Notes › EENG 3345: AC Circuit Analysis Lecture 6

Three-Phase Circuits

calendar_today July 2, 2025 article 398 words access_time 3 min replay Modified July 7, 2025

Table of Contents

Three-Phase Circuits

Definition 06.1 (Three-Phase Voltages).

Consists of three sinusoidal voltages of equal magnitude but phase-shifted by $120^{\circ}$. Advantage: constant power transmission.

Balanced condition: $V_{aa^{'}} + V_{bb^{'}} + V_{cc^{'}} = 0$
- This also implies the resistive loads in each branch are the same, since the potential at each point is identical
For notational simplicity: $V_{aa^{'}} = V_a$ and so forth

For any balanced bridge circuit, the phase voltages should be the same

Current Relationships

Line currents: $I_a, I_b, I_c$
Phase currents: $I_{ab}, I_{bc}, I_{ca}$
Phase to line current: $I_{m}\sqrt{3} \angle -30^{\circ}$

$$\begin{align} I_{ab} &= I_{m} \cos \omega t \\ &= I_{m} \angle 0^{\circ} \\ I_{bc} &= I_{m} \cos(\omega t - 120^{\circ}) \\ &= I_{m} \angle -120^{\circ} \\ I_{ca} &= I_{m} \cos(\omega t + 120^{\circ}) \\ &= I_{m} \angle 120^{\circ} \end{align}$$

For Delta Configuration:

$$I_{a} = I_{ab} - I_{ca} = \sqrt{3} I_{m} \angle -30^{\circ}$$

Voltage Relationships

Line voltages: $V_{ab}, V_{bc}, V_{ca}$
Phase voltages: $V_a, V_b, V_c$
Phase to line voltage: $V_{m} \sqrt{3} \angle 30^{\circ}$

Line voltages:

$$\begin{align} V_{aa^{'}} &= V_{m}\cos \omega t \\ &= Vm \angle 0^{\circ}\\ V_{bb^{'}} &= V_{m}\cos(\omega t - 120^{\circ}) \\ &= Vm \angle -120^{\circ}\\ V_{cc^{'}} &= V_{m}\cos(\omega t - 240^{\circ}) \\ &= Vm \angle 120^{\circ}\\ \end{align}$$

For Delta Configuration:

$$V_{ab} = V_{a} - V_{b} = \sqrt{3} V_{m} \angle 30^{\circ}$$

Bridge Circuits

Y-to-Y Circuit

Three voltage sources connected in a Y configuration to three impedances in a Y configuration
Assuming the voltage sources are three-phase and the impedances match, then the neutral current in the middle $i_N$ is zero
Otherwise, use nodal analysis to determine the currents
Phase currents equal line currents

D-to-D Circuit

Three voltage sources connected in a D configuration to three impedances in a D configuration
The current through each impedance is the corresponding voltage source times the admittance

Y-to-D Circuit

Use KCL and KVL

Power

Total real power (balanced load)

$$P_{\text{total}} = \sqrt{3} V_{L} I_{L} \cos(\phi)$$

Apparent power:

$$ |S| = \sqrt{3} V_{L} I_{L}$$

Reactive power:

$$ Q = \sqrt{3} V_{L} I_{L}\sin(\phi)$$

Examples

Example 06.2.

In a purely resistive circuit, the voltage and current are:

$$v(t) = 230\cos(\omega t), i(t) = 10\cos(\omega t)$$

Calculate the real power consumed by the circuit
Determine the apparent power
What is the power factor?

06 Three-Phase Circuits 2025-07-02 11.27.11

Example 06.3.

A balanced Y-Y system has a phase voltage $V_{ph} = 120V$ and resistive load $Z = 10\Omega$. Find the line current and the total power consumed.

06 Three-Phase Circuits 2025-07-03 09.46.08

References

Three-Phase Circuits (course handout)

Sources

Three-Phase Circuits

Graph

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