# What are the Steps for Solving the Two-Wattmeter Method Problem

While accurately measuring a balanced three-phase load’s absolute power and power factor, the two-wattmeter method always comes in handy. Learning how to solve the 2-wattmeter method problem is essential to PE Power and FE Electrical exam preparation per the NCEES® examination guidelines.

In this guide, we’ll walk you through the step-by-step process of solving a two-wattmeter problem, from understanding wattmeter circuits to obtaining those critical power readings. So, buckle up to discover the steps for solving the two-wattmeter method problem.

## Introduction to Wattmeter

A wattmeter is an instrument used to measure electrical power (real power) in a circuit. It works based on the principle of electromagnetic induction and utilizes both current and voltage measurements.

### Wattmeter Circuit Diagram

The circuit diagram of a wattmeter typically consists of two main components: the current coil (CC) and the potential coil (PC). These coils are connected in series and parallel, respectively, with the load for power measurement.

#### Current Coil (CC)

The current coil is designed with low resistance and is connected in series with the load. It carries the load current (I) and produces a magnetic field (Bc) proportional to the current passing through it.

#### Potential Coil (PC)

The potential coil is designed with high resistance and is connected in parallel to the load. It is across the bag and exposed to the load voltage (V). The potential coil generates a magnetic field (Bp) proportional to the voltage applied across it.

#### Moving Element

Within the wattmeter is a moving element, typically a coil or an iron vane. The moving element is placed so that it experiences the combined magnetic fields (Bc and Bp) from both the current and potential coils.

#### Force and Deflection

When the load is energized, the interaction of the magnetic fields (Bc and Bp) creates a mechanical force on the moving element. The force is proportional to the product of the load current (I) and voltage (V), which corresponds to the real power (P = VI) consumed by the load.

#### Pointer or Display

The force causes the moving element to deflect. The deflection of the moving element is directly proportional to the real power consumed by the load. The movement of the moving element is linked to a pointer, which indicates the power value on a calibrated scale.

#### Calibration

Wattmeters are calibrated by passing known currents and voltages through the current and potential coils to ensure accurate measurements. The scale is adjusted accordingly to represent the actual power values accurately.

#### Wattmeter Circuit Connections

Two wattmeters are typically connected for power measurement in a balanced three-phase circuit. One wattmeter is connected between any two line phases, and the other is connected between one line phase and the neutral point (or another line phase in a delta-connected load).

The wattmeters measure the instantaneous power, the product of each phase’s voltage and current. The total real power (P) consumed by the load is the sum of the readings from the two wattmeters.

## How Does a Wattmeter Work?

A wattmeter consists of two main components: the current coil (CC) and the potential coil (PC). The CC is connected in series with the load, carrying the load current (I), while the PC is connected in parallel to the load, across which the load voltage (V) is applied.

Let’s discuss in detail how a wattmeter works.

### Magnetic Fields and Force

When current flows through the CC, it creates a magnetic field (Bc) proportional to the current. Similarly, when voltage is applied to the PC, it generates a magnetic field (Bp) proportional to the voltage. These magnetic fields interact, resulting in the deflection of a moving coil or iron element within the wattmeter.

#### Mathematical Representation

Let’s represent the magnetic fields the CC and PC generated as Bc = k1 * I and Bp = k2 * V, respectively, where k1 and k2 are proportionality constants.

### Force on the Moving Element

The force (F) on the moving element is given by the cross product of the magnetic fields and the angle (θ) between them:

F = Bc * Bp * sin(θ)

### Power Measurement

Since the wattmeter measures real power (P) in the circuit, the force experienced by the moving element is proportional to the product of current and voltage (P = VI). Thus, the force can be expressed as:

F = k * VI * sin(θ)

### Deflection of the Moving Element

The force F causes the moving element to deflect, and the deflection is proportional to the force. By proper calibration, the deflection of the moving element is directly related to the load’s power (P) consumed.

### Power Calculation

Two-wattmeter circuits measure the real power in a balanced three-phase circuit. The total real power (P) consumed by the load is the sum of the readings from the two wattmeters: P = W1 + W2.

## Fundamentals of Solving the Two-Wattmeter Method Problem

Before going into the practical implementation of 2 wattmeter method, let’s look at the industry-adapted fundamental process of solving the two-wattmeter method problem.

### Start – Problem Statement

Clearly state the problem you are trying to solve, including the type of load (balanced three-phase), the given values (voltages and currents), and what you are asked to find (total real power, power factor, etc.).

### Step 1 – Identity Given Values

List all the given values from the problem statement, such as RMS phase voltages (VRN, VYN, VBN), RMS phase currents (IR, IY, IB), and other relevant information.

### Step 2 – Phasor Diagram

Draw the phasor diagram using the given values to represent the voltages and currents in a balanced three-phase load. This will help you visualize the angles between the voltage and current phasors.

