\times I_L\)

If we multiply both W₁ and W₂ by √3, then:

\(W_1={\sqrt{3}\over 2}\times V_L\times I_L\) and \(W_2=-{\sqrt{3}\over 2}\times V_L\times I_L\)

Concept

The reactive power in power measurement of a three-phase using 2 wattmeters is given by:

\(Q={\sqrt{3}(W_1-W_2 )}\)

where, W₁ and W₂ are wattmeter readings

Calculation

Given, W₁ = 2000 W

W₂ = 1000 W

\(Q=\sqrt{3}(2000-1000)\)

Q = 1000√3 VAr

In a two-wattmeter method,

The reading of first wattmeter (W₁) = V_L I_L cos (30 – ϕ)

The reading of second wattmeter (W₂) = V_L I_L cos (30 + ϕ)

Total power in the circuit (P) = W₁ + W₂ = √3 V­­_LI_L cos ϕ

Total reactive power in the circuit \(Q = √ 3 \left( {{W_1} - {W_2}} \right) =\sqrt3 {V_L}{I_L}\sin \phi\)

Power factor = cos ϕ

\(\phi = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{√ 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Measurement of polyphase power: 3 - ϕ

1) For 3ϕ – 4 wire → neutral is present

2) As per Blondel’s theorem, if neutral is not present for ‘n’ wire system, then ‘n-1’ number of wattmeter’s are required for ‘n’ wire system.

• The one-wattmeter method is used for the measurement of three-phase power in case balanced loads only
• It is used for both star and delta connected loads
• Two wattmeter method and three wattmeter methods are used for both balanced and unbalanced loads
• Two wattmeter method is best suitable for three-phase three-wire system and three wattmeter method is suitable for the three-phase four-wire system

Note: If neutral is available we shouldn’t use a 2 – wattmeter method.

Concept:

Wattmeter reads the power and it is given by

P = V_PC I_CC cos ϕ

V_PC is the voltage across pressure coil

I_CC is current flows through the current coil

ϕ is the phase angle between V_PC and I_CC

Application:

The circuit representation of the given question is as shown below.

• The potential coil is connected across Z₂.
• It reads the voltage across Z₂ only.
• So, Wattmeter reads only power consumed by Z₂.

Concept:

In a two-wattmeter method, for lagging load

The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q=\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)\)

Power factor = cos ϕ

\(\phi =\text{ta}{{\text{n}}^{-1}}\left( \frac{\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)}{{{W}_{1}}+{{W}_{2}}} \right)\)

Calculation:

Given that, W1 = 2 W2

\(\tan \phi = \frac{{\sqrt 3 \left( {2{W_2} - {W_2}} \right)}}{{\left( {2{{\rm{W}}_2} + {W_2}} \right)}} = \frac{1}{{\sqrt 3 }}\)

⇒ ϕ = 30°

Power factor = cos 30° = 0.866

Important Point:

According to Blondell’s theorem, the no. of watt meters to be required to measure the total power in n-phase system is either N (or) N–1.

When a separate neutral wire is available in the system the no. of watt meters to be required is N.

When the neutral wire is not available in the system, then the no. of watt meters to be required is (N–1).

Power is measured according to Blondel's theorem

One line is acting as a common line for the return path. Hence the minimum number of wattmeters required is 2.

Additional Information

Measurement of polyphase power: 3 - ϕ

1) For 3ϕ – 4 wire → neutral is present

2) As per Blondel’s theorem, if neutral is not present for the ‘n’ wire system, then the ‘n-1’ number of wattmeter’s are required for the ‘n’ wire system.

Concept:

Dynamometer type 3-phase power factor meter:

• The power factor meter circuit includes two fixed coils namely pressure coil and current coil.
• Pressure coil is connected across the circuit while the current coil is connected such that it can carry a circuit current or a definite fraction of current.
• The power factor meter circuit include two moving coils. These are rigidly connected to each other so that their axes are at 120° to each other.
• These coils move together and carry the pointer which indicates the power factor of the circuit.
• Unlike, the single-phase dynamometer type power factor meter, it has both voltage coils in series with pure resistances. Instead of fixing the voltage coils 90° apart, we fix those coils 120° apart. 

Concept:

3-phase power measurement: for a balanced load 3-phase power measurement one wattmeter required

3-phase power measurement of a balanced load by 1 wattmeter method

P_3ϕ = 3 V_ph I_ph cosϕ 

V_ph = phase voltage

I_ph = phase current

Calculation:

As per the given circuit diagram, a balanced load wattmeter reading will be

I_ph = 10 A, V_ph = V_l /√ 3 = 200/ √3 = 115.47

\(W = {V_{ph}}{I_L}\cos ϕ \)

\(1000 = 115.47 \times 10 \times \cos ϕ \)

\( \Rightarrow \cos ϕ = 0.866\)

Power factor \(\cos ϕ = 0.866\)

Important Point:

Power is measured according to Blondel's theorem

Concept:

In two wattmeter method, the power factor is given by

\(cos\phi = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)

Explanation:

Given that power factor = 0.866

Here, \(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

\(\begin{array}{l} \cos \phi = 0.866 \Rightarrow \phi = 30^\circ \\ 30^\circ = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\\ \frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}} = \frac{1}{{\sqrt 3 }} \end{array}\)

Given the power input is 5 kW, i.e. (W₁ + W₂) = 5

\(\begin{array}{l} \frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{5} = \frac{1}{{\sqrt 3 }}\\ \Rightarrow \left( {{W_1} - {W_2}} \right) = \frac{5}{3}\\ \Rightarrow {W_1} = \frac{{10}}{3}kW,{W_2} = \frac{5}{3}kW \end{array}\)

Wattmeter:

A wattmeter is an electrical instrument that is used to measure the electric power of any electrical circuit.

The internal construction of a wattmeter is such that it consists of two coils. One of the coils is in series and the other is connected in parallel.

The coil that is connected in series with the circuit is known as the current coil and the one that is connected in parallel with the circuit is known as the voltage coil

Wattmeter measures the average power.

The reading of wattmeter will be,

Average power = Vpc Icc cos ϕ

Where,

Vpc is the voltage across pressure coil

Icc is current flows through the current coil

ϕ is the phase angle between Vpc and Icc

So, wattmeter is a device capable of detecting voltage, current and the angle between the voltage and the current to provide power readings.

Calculation:

Given,

Vpc = V = 200 ∠0° V

Z = Impedance = 4 +j3 = 5∠36.87° 

I = Current from supply = \({V\over Z}={200\angle0 \over 5\angle 36.87}=40\angle -36.87\)

Icc = 40 × (5/50) = 4 A

ϕ = 36.87°

From concept

Average power = 200 ×  4 ×  cos 36.87° = 640 W 

Concept:

Wattmeter reads the power and it is given by

P = VPC ICC cos ϕ

VPC is the voltage across pressure coil

ICC is current flows through the current coil

ϕ is the phase angle between VPC and ICC

Application:

The circuit representation of the given question is as shown below.

• The potential coil is connected across Z2.
• It reads the voltage across Z2 only.
• So, Wattmeter reads only power consumed by Z2.