Reactive Power Formula in 3 Phase: Derivation, Tables, and Worked Examples

The total reactive power ($Q$) in a balanced three-phase AC system is calculated using the formula: $Q = \sqrt{3} \times V_L \times I_L \times \sin(\theta)$. Measured in Volt-Amperes Reactive (VAR) or kilovars (kVAR), this value represents the energy that oscillates between the source and the load's magnetic or electric fields, performing no real work but dictating the physical sizing of your conductors, transformers, and switchgear.

Whether you are sizing a capacitor bank for power factor correction or troubleshooting a tripped main breaker on a VFD panel, understanding how to manipulate and apply this formula is non-negotiable. Below is the complete derivation, real-world magnitude data, and step-by-step worked examples with strict unit tracking.

The Core Reactive Power Formula in 3 Phase Systems

The standard equation assumes a balanced three-phase load with sinusoidal voltage and current waveforms. Here is the exact mathematical representation and the definition of every variable involved.

$$Q = \sqrt{3} \times V_L \times I_L \times \sin(\theta)$$

Symbol	Parameter	Standard Unit	Practical Notes
$Q$	Total Reactive Power	VAR or kVAR	Positive for inductive loads (lagging); negative for capacitive loads (leading).
$\sqrt{3}$	Three-Phase Multiplier	Dimensionless	Approx. 1.732. Derived from the $120^\circ$ phase shift geometry between line and phase vectors.
$V_L$	Line-to-Line Voltage	Volts (V)	Must be the potential difference between two phases (e.g., 480V), NOT line-to-neutral (277V).
$I_L$	Line Current	Amperes (A)	The current measured on any single phase conductor feeding the load.
$\theta$	Phase Angle	Degrees ($^\circ$) or Radians	The angular difference between the voltage and current waveforms. $\cos(\theta)$ is the Power Factor.
$\sin(\theta)$	Reactive Factor	Dimensionless	Derived from the power triangle. If PF is known, $\sin(\theta) = \sqrt{1 - \text{PF}^2}$.

Rearranged Forms for Field Calculations

On the jobsite, you rarely solve for $Q$ directly. More often, you are troubleshooting a known reactive draw or sizing a component. Here are the algebraic rearrangements solving for each remaining variable:

• Solve for Line Voltage ($V_L$): $$V_L = \frac{Q}{\sqrt{3} \times I_L \times \sin(\theta)}$$
• Solve for Line Current ($I_L$): $$I_L = \frac{Q}{\sqrt{3} \times V_L \times \sin(\theta)}$$
• Solve for Phase Angle ($\theta$): $$\theta = \arcsin\left(\frac{Q}{\sqrt{3} \times V_L \times I_L}\right)$$

Real-World Magnitudes: 3-Phase Motor Reactive Power Data

What does a realistic answer magnitude look like? In industrial settings, reactive power is dominated by induction motors. A common mistake among junior engineers is underestimating the kVAR draw of large motors, leading to undersized capacitor banks and utility penalty fees. According to US Department of Energy motor systems data, standard efficiency motors exhibit significant reactive draws.

The table below provides real-world baseline data for NEMA Design B three-phase induction motors operating at 480V. This data is critical for estimating baseline $Q$ before power factor correction.

Motor Rating	Nominal $V_L$	Full Load $I_L$	Uncorrected PF ($\cos\theta$)	Calculated $Q$ (kVAR)
10 HP	480 V	14.0 A	0.82 (Lagging)	6.6 kVAR
50 HP	480 V	65.0 A	0.86 (Lagging)	27.5 kVAR
100 HP	480 V	124.0 A	0.88 (Lagging)	48.9 kVAR
250 HP	480 V	302.0 A	0.89 (Lagging)	114.3 kVAR

Benchmark Note: If your calculation for a 100 HP motor yields a reactive power of 5 kVAR, you have made a decimal or unit error. A 100 HP motor physically requires roughly 45 to 55 kVAR to maintain its magnetic field at full load. To correct a 100 HP motor from 0.88 to 0.95 PF, you would need to install a parallel capacitor bank rated for approximately 20 kVAR.

