3-Phase AC Circuits 1

LessonAbout 2 h1 of 2 in 3P AC Circuits

3P AC Circuits 1

3-phase AC circuit is a polyphase electrical system powered by three sinusoidal voltages of equal frequency and magnitude with \(120^\circ\) phase displacement.

1. Polyphase System

Polyphase system consists of two or more similar single-phase systems with voltage maximums occurring at different times.

  • Polyphase machines are simpler, cheaper, and more efficient
  • Polyphase transmission lines deliver more power per conductor weight
  • Voltage regulation is inherently superior to single-phase lines

2. n-Phase Star Connected

n-phase star connected system connects one terminal of each phase winding together at a common neutral point.

n-Phase Star Line Voltage

\[ V_L = 2 V_P \sin\left(\dfrac{180^\circ}{n}\right) \]
\[ I_L = I_P \]

Where: \(V_L\) = line-to-line voltage (V), \(V_P\) = phase voltage (V), \(I_L\) = line current (A), \(I_P\) = phase current (A), \(n\) = number of phases

Example. A 4-phase star-connected system is connected across a \(220\ \text{V}\) supply. Calculate the phase voltage of the system.

Solution.

\[ V_P = \dfrac{V_L}{2\sin\left(\dfrac{180^\circ}{n}\right)} = \dfrac{220}{2\sin\left(\dfrac{180^\circ}{4}\right)} = 155.56\ \text{V} \]

Answer: \(155.56\ \text{V}\)

3. n-Phase Mesh Connected

n-phase mesh connected system connects phase windings end-to-end to form a closed geometric polygon.

n-Phase Mesh Line Current

\[ I_L = 2 I_P \sin\left(\dfrac{180^\circ}{n}\right) \]
\[ V_L = V_P \]

Where: \(I_L\) = line current (A), \(I_P\) = phase current (A), \(V_L\) = line voltage (V), \(V_P\) = phase voltage (V), \(n\) = number of phases

4. n-Phase Active Power Formula

n-phase active power is the total real electric power delivered by an \(n\)-phase polyphase system.

n-Phase Active Power

\[ P = n V_P I_P \cos\theta \]

Where: \(P\) = active power (W), \(n\) = number of phases, \(V_P\) = phase voltage (V), \(I_P\) = phase current (A), \(\cos\theta\) = load power factor

Example. A 4-phase star-connected system with a phase voltage of \(155.56\ \text{V}\) supplies a phase current of \(50\ \text{A}\) at \(0.85\) power factor. Calculate the active power drawn.

Solution.

\[ P = n V_P I_P \cos\theta = 4(155.56)(50)(0.85) = 26445.2\ \text{W} \]

Answer: \(26445.2\ \text{W}\)

5. Three-Phase System

Three-phase system is an interconnection of three single-phase systems with voltages displaced \(120^\circ\) apart.

  • Phase displacement is \(120^\circ\) electrical (\(2\pi/3\ \text{rad}\))
  • Equal amplitudes for balanced generated voltages
  • Zero instantaneous sum of line voltages (\(v_A + v_B + v_C = 0\))

6. Phase Sequence

Phase sequence is the time order in which phase voltages reach their maximum positive values.

  • Positive sequence is \(a\text{-}b\text{-}c\) or 1-2-3 rotation
  • Negative sequence is \(a\text{-}c\text{-}b\) or 1-3-2 rotation
  • Counterclockwise rotation is assumed standard vector rotation

7. Wye or Star-Connected Balanced System

Wye-connected balanced system connects three phase windings to a common neutral point \(n\).

Wye Balanced System

\[ V_L = \sqrt{3} V_P \]
\[ I_L = I_P \]
\[ V_{12} = \sqrt{3} V_{1n} \angle 30^\circ \]

Where: \(V_L\) = line-to-line voltage (V), \(V_P\) = phase-to-neutral voltage (V), \(I_L\) = line current (A), \(I_P\) = phase current (A)

Example. A balanced Wye-connected load with phase impedance \(Z_P = 5\ \Omega\) is connected across a \(220\ \text{V}\), 3-phase supply. Calculate the line current.

Solution.

\[ I_L = I_P = \dfrac{E_L / \sqrt{3}}{Z_P} = \dfrac{220 / \sqrt{3}}{5} = 25.4\ \text{A} \]

Answer: \(25.4\ \text{A}\)

8. Delta or Mesh-Connected Balanced System

Delta-connected balanced system connects three phase windings end-to-end in a closed triangular loop.

Delta Balanced System

\[ V_L = V_P \]
\[ I_L = \sqrt{3} I_P \]
\[ I_P = \dfrac{V_L}{Z_P} \]

Where: \(V_L\) = line voltage (V), \(V_P\) = phase voltage (V), \(I_L\) = line current (A), \(I_P\) = phase current (A), \(Z_P\) = phase impedance (\(\Omega\))

Example. A balanced delta-connected load has a phase impedance of \(20\ \Omega\) connected across a \(230\ \text{V}\), 3-phase supply. Determine the line current.

Solution.

\[ I_P = \dfrac{V_L}{Z_P} = \dfrac{230}{20} = 11.5\ \text{A} \]
\[ I_L = \sqrt{3} I_P = \sqrt{3}(11.5) = 19.918\ \text{A} \]

Answer: \(19.918\ \text{A}\)

9. Wye Connected Balanced Loads

Wye-connected balanced load consists of three equal impedances connected in star configuration.

Wye Balanced Load Current

\[ Z_1 = Z_2 = Z_3 = Z \]
\[ I_P = \dfrac{V_P}{Z} = \dfrac{V_L}{\sqrt{3} Z} \]

Where: \(Z\) = load impedance per phase (\(\Omega\)), \(V_P\) = phase voltage (V), \(V_L\) = line voltage (V), \(I_P\) = phase current (A)

Example. A balanced Wye-connected load with phase impedance \(Z = 20\ \Omega\) is supplied by a \(240\ \text{V}\) phase voltage. Calculate the phase current.

Solution.

\[ I_P = \dfrac{V_P}{Z} = \dfrac{240}{20} = 12\ \text{A} \]

Answer: \(12\ \text{A}\)

10. Neutral Current

Neutral current is the current returning through the neutral wire equal to the phasor sum of line currents.

Neutral Current

\[ I_N = I_1 + I_2 + I_3 \]
\[ I_1 + I_2 + I_3 = 0 \]

Where: \(I_N\) = neutral current (A), \(I_1, I_2, I_3\) = line currents (A)

EE Board Exam October 1998

Example. The phase currents of a Wye-connected transformer are \(I_a = 10\angle -30^\circ\ \text{A}\), \(I_b = 12\angle 215^\circ\ \text{A}\), and \(I_c = 15\angle 82^\circ\ \text{A}\). Calculate the neutral current.

Solution.

\[ I_N = I_a + I_b + I_c = 10\angle -30^\circ + 12\angle 215^\circ + 15\angle 82^\circ = 3.11\angle 72.8^\circ\ \text{A} \]

Answer: \(3.11\angle 72.8^\circ\ \text{A}\)

11. Delta Balanced or Unbalanced Load

Delta-connected load consists of three impedances connected across pairs of supply lines.

Unbalanced Delta Line Currents

\[ I_{12} = \dfrac{V_{12}}{Z_1} \qquad\qquad I_{23} = \dfrac{V_{23}}{Z_2} \qquad\qquad I_{31} = \dfrac{V_{31}}{Z_3} \]
\[ I_1 = I_{12} - I_{31} \qquad\qquad I_2 = I_{23} - I_{12} \qquad\qquad I_3 = I_{31} - I_{23} \]

Where: \(I_{12}, I_{23}, I_{31}\) = phase currents (A), \(V_{12}, V_{23}, V_{31}\) = line voltages (V), \(Z_1, Z_2, Z_3\) = phase impedances (\(\Omega\)), \(I_1, I_2, I_3\) = line currents (A)

EE Board Exam April 1982

Example. An unbalanced delta load has branch impedances \(Z_{ab} = 10\angle 53.13^\circ\ \Omega\), \(Z_{bc} = 10\angle 0^\circ\ \Omega\), and \(Z_{ca} = 10\angle 30^\circ\ \Omega\) across a \(200\ \text{V}\) 3-phase supply. Calculate the current in line b.

Solution.

\[ I_{ab} = \dfrac{200\angle 0^\circ}{10\angle 53.13^\circ} = 20\angle -53.13^\circ\ \text{A} \qquad\qquad I_{bc} = \dfrac{200\angle 240^\circ}{10\angle 0^\circ} = 20\angle 240^\circ\ \text{A} \]
\[ I_b = I_{bc} - I_{ab} = (-10 - j17.32) - (12 - j16) = 22.04\angle -176.57^\circ\ \text{A} \]

Answer: \(22.04\ \text{A}\)

12. Three-Phase Power Formulas

Three-phase power formulas compute active power, apparent power, and reactive power in balanced 3-phase circuits.

