# Three-Phase Circuits in PE Power Exam

Three-phase power systems form the backbone of modern electrical infrastructure, powering industries, commercial establishments, and even our homes. With their efficiency, reliability, and ability to handle immense power loads, three-phase circuits are paramount in power engineering.

In a three-phase power system, electrical energy is distributed using three alternating current (AC) waveforms 120 degrees out of phase. The symmetrical and balanced nature of three-phase circuits results in smoother power delivery and efficient use of conductors and equipment.

Phasor representation allows us to visualize the magnitude and phase relationships between the voltages and currents in a three-phase system, aiding in their analysis and design.

This detailed guide on Three-Phase Circuits in PE Power Exam will explore various aspects of three-phase circuits, including impedance and admittance calculations, voltage drop analysis, unbalanced load considerations, and fault analysis.

By understanding the three-phase power question discussed in this guide, you can effectively design, analyze, and troubleshoot three-phase systems, ensuring reliable and optimized power distribution.

## Basics of Three-Phase Power Systems

Three-Phase Circuits in the PE Power Exam is a crucial topic from an academic and career perspective. Three-phase power systems are widely used in electrical power distribution systems and industrial applications due to their numerous advantages over single-phase methods.

Three-phase power is a system that utilizes three alternating currents (AC) 120 degrees out of phase with each other. This arrangement provides a more efficient and balanced power distribution than single-phase systems.

The table below outlines the key differences and highlights the advantages and characteristics Single Phase vs. Three Phase power circuits in terms of power delivery, efficiency, transmission, equipment size, motor operation, and fault tolerance.

### Phasor Representation

Phasor representation is a mathematical technique for analyzing and visualizing three-phase power systems. It represents the quantities in a three-phase system, such as voltage and current, as vectors or phasors in a complex plane. Phasors represent the magnitude, phase angle, and frequency of the quantities.

#### Voltage Phasors

In a three-phase system, three voltage phasors represent the three voltages. These phasors are typically labeled as Va, Vb, and Vc and are displaced from each other by 120 degrees (electrical degrees). The magnitude and phase angle of each voltage phasor corresponds to the instantaneous values of the voltages at any given time.

#### Current Phasors

Similarly, three current phasors represent the three currents in a three-phase system. These phasors are usually denoted as Ia, Ib, and Ic and are separated by 120 degrees. The magnitude and phase angle of each current phasor corresponds to the instantaneous values of the currents.

### Phasor Relationships

Specific relationships relate to phasors in a three-phase power system. These relationships form the basis for analyzing and understanding three-phase circuits.

Before moving further, it is recommended to explore balanced & unbalanced circuits and the two standard layouts of three-phase circuits as Delta and Star (Wye) Circuits. Read our detailed on Circuits in the PE Power exam for a detailed account of these different three-phase complex power circuits.

Below are some critical Phasor relationships you must know for preparing three-phase circuits in the PE Power exam.

#### Phase Sequence

The phase sequence refers to the order in which the voltage and current phasors follow each other. In a balanced system, the commonly used sequences are

• Phase sequence RYB – Va leads Vb, which leads Vc by 120 degrees. Similarly, Ia leads Ib, which leads Ic by 120 degrees.
• Phase sequence RBY – Va leads Vc, which leads Vb by 120 degrees. Similarly, Ia leads Ic, which leads Ib by 120 degrees.

#### Line Voltages and Phase Voltages

A three-phase system has two types of voltages: line and phase.

• Line voltages (VL) are those between the three conductors or lines, such as Va-Vb, Vb-Vc, and Vc-Va.
• Phase voltages (Vph) are the voltages between each line and the neutral point, such as Va-neutral, Vb-neutral, and Vc-neutral.

The relationship between line voltages and phase voltages depends on the type of connection:

• In a delta (Δ) connection, VL = Vph.
• In a star (Y) connection, VL = √3 * Vph.

