Theory Notes
$P = VI\cos\phi$ (active), $Q = VI\sin\phi$ (reactive), $S = VI$ (apparent)Power triangle: $S^2 = P^2 + Q^2$, $\cos\phi = P/S$
3-phase: $P_{3\phi}= \sqrt{3}\,V_L I_L \cos\phi$
Adjust voltage, current, and power factor angle to see the real-time phasor diagram and P-Q-S power triangle. Covers resistive, inductive, capacitive, RL, RC, RLC, and 3-phase circuits.
The core relationship is the power triangle, which separates the total apparent power into its real (useful) and reactive (oscillating) components.
$$S = V_{rms}\cdot I_{rms}$$ $$P = S \cdot \cos\phi = V_{rms}I_{rms}\cos\phi$$ $$Q = S \cdot \sin\phi = V_{rms}I_{rms}\sin\phi$$Where $S$ is Apparent Power (VA), $P$ is Real Power (W), $Q$ is Reactive Power (VAR), $V_{rms}$ and $I_{rms}$ are RMS voltage and current, and $\phi$ is the phase angle between them. $\cos\phi$ is the Power Factor.
For three-phase systems, which are the backbone of power distribution, the calculation scales. The formula accounts for the phase relationships between the three lines.
$$P_{3\phi}= \sqrt{3}\cdot V_L \cdot I_L \cdot \cos\phi$$Here, $V_L$ is the line-to-line voltage, $I_L$ is the line current, and $\phi$ is the same phase angle. The $\sqrt{3}$ factor arises from the 120° separation between phases in a balanced system. This is why three-phase power is more efficient for delivering large amounts of energy.
Industrial Motor Loads: Large induction motors in factories are highly inductive. At low load, their power factor can be very poor (e.g., 0.3). This causes high reactive power flow, increasing energy losses in plant wiring and potentially incurring penalty fees from the utility company. Engineers install capacitor banks to correct the power factor closer to 1.
Power Utility Billing: Utilities charge large industrial customers not just for the real energy (kWh) they consume, but also for the peak apparent power (kVA) demand. A low power factor means a higher kVA demand for the same kW of work, so improving the power factor directly reduces electricity costs for factories and data centers.
Residential Solar Inverters: Modern grid-tied solar inverters can do more than just push out real power (P). They can be programmed to also supply or absorb reactive power (Q) to help stabilize the local grid voltage, a feature called "volt-var control." This turns homes into active participants in grid management.
Data Center Design: The power capacity of a data center's Uninterruptible Power Supply (UPS) systems and distribution wiring is rated in kVA (apparent power). If the servers and cooling units have a poor collective power factor, the data center can hit its kVA limit before using its full intended real power (kW), wasting expensive infrastructