Three-phase systems and power measurement notes — Unit 3

A single-phase system uses just two wires (phase and neutral) to transmit power. A polyphase system uses multiple identical single-phase systems with a fixed phase difference between them. The most universally adopted polyphase system is the Three-Phase System.

⇒Why Three-Phase over Single-Phase?

• More Power: A 3-phase machine produces 1.5 times more output than a single-phase machine of the same physical size.
• Constant Power: In single-phase, power pulses (goes to zero twice per cycle). In 3-phase, the total power delivered to a balanced load is perfectly constant, causing less vibration in motors.
• Self-Starting Motors: 3-phase systems naturally produce a Rotating Magnetic Field, making 3-phase induction motors self-starting. Single-phase motors are not.
• Copper Saving: For transmitting the same amount of power over a given distance at the same voltage, a 3-phase system requires roughly 25% less copper wire than a single-phase system.

An alternator (AC generator) generates 3-phase voltage by rotating three identical sets of coils (armature windings) within a magnetic field.

Crucially, these three coils (let's call them R, Y, and B) are physically displaced by exactly 120 degrees from each other in space.

As a result, the three induced EMFs have the exact same magnitude and frequency, but are displaced in time phase by 120°.

Let the three phase voltages be $V_R, V_Y, V_B$. Taking the R-phase as the reference (0°):

v_R = V_m sin(ωt)
v_Y = V_m sin(ωt - 120°)
v_B = V_m sin(ωt - 240°) = V_m sin(ωt + 120°)

The Phase Sequence is the order in which the three phases attain their positive maximum values.

By standard convention, the sequence is named after colors: Red (R), Yellow (Y), Blue (B).

• Positive Sequence (R-Y-B): R peaks first, then Y 120° later, then B 120° after that.
• Negative Sequence (R-B-Y): R peaks first, then B, then Y.

⇒Why it matters

The phase sequence strictly determines the direction of rotation in 3-phase induction motors. Reversing the sequence (swapping any two wires) will instantly reverse the motor's direction of rotation.

Since there are 3 coils, there are 6 wire ends. Running 6 wires to transmit power is uneconomical. We interconnect the coils to reduce the number of wires to 3 (or 4). There are two standard connections:

• 1. Star (Y) Connection: Similar ends (all starting ends or all finishing ends) of the three coils are joined together at a common point called the Neutral point.
• 2. Delta (Δ) Connection: The finishing end of the first coil is connected to the starting end of the second, and so on, forming a closed triangular loop.

A Star system usually employs 4 wires: 3 phase wires (Lines R, Y, B) and 1 neutral wire (N) originating from the center star point.

• Phase Voltage ($V_{ph}$): The voltage between any ONE line (R, Y, or B) and the Neutral point (N).
• Line Voltage ($V_L$): The voltage between any TWO lines (e.g., between R and Y).
• Phase Current ($I_{ph}$): The current flowing inside the individual coil.
• Line Current ($I_L$): The current flowing in the external line wires.

Look at the topology of a Star connection. The coil is directly in series with the outgoing line wire.

Because they are in series, whatever current flows through the phase coil MUST be the exact same current that flows out into the line.

Line Current = Phase Current
I_L = I_{ph}

However, the voltages are different. The voltage between two lines spans across two coils.

Let the phase voltages be $V_{RN}, V_{YN}, V_{BN}$. By Kirchhoff's Voltage Law, the Line voltage $V_{RY}$ (voltage between line R and line Y) is:

V_{RY} = V_{RN} - V_{YN}

This is a phasor subtraction, not simple arithmetic. On a phasor diagram, $V_{RN}$ is at 0° and $V_{YN}$ is at -120°.

-V_{YN} is a vector equal and opposite to V_{YN} (at +60°).
The resultant V_{RY} bisects the 60° angle between V_{RN} and -V_{YN}.
By parallelogram law: |V_{RY}| = √[ V_{RN}² + V_{YN}² + 2(V_{RN})(V_{YN})cos(60°) ]
Since V_{RN} = V_{YN} = V_{ph}:
|V_L| = √[ V_{ph}² + V_{ph}² + 2V_{ph}²(1/2) ] = √(3V_{ph}²)
V_L = √3 V_{ph}

A Delta system forms a closed loop. There is NO neutral point. It is strictly a 3-wire system.

• Phase Voltage ($V_{ph}$): Voltage across the individual coil.
• Line Voltage ($V_L$): Voltage between the two external lines.
• Phase Current ($I_{ph}$): Current inside the closed loop coil.
• Line Current ($I_L$): Current flowing in the external line wires.

Look at the topology of a Delta connection. The two ends of a phase coil are connected directly to the two external line wires.

Because they are in parallel (connected to the same two physical nodes), the voltage across the coil is identical to the voltage between the lines.

Line Voltage = Phase Voltage
V_L = V_{ph}

However, the currents are different. A line current splits into two phase currents at the junction node.