#### Power Relationships

In three-phase power systems, power is commonly measured using complex power (S) and power factor (PF). Some meaningful relationships are:

• Apparent power (S) is the product of the line voltage (VL) and the line current (IL): S = VL * IL.
• The Real power (P) is the actual power consumed by the load and is the product of the line voltage (VL), line current (IL), and power factor (PF): P = VL * IL * PF.
• Reactive power (Q) is the power that oscillates between the source and load due to inductive or capacitive elements. It is the product of the line voltage (VL), line current (IL), and the sine of the phase angle difference between them: Q

## Three-Phase Circuit Connections

Let’s explore the two crucial types of connections and their respective analysis in the three-phase circuits in the PE Power Exam.

### Three-Phase Delta (Δ) Circuit Analysis

Consider a three-phase delta-connected circuit with line voltages (VL) 400V and line currents (IL) of 10A. Let’s determine the phase voltages (Vph), total power (S), real power (P), and reactive power (Q) in the circuit.

Step 1 – Determine the Phase Voltages (Vph)

The line voltages (VL) and phase voltages (Vph) are equal in a delta-connected circuit. Therefore, Vph = VL.

Thus, Vph = 400V.

Step 2 – Calculate the Apparent Power (S)

The apparent power (S) is given by the product of line voltage (VL) and line current (IL):

S = VL * IL

= 400V * 10A

= 4000 VA

Step 3 – Calculate the Real Power (P)

The real power (P) is the actual power the load consumes. The real three-phase power formula is given by the product of apparent power (S) and power factor (PF):

P = S * PF

We need to know the load’s impedance or power factor angle to calculate the power factor. Assuming a power factor angle of 30 degrees, we can calculate the power factor (PF) using the cosine of the angle:

PF = cos(30°)

= 0.866

Now, we can calculate the real power:

P = 4000 VA * 0.866

= 3464 W

Step 4 – Calculate the Reactive Power (Q)

The reactive power (Q) is the power that oscillates between the source and load due to inductive or capacitive elements. It is given by the product of apparent power (S) and the sine of the power factor angle (θ):

Q = S * sin(θ)

For a power factor angle of 30 degrees, we have:

Q = 4000 VA * sin(30°)

= 2000 VAR

Therefore, the phase voltages (Vph) are 400V, the total power (S) is 4000 VA, the real power (P) is 3464 W, and the reactive power (Q) is 2000 VAR in the delta-connected circuit.

### Three-Phase Wye (Star) Circuit Analysis

Consider a three-phase wye-connected circuit with phase voltages (Vph) of 230V and line currents (IL) of 15A. Let’s determine the line currents (IL), total power (S), real power (P), and reactive power (Q) in the circuit.

Step 1 – Determine the Line Currents (IL)

In a wye-connected circuit, the line currents (IL) are related to the phase currents (Iph) by the square root of 3 (√3). Therefore, IL = √3 * Iph.

Given that Vph = 230V and IL = 15A, we can calculate Iph as follows:

IL = √3 * Iph

15A = √3 * Iph

Iph = 15A / √3

Iph ≈ 8.66 A

Step 2 – Calculate the Apparent Power (S)

The apparent power (S) is given by the product of line voltage (VL) and line current (IL):

S = VL * IL

= 230V * 15A

= 3450 VA

Step 3 – Calculate the Real Power (P)

Assuming a power factor angle of 45 degrees, we can calculate the power factor (PF) using the cosine of the angle:

PF = cos(45°)

= 0.707

Now, we can calculate the real power:

P = 3450 VA * 0.707

= 2439 W

Step 4 – Calculate the Reactive Power (Q)

For a power factor angle of 45 degrees, we have:

Q = 3450 VA * sin(45°)

= 2439 VAR

Therefore, the line currents (IL) are approximately 15A, the total power (S) is 3450 VA, the real power (P) is 2439 W, and the reactive power (Q) is 2439 VAR in the wye-connected circuit.

*Power factor (PF) – It represents the ratio of real power (P) to the apparent power (S) in an AC circuit. It indicates the phase difference between voltage and current waveforms.