Step-by-Step Solved Problems with Unit Tracking

Abstract formulas fail when units are mixed. The following examples enforce strict unit tracking from premise to conclusion. For deeper theoretical background on three-phase vector geometry, refer to the All About Circuits three-phase systems guide.

Problem 1: Calculating kVAR from Nameplate and Meter Data

Scenario: You are auditing a 480V 3-phase air compressor. Your clamp meter reads a balanced line current of 60 A. The facility power quality meter logs a displacement power factor of 0.85 lagging. Calculate the total reactive power ($Q$) in kVAR.

Step 1: Identify and standardize known variables.

• $V_L = 480 \text{ V}$
• $I_L = 60 \text{ A}$
• $\text{PF} = \cos(\theta) = 0.85$

Step 2: Derive the phase angle ($\theta$) and the reactive factor ($\sin\theta$).

• $\theta = \arccos(0.85) = 31.79^\circ$
• $\sin(31.79^\circ) = 0.5268$
• Shortcut check: $\sin(\theta) = \sqrt{1 - 0.85^2} = \sqrt{1 - 0.7225} = \sqrt{0.2775} = 0.5268$.

Step 3: Apply the formula with explicit units.

• $Q = 1.732 \times 480 \text{ V} \times 60 \text{ A} \times 0.5268$
• $Q = 49,881.6 \text{ VA} \times 0.5268$
• $Q = 26,277.6 \text{ VAR}$

Step 4: Convert to standard engineering units.

• $Q_{\text{kVAR}} = \frac{26,277.6 \text{ VAR}}{1000} = \mathbf{26.3 \text{ kVAR}}$

Problem 2: Sizing Conductor for a Known Reactive Load

Scenario: A Canadian industrial facility operates a 600V 3-phase system. A specific synchronous condenser is injecting exactly 45 kVAR of leading reactive power into the bus to correct grid sag. The phase angle between voltage and current is measured at $25^\circ$. What is the line current ($I_L$) flowing through the feeder?

Step 1: Identify and standardize known variables.

• $Q = 45 \text{ kVAR} = 45,000 \text{ VAR}$ (Must convert to base units first)
• $V_L = 600 \text{ V}$
• $\theta = 25^\circ$

Step 2: Calculate $\sin(\theta)$.

• $\sin(25^\circ) = 0.4226$

Step 3: Rearrange the formula to solve for $I_L$.

• $I_L = \frac{Q}{\sqrt{3} \times V_L \times \sin(\theta)}$

Step 4: Substitute values and track units.

• $I_L = \frac{45,000 \text{ VAR}}{1.732 \times 600 \text{ V} \times 0.4226}$
• $I_L = \frac{45,000}{1,039.2 \text{ V} \times 0.4226}$
• $I_L = \frac{45,000}{439.16 \text{ V}}$
• $I_L = \mathbf{102.5 \text{ A}}$

Result: The feeder must be sized to handle at least 102.5 A continuously, requiring a minimum of 3 AWG THHN copper wire (rated 100A at 75°C, so you must step up to 2 AWG or apply specific derating allowances per NEC 310.16).

Boundary Conditions: Assumptions and Fatal Unit Mistakes

The $\sqrt{3}$ formula is an elegant shortcut, but it is mathematically fragile. It only holds true under specific physical conditions. Violating these assumptions or mismanaging units will yield dangerously incorrect results.

When the Formula Applies (and When it Fails)

The formula assumes a perfectly balanced load. This means the impedance (and therefore the current and phase angle) is identical across Phase A, Phase B, and Phase C. If you are measuring a system with heavy single-phase line-to-neutral loads (like a commercial building with 277V lighting and 480V HVAC), the system is inherently unbalanced.

In unbalanced scenarios, the $\sqrt{3}$ shortcut fails. You must calculate the reactive power per phase using line-to-neutral voltage ($V_{LN}$) and sum them: $$Q_{\text{total}} = (V_{an} \times I_a \times \sin\theta_a) + (V_{bn} \times I_b \times \sin\theta_b) + (V_{cn} \times I_c \times \sin\theta_c)$$

Additionally, the formula assumes pure sinusoidal waveforms. If the load is driven by a 6-pulse VFD without line reactors, harmonic distortion (THDi) creates non-fundamental reactive power components that standard power meters and this basic formula cannot accurately resolve.