Three-Phase Power

\[ P = 3 V_P I_P \cos\theta = \sqrt{3} V_L I_L \cos\theta \]
\[ S = 3 V_P I_P = \sqrt{3} V_L I_L \]
\[ Q = 3 V_P I_P \sin\theta = \sqrt{3} V_L I_L \sin\theta \]

Where: \(P\) = active power (W), \(S\) = apparent power (VA), \(Q\) = reactive power (VAR), \(V_P\) = phase voltage (V), \(I_P\) = phase current (A), \(V_L\) = line voltage (V), \(I_L\) = line current (A), \(\cos\theta\) = power factor

EE Board Exam April 1985

Example. A balanced 3-phase load draws \(120\ \text{A}\) line current at \(230\ \text{V}\) line-to-line with \(0.848\) lagging power factor. Solve for the real power.

Solution.

\[ P = \sqrt{3} V_L I_L \cos\theta = \sqrt{3}(230)(120)(0.848) = 40540\ \text{W} = 40.54\ \text{kW} \]

Answer: \(40.54\ \text{kW}\)

13. Formula Summary

Formula NameFormulaVariables
n-Phase Star Line Voltage\(V_L = 2 V_P \sin\left(\dfrac{180^\circ}{n}\right)\)
\(I_L = I_P\)		\(V_L\) = line voltage (V), \(V_P\) = phase voltage (V), \(I_L\) = line current (A), \(I_P\) = phase current (A), \(n\) = number of phases
n-Phase Mesh Line Current\(I_L = 2 I_P \sin\left(\dfrac{180^\circ}{n}\right)\)
\(V_L = V_P\)		\(I_L\) = line current (A), \(I_P\) = phase current (A), \(V_L\) = line voltage (V), \(V_P\) = phase voltage (V), \(n\) = number of phases
n-Phase Active Power\(P = n V_P I_P \cos\theta\)\(P\) = active power (W), \(n\) = number of phases, \(V_P\) = phase voltage (V), \(I_P\) = phase current (A), \(\cos\theta\) = power factor
Wye Balanced System\(V_L = \sqrt{3} V_P\)
\(I_L = I_P\)
\(V_{12} = \sqrt{3} V_{1n} \angle 30^\circ\)		\(V_L\) = line voltage (V), \(V_P\) = phase voltage (V), \(I_L\) = line current (A), \(I_P\) = phase current (A)
Delta Balanced System\(V_L = V_P\)
\(I_L = \sqrt{3} I_P\)
\(I_P = \dfrac{V_L}{Z_P}\)		\(V_L\) = line voltage (V), \(V_P\) = phase voltage (V), \(I_L\) = line current (A), \(I_P\) = phase current (A), \(Z_P\) = phase impedance (\(\Omega\))
Wye Balanced Load Current\(I_P = \dfrac{V_P}{Z} = \dfrac{V_L}{\sqrt{3} Z}\)\(I_P\) = phase current (A), \(V_P\) = phase voltage (V), \(V_L\) = line voltage (V), \(Z\) = phase impedance (\(\Omega\))
Neutral Current\(I_N = I_1 + I_2 + I_3\)
\(I_1 + I_2 + I_3 = 0\)		\(I_N\) = neutral current (A), \(I_1, I_2, I_3\) = line currents (A)
Unbalanced Delta Line Currents\(I_{12} = \dfrac{V_{12}}{Z_1}\)
\(I_{23} = \dfrac{V_{23}}{Z_2}\)
\(I_{31} = \dfrac{V_{31}}{Z_3}\)
\(I_1 = I_{12} - I_{31}\)		\(I_{12}, I_{23}, I_{31}\) = phase currents (A), \(V_{12}, V_{23}, V_{31}\) = line voltages (V), \(Z_1, Z_2, Z_3\) = phase impedances (\(\Omega\)), \(I_1\) = line current (A)
Three-Phase Active Power\(P = 3 V_P I_P \cos\theta\)
\(P = \sqrt{3} V_L I_L \cos\theta\)		\(P\) = active power (W), \(V_P\) = phase voltage (V), \(I_P\) = phase current (A), \(V_L\) = line voltage (V), \(I_L\) = line current (A), \(\cos\theta\) = power factor
Three-Phase Apparent Power\(S = 3 V_P I_P\)
\(S = \sqrt{3} V_L I_L\)		\(S\) = apparent power (VA), \(V_P\) = phase voltage (V), \(I_P\) = phase current (A), \(V_L\) = line voltage (V), \(I_L\) = line current (A)
Three-Phase Reactive Power\(Q = 3 V_P I_P \sin\theta\)
\(Q = \sqrt{3} V_L I_L \sin\theta\)		\(Q\) = reactive power (VAR), \(V_P\) = phase voltage (V), \(I_P\) = phase current (A), \(V_L\) = line voltage (V), \(I_L\) = line current (A), \(\theta\) = power factor angle (\(^\circ\))