Three-Phase AC Power & Machine Analyzer

Interactive visualization of Balanced and Unbalanced 3-Phase Voltage, Current, and Power Flows

Leading (Capacitive)ResistiveLagging (Inductive)
Total Active (P_tot)
0.000 W
Total Reactive (Q_tot)
0.000 VAR
Total Apparent (S_tot)
0.000 VA
Total Power Factor
0.000
1. Three-Phase Voltages & Currents
2. Instantaneous Powers & The Constant Power Theorem

Note: In a balanced system, notice how the individual oscillating phase powers (dotted lines) sum perfectly to a flat, constant total power line (solid green). This is the fundamental reason 3-phase machines produce smooth, non-pulsating torque!

3. Phasor Diagram (Voltages & Currents)

Solid lines represent Voltages ($V_a, V_b, V_c$), Dashed lines represent Currents ($I_a, I_b, I_c$).

4. Average Power Components (Macroscopic)
Live Calculation Example: Total Active Power (P_tot)

Assuming balanced source voltages: V_a = V_b = V_c = 1.0 V (RMS).

P_a = V_a × I_a × cos(θ_a) = 1.0 × 1.0 × cos(30°) = 0.866 W
P_b = V_b × I_b × cos(θ_b) = 1.0 × 1.0 × cos(30°) = 0.866 W
P_c = V_c × I_c × cos(θ_c) = 1.0 × 1.0 × cos(30°) = 0.866 W
---------------------------------------------------------
P_tot = P_a + P_b + P_c = 2.598 W

Mathematical Derivation of Three-Phase Power

In a three-phase system, we define the instantaneous phase voltages displaced by $120^\circ$ ($\frac{2\pi}{3}$ radians) in time. Let $V$ and $I$ denote RMS values, making peak amplitudes $\sqrt{2}V$ and $\sqrt{2}I$.

$$ v_a(t) = \sqrt{2} V_a \sin(\omega t) $$ $$ v_b(t) = \sqrt{2} V_b \sin(\omega t - 120^\circ) $$ $$ v_c(t) = \sqrt{2} V_c \sin(\omega t + 120^\circ) $$

The phase currents lag their respective voltages by angles $\theta_a, \theta_b, \theta_c$:

$$ i_a(t) = \sqrt{2} I_a \sin(\omega t - \theta_a) $$ $$ i_b(t) = \sqrt{2} I_b \sin(\omega t - 120^\circ - \theta_b) $$ $$ i_c(t) = \sqrt{2} I_c \sin(\omega t + 120^\circ - \theta_c) $$

1. Instantaneous Phase Powers

Similar to the single-phase derivation, the instantaneous power for each phase is $p(t) = v(t)i(t)$. By applying the identity $\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]$, we get:

$$ p_a(t) = V_a I_a \cos(\theta_a) - V_a I_a \cos(2\omega t - \theta_a) $$ $$ p_b(t) = V_b I_b \cos(\theta_b) - V_b I_b \cos(2\omega t - 240^\circ - \theta_b) $$ $$ p_c(t) = V_c I_c \cos(\theta_c) - V_c I_c \cos(2\omega t + 240^\circ - \theta_c) $$

Notice that each phase power contains a constant term (the Active Power, $P$) and a term oscillating at twice the fundamental frequency ($2\omega$).

2. Total Instantaneous Power ($p_{total}$) & The Constant Power Theorem

The total instantaneous power delivered to the 3-phase machine or load is the sum of the individual phase powers: $p_{total}(t) = p_a(t) + p_b(t) + p_c(t)$.

Balanced Condition

If the system is perfectly balanced ($V_a=V_b=V_c=V$, $I_a=I_b=I_c=I$, and $\theta_a=\theta_b=\theta_c=\theta$), the constant terms sum to $3VI\cos(\theta)$. But look at the double-frequency oscillatory terms:

$$ p_{osc}(t) = -VI \Big[ \cos(2\omega t - \theta) + \cos(2\omega t - 240^\circ - \theta) + \cos(2\omega t + 240^\circ - \theta) \Big] $$

A fundamental property of three-phase mathematics is that the sum of three balanced, $120^\circ$-shifted sinusoids is exactly zero. Therefore, the oscillatory terms completely cancel each other out!

$$ p_{total}(t) = 3 V I \cos(\theta) = P_{total} \quad \text{(A Constant Value)} $$

Engineering Implication: Because $p_{total}(t)$ is strictly constant in a balanced system, the electromagnetic torque developed in a three-phase motor is smooth and constant. In contrast, single-phase motors suffer from torque pulsations at $2\omega$, causing vibrations.

Unbalanced Condition

If the load is unbalanced (currents or angles differ), the double-frequency terms do not cancel. This results in ripples in $p_{total}(t)$, which physically manifest as harmful torque pulsations and vibrations in electrical machines. You can observe this by switching the interactive simulator above to "Unbalanced" and varying the load parameters.

3. Macroscopic Total Powers & Power Factor

Whether balanced or unbalanced, the total active and reactive powers are the algebraic sums of the individual phases. The overall Power Factor indicates how effectively the total apparent power is being converted into real work.

$$ P_{total} = P_a + P_b + P_c = V_a I_a \cos(\theta_a) + V_b I_b \cos(\theta_b) + V_c I_c \cos(\theta_c) $$ $$ Q_{total} = Q_a + Q_b + Q_c = V_a I_a \sin(\theta_a) + V_b I_b \sin(\theta_b) + V_c I_c \sin(\theta_c) $$ $$ S_{total} = \sqrt{P_{total}^2 + Q_{total}^2} $$ $$ PF_{total} = \frac{P_{total}}{S_{total}} $$