### Step 3 – Calculate Wattmeter Readings (W1 and W2)

Use the phasor diagram and the formulas for W1 and W2 to calculate the wattmeter readings.

### Step 4 – Calculate Total Real Power (P)

Find the total real power the load consumes by adding the wattmeter readings: P = W1 + W2.

### Step 5 – Calculate Power Factor Angle (ϕ)

After simplification, determine the power factor angle (ϕ) by taking the inverse tangent (tan-1). Simplification is done by dividing the value obtained in Step 5 with [W2-W1].

### Step 7 – Check for Inductive or Capacitive Load

Based on the calculated value of ϕ, determine whether the load is inductive or capacitive. If ϕ is positive, the load is inductive; if ϕ is negative, the load is capacitive.

### End – Finalize the Solution

Present your final results, including the total real power (P) and power factor angle (ϕ). Include units for all quantities, and check for signs indicating the load type. Do not forget to review your calculations and results to ensure accuracy and consistency.

Verify that the power factor angle (ϕ) matches the type of load (inductive or capacitive) you determined in Step 7.

**Remember to follow the correct formulas and trigonometric identities when calculating. This basic template will help you organize your approach and efficiently solve two-wattmeter method problems for balanced three-phase loads.*

## Steps for Solving the Two-Wattmeter Method Problem

The Two-Wattmeter Method is a technique used to measure the power consumed by a balanced three-phase load. It is commonly employed in power system analysis to determine the load’s real (active) and reactive (inductive or capacitive) power.

This method utilizes two wattmeters connected to the load so that their readings yield accurate measurements of the active and reactive power.

### Phasor Diagram

Before we proceed with the derivation, let’s understand the phasor diagram for the two-wattmeter method. The phasor diagram represents the relationship between the balanced three-phase load’s phase voltages and phase currents.

In a balanced load, the magnitudes of all phase voltages and phase currents are equal, and they are phase-shifted by the power factor angle (ϕ) in an inductive circuit.

**Definition of Quantities:**

- VRN, VYN, VBN: RMS values of phase voltages in a balanced three-phase load.
- IR, IY, IB: RMS values of phase currents in a balanced three-phase load.
- W1, W2: Power readings of the first and second wattmeters, respectively.
- ϕ (phi): The phase angle between phase voltages and phase currents, known as the power factor angle.
- P: The total real power (active power) the load consumes.

### Expression for Total Real Power (P)

The total real power consumed by the balanced three-phase load is the sum of the natural powers measured by both wattmeters. In a balanced load, the total real strength is the sum of the two-wattmeter readings.

P = (W1 + W2)

### Expression for Power Factor Angle (Q)

The power factor angle can be found by dividing W2-W1 with W1+W2.

### Step by Step Derivation

#### Step 1 – Derivations for Wattmeter Readings

The wattmeter readings, W1 and W2, can be found by the phasor calculations:

The voltage across Wattmeter W1 will be given by:

VRY = VRN – VYN

From the given phasor diagram, it is evident that: The phase angle between VRY and IR equals (30+ϕ). Therefore, the power will be given as follows:

W1 = VRN * IR * cos(30+ϕ) [Real power measured by the first wattmeter]

Similarly, the current flowing through the second wattmeter is given as:

VBY = VBN – VYN

The phasor diagram shows that the voltage VBY and IB phase angle equals (30-ϕ). Therefore, the power will be expressed as:

W2 = VBY * IB * cos(30-ϕ) [Real power measured by the second wattmeter]

Since VRY = VBY = VL (line voltage)

And

IR = IB = IL (line current)

Hence, the expressions for W1 and W2 can be written as

**W1 = VL * IL * cos(30+ϕ)**

**W2 = VL * IL * cos(30-ϕ)**

#### Step 2 – Total Real Power (P)

The total real power consumed in the circuit is the algebraic sum of the wattmeter readings. Therefore, the total real power (P) is:

P = W1 + W2

Substituting the expressions for W1 and W2 from steps 1 and 2:

P = VL * IL * cos(30+ϕ) + VL * IL * cos(30-ϕ)

Taking VL and IL as common factors

P = VL * IL *[ cos(30+ϕ) + cos(30-ϕ)]

Since, cosA + cosB = 2cos (A+B)/2 cos (A-B)/2

Therefore,

P = VL * IL * 2 * cos(30) * cos(ϕ)

Since cos(30) = √3 / 2

So

P = VL * IL * 2 * (√3 / 2) * cos(ϕ)

The total power will be:

**P = √3 * VL * IL * cos(ϕ)**

#### Step 3 – Power Factor Calculation

To find the power factor angle, we have:

W1 + W2 = √3 * VL * IL * cos(ϕ)

For W2-W1, we can write:

W2 – W1 = VL * IL * [cos(30-ϕ)-cos(30+ϕ)]

As we know, cosA – cosB = 2sin (A+B)/2 sin (B-A)/2

Applying the formula, we get:

W2 – W1 = VL * IL * [2sin(30-ϕ+30+ϕ)/2 sin (30+ϕ-30+ϕ)/2]

W2 – W1 = VL * IL * [2sin(60)/2 sin (2ϕ)/2]

W2 – W1 = VL * IL * [2sin(30) sin (ϕ)]

since , sin(30) = ½

Therefore,

W2 – W1 = VL * IL * [2 *½ sin (ϕ)]

**W2 – W1 = VL * IL * sin(ϕ)**

Divide by W1+W2 on both sides

**[W2 – W1]/[W1+W2] **= [VL * IL * sin(ϕ)] **/** [W1+W2]

Since, W1 + W2 = √3 * VL * IL * cos(ϕ)

So, put its value on the right-hand-side:

[W2 – W1]**/**[W1+W2] = [VL * IL * sin(ϕ)] **/** [√3 * VL * IL * cos(ϕ)]

After simplification, we will get:

[W2 – W1]**/**[W1+W2] = 1/√3 * sin(ϕ)**/**cos(ϕ)

Since sin(ϕ)/cos(ϕ) = tan(ϕ)

[W2 – W1]**/**[W1+W2] = 1/√3 * tan(ϕ)

Rearranging the terms, we will get the following:

√3 * [W2 – W1]**/**[W1+W2] = tan(ϕ)

tan(ϕ) = √3 * [W2 – W1]**/**[W1+W2]

To find the angle:

**(ϕ) = tan-1 { √3 * [W2 – W1]/[W1+W2] }**

## Practical Tips and Considerations

The wattmeter parameters, quality, and accuracy are significant for the correct reading and calculations. Below are helpful tips for solving two Wattmeter method problems without errors or mistakes.

### Best Practices for Accurate Measurements

**Calibration**– Ensure that the wattmeters used in the measurement are calibrated regularly to maintain accuracy. A certified calibration facility should calibrate industrial and commercial rules to avoid costly mistakes.**Proper Connections**– Double-check all connections in the circuit to ensure they are secure and correctly made. Loose circuit connections can lead to inaccurate readings.**Balanced Load**– Ensure the three-phase load is balanced, i.e., the line currents and voltages are equal in magnitude and 120 degrees apart. An unbalanced load can lead to inaccurate results.**Steady-State Conditions**– Perform measurements under steady-state conditions, as transient or dynamic behavior can affect the readings.**Zero Phase Shift**– Ideally, use wattmeters with negligible phase shifts. This ensures accurate measurement of true power.**Average Readings**– For highly fluctuating loads, average multiple readings to obtain more accurate results.

### Safety Precautions and Practices

**Electrical Safety Gear**– When working with electrical circuits, use appropriate personal protective equipment (PPE) such as insulated gloves, safety glasses, and non-conductive footwear.**De-Energize the Circuit**– De-energizing the circuit before making any connections or adjustments. Follow lockout/tagout procedures when necessary.**Appropriate Voltage and Current Ranges**– Ensuring that the wattmeters’ selected voltage and current ranges are suitable for the measurements to prevent damage to the instruments.**Grounding**– Ensuring the circuit is properly grounded to avoid potential faults or electrical hazards.**Avoid Touching Live Parts**– Refrain from touching live parts of the circuit, and always use insulated gloves.

### Troubleshooting Errors to Avoid Costly Mistakes

**Zero Reading Errors**– If the wattmeters show zero readings, check for loose connections, blown fuses, or faulty instruments.**Unstable Reading Errors**– Fluctuating or unstable readings may be due to unbalanced loads, poor connections, or interference from other electrical equipment. Recheck the connections and the load balance.**Incorrect Phase Shift**– If the phase shift between voltage and current is not exactly 30 degrees or the system is not balanced, verify the load connections and consider using correction factors if applicable.**Mismatched Reading Errors**– If the readings from the two wattmeters do not match, it could be due to phase angle errors or unbalanced loads. Review the circuit connections and load conditions.**Overload Protection Technique**– Ensure the wattmeters have adequate overload protection to avoid instrument damage during high-current or high-voltage conditions.**Thermal Factor**– Temperature variations can impact instrument accuracy. Keep the wattmeters within their specified temperature range for accurate readings.**Circuit Resonance**– In the presence of resonance, the wattmeters may not give accurate readings. Consider using power quality analyzers to analyze harmonics in the system.

### Conclusion

The two-wattmeter method is robust for accurate power measurement and power factor determination in balanced three-phase loads. The technique is helpful in the PE Power exam preparation and FE Electrical exam preparation per the NCEES® exam guidelines.

Utilizing the phasor diagram and mathematical expressions for wattmeter readings, we can calculate the total real power (P) and power factor angle to unveil valuable insights into the load’s power consumption.

Embracing this methodology ensures precise power analysis and empowers engineers and technicians to optimize electrical systems and enhance energy efficiency. Explore Study for FE to find comprehensive and efficient PE and FE exam preparation courses tailored to your needs.