*Reactive power (Q) – It is the power that oscillates between the source and load due to inductive or capacitive elements. It is responsible for exchanging energy between the source and the load.

## Analysis and Modeling of Three-Phase Circuits

Let’s analyze three-phase circuits for some other critical parameters that significantly impact the performance, efficiency, and reliability of your three-phase circuits.

### Impedance Calculation

Impedance measures the total opposition to the current flow in an AC circuit. It includes both resistance and reactance components. Impedance is represented by the Z and measured in ohms (Ω). It is calculated by dividing the voltage (V) by the current (I) in an AC circuit.

Consider a three-phase circuit with a delta (Δ) connection. The line currents (IL) are 10A, and the line voltages (VL) are 400V. Let’s calculate the impedance (Z) of the circuit.

Step 1 – Determine the Phase Currents (Iph)

In a delta-connected circuit, the line currents (IL) and phase currents (Iph) are equal. Therefore, Iph = IL.

Thus, Iph = 10A.

Step 2 – Calculate the Impedance (Z)

Impedance (Z) represents the total opposition to the current flow in an AC circuit. It comprises resistance (R) and reactance (X) components. The impedance can be calculated using Ohm’s law:

Z = VL / Iph

Given that VL = 400V and Iph = 10A, we can calculate the impedance:

Z = 400V / 10A

= 40 Ω

Therefore, the impedance (Z) of the circuit is 40 Ω.

Admittance (Y) is another crucial topic you must prepare while studying the Three-Phase Circuits in PE the Power Exam. Admittance measures the ease with which current flows in an AC circuit.

It is the reciprocal of impedance (Y = 1 / Z). Admittance is represented by the symbol Y and is measured in Siemens (S). It is calculated by dividing the current (I) by the voltage (V) in an AC circuit.

Consider a three-phase circuit with a wye (star) connection. The line current (IL) is 15A, and the line voltage (VL) is 230V. Let’s see how to calculate the admittance (Y) of the circuit.

Step 1 – Determine the Phase Currents (Iph)

In a wye-connected circuit, the line currents (IL) are related to the phase currents (Iph) by the square root of 3 (√3). Therefore, Iph = IL / √3.

Given that IL = 15A, we can calculate Iph as follows:

Iph = 15A / √3

Iph ≈ 8.66 A

Step 2 – Calculate the Admittance (Y)

Admittance (Y) represents the ease of current flow in an AC circuit. It is the reciprocal of impedance (Y = 1 / Z). The admittance can be calculated using Ohm’s law:

Y = IL / VL

Given that IL = 15A and VL = 230V, we can calculate the admittance:

Y = 15A / 230V

≈ 0.0652 S

Therefore, the admittance (Y) of the circuit is approximately 0.0652 S (Siemens).

### Voltage Drop Analysis

A voltage drop of the three-phase circuits in the PE Power Exam refers to the reduction in voltage that occurs when current flows through an impedance or resistance in a circuit.

It is calculated using Ohm’s law, which states that voltage drop (Vd) is equal to the product of current (I) and impedance (Z). Voltage drop is an essential consideration for proper circuit analysis and design.

Consider a three-phase circuit with a line current (IL) of 20A and an impedance (Z) of 8 Ω. Let’s discover how to calculate the voltage drop (Vd) across the impedance.

Step 1 – Calculate the Voltage Drop (Vd)

Voltage drop (Vd) across an impedance (Z) can be calculated using Ohm’s law:

Vd = IL * Z

Given that IL = 20A and Z = 8 Ω, we can calculate the voltage drop:

Vd = 20A * 8 Ω

= 160 V

Therefore, the voltage drop (Vd) across the impedance is 160V.

## Analyzing Unbalanced Three-Phase Circuits

Analyzing and calculating loads and faults in unbalanced circuits is necessary to ensure electrical systems’ proper functioning, reliability, and safety. Unbalanced loads can lead to uneven current, voltage, and power distribution, causing inefficient operation and potential equipment damage.

Engineers can identify potential issues, design appropriate protection systems, and take corrective measures by understanding the characteristics of unbalanced loads and faults, such as uneven current distribution and short circuits.

These measures help maintain system stability, optimize performance, and prevent hazards, ultimately enhancing the overall efficiency and safety of the electrical system.

Consider a three-phase system with a wye (star) connected load. The phase currents are Ia = 10A, Ib = 12A, and Ic = 15A. The line voltages are Vab = 400V, Vbc = 415V, and Vca = 380V. Let’s determine the unbalanced load’s neutral current and the total apparent power (S).

Step 1 – Calculate the Neutral Current (In)

In a wye-connected load, the neutral current (In) can be calculated by summing the phase currents:

In = Ia + Ib + Ic

= 10A + 12A + 15A

= 37A

Step 2 – Calculate the Total Apparent Power (S)

The total apparent power (S) is the sum of the apparent powers of individual phases. The apparent power (S) of each phase can be calculated using the formula:

S = VL * IL

For Phase A:

Sa = Vab * Ia

= 400V * 10A

= 4000 VA

For Phase B:

Sb = Vbc * Ib

= 415V * 12A

= 4980 VA

For Phase C:

Sc = Vca * Ic

= 380V * 15A

= 5700 VA

Now, we can calculate the total apparent power (S) by summing the individual apparent powers:

S = Sa + Sb + Sc

= 4000 VA + 4980 VA + 5700 VA

= 14680 VA

Therefore, the neutral current (In) is 37A, and the unbalanced load’s total apparent power (S) is 14680 VA.

### Fault Analysis

Consider a three-phase system with a delta (Δ) connected load. A fault occurs between phases A and B, resulting in a short circuit with a fault impedance (Zf) of 0.1 + j0.2 Ω.

The line voltage is 400V, and the line current before the fault is 15A. Let’s see how to calculate the fault current (If), and the fault power (Pf) dissipated in the fault.

Step 1 – Calculate the Fault Current (If)

The fault current (If) can be calculated using Ohm’s Law:

If = Vf / Zf

where Vf is the fault voltage across the fault impedance (Zf).

Given that Vf = 400V, and Zf = 0.1 + j0.2 Ω, we can calculate the fault current:

If = 400V / (0.1 + j0.2) Ω

≈ 400V / 0.223 Ω

≈ 1793.72 A ∠ 71.57°

Therefore, the fault current (If) is approximately 1793.72 A with an angle of 71.57 degrees.

Step 2 – Calculate the Fault Power (Pf)

The fault power (Pf) dissipated in the fault can be calculated as the product of the fault current (If) and the fault voltage (Vf):

Pf = |If|^2 * Re(Zf)

where |If| denotes the magnitude of the fault current and Re(Zf) represents the real part of the fault impedance.

Given that |If| = 1793.72 A and Re(Zf) = 0.1 Ω, we can calculate the fault power:

Pf = (1793.72 A)^2 * 0.1

Pf = 3.2153725984 × 10^5 W

Therefore, the fault power (Pf) is 3.2153725984 × 10^5 watts.

## Symbols and Representations

The table below outlines all the key symbols we used for different three-phase quantity measurements in this guide.

Note these three-phase quantity measurements & symbols and practice with different Three Phase Power Formulas to master the complex mathematics associated with Three-Phase Circuits in PE Power Exam.

### Conclusion

Now you have a rich understanding of all the key aspects of Three-Phase Circuits in PE Power Exam. By mastering these concepts and calculations involved in three-phase power questions and problems, you can unlock new opportunity corridors for success not only in your academic studies but also in your future career as a professional power engineer.

The knowledge and skills acquired in understanding three-phase circuits will prove invaluable as you tackle three-phase complex power systems, analyze power distribution networks, and ensure the efficient and reliable operation of critical power infrastructures.

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Licensed Professional Engineer in Texas (PE), Florida (PE) and Ontario (P. Eng) with consulting experience in design, commissioning and plant engineering for clients in Energy, Mining and Infrastructure.