By Kirchhoff's Current Law at node R, the line current $I_R$ is the phasor difference of the two phase currents meeting at that node:

I_R = I_{RY} - I_{BR}

Similar to the Star voltage derivation, this is a phasor subtraction of two vectors separated by 120°.

|I_L| = √[ I_{ph}² + I_{ph}² + 2(I_{ph})(I_{ph})cos(60°) ]
I_L = √3 I_{ph}

• Balanced Load: The impedances of all three phases are identical in both magnitude and phase angle ($Z_R = Z_Y = Z_B$). In a balanced Star system, the current in the neutral wire is exactly zero.
• Unbalanced Load: The impedances are different. In an unbalanced Star system, a neutral current flows ($I_N = I_R + I_Y + I_B \neq 0$).

Most heavy industrial loads (like 3-phase motors) are perfectly balanced. Residential lighting loads distributed across phases are often unbalanced.

Total active power ($P$) is the sum of the power consumed by each of the three phases: $P = 3 \times P_{ph}$.

P = 3 × (V_{ph} I_{ph} cosφ)

Let's convert this to Line values (which are what we actually measure with meters). For Star: $V_{ph} = V_L/\sqrt{3}$ and $I_{ph} = I_L$. For Delta: $V_{ph} = V_L$ and $I_{ph} = I_L/\sqrt{3}$.

Substituting either case yields the exact same universal formula for 3-phase Active Power:

Active Power (P) = √3 V_L I_L cosφ  (Watts)
Reactive Power (Q) = √3 V_L I_L sinφ  (VAR)
Apparent Power (S) = √3 V_L I_L  (VA)

Blondel's Theorem states that to measure the total power in an $N$-wire system, you need exactly $N-1$ wattmeters.

• Three-Wattmeter Method: Used for 4-wire unbalanced Star systems. One wattmeter in each phase. $P_{Total} = W_1 + W_2 + W_3$.
• Two-Wattmeter Method: Used for 3-wire systems (both Star and Delta), balanced OR unbalanced. $P_{Total} = W_1 + W_2$.
• One-Wattmeter Method: Can only be used for perfectly balanced circuits by measuring one phase and multiplying by 3.

This is the most important method for exams. Two wattmeters are used to measure the total power of any 3-wire, 3-phase load.

⇒Connection Setup:

• The Current Coil (CC) of Wattmeter 1 ($W_1$) is connected in Line R. It measures $I_R$.
• The Pressure Coil (PC) of $W_1$ is connected between Line R and Line B. It measures $V_{RB}$.
• The CC of Wattmeter 2 ($W_2$) is connected in Line Y. It measures $I_Y$.
• The PC of $W_2$ is connected between Line Y and Line B. It measures $V_{YB}$.

Instantaneous readings of the wattmeters:

W₁ = v_{RB} · i_R = (v_R - v_B) · i_R
W₂ = v_{YB} · i_Y = (v_Y - v_B) · i_Y

Adding them together:

W₁ + W₂ = v_R(i_R) - v_B(i_R) + v_Y(i_Y) - v_B(i_Y)
W₁ + W₂ = v_R(i_R) + v_Y(i_Y) - v_B(i_R + i_Y)

By KCL in a 3-wire system, $i_R + i_Y + i_B = 0 \implies -(i_R + i_Y) = i_B$. Substituting this back:

W₁ + W₂ = v_R(i_R) + v_Y(i_Y) + v_B(i_B) = Total Instantaneous Power

This holds true for both balanced and unbalanced loads!

For a balanced lagging load (power factor angle $\phi$), we can use phasors to find the exact formulas for $W_1$ and $W_2$.

$W_1$ measures $V_{RB}$ and $I_R$. The phase angle between $V_{RB}$ and $I_R$ is $(30^\circ - \phi)$.

W₁ = V_L I_L cos(30° - φ)

$W_2$ measures $V_{YB}$ and $I_Y$. The phase angle between $V_{YB}$ and $I_Y$ is $(30^\circ + \phi)$.

W₂ = V_L I_L cos(30° + φ)

Because we have equations for $W_1$ and $W_2$, we can calculate the phase angle $\phi$ of the load just by reading the two meters.

W₁ + W₂ = √3 V_L I_L cosφ  (Total Active Power)
W₁ - W₂ = V_L I_L sinφ

Dividing $(W_1 - W_2)$ by $(W_1 + W_2)$:

tanφ = √3 [ (W₁ - W₂) / (W₁ + W₂) ]
Power Factor = cosφ

Depending on the power factor ($\cos\phi$), the two wattmeters will read differently:

⇒Unit 3 Mastery

• What are the advantages of a 3-phase system over single-phase?
• Derive the relationship between Line and Phase voltages in a Star connection.
• Derive the relationship between Line and Phase currents in a Delta connection.
• Prove that $W_1 + W_2$ equals total power in the Two-Wattmeter method.
• Derive the formula $\tan\phi = \sqrt{3}[(W_1-W_2)/(W_1+W_2)]$.
• What happens to wattmeter readings if the power factor is exactly 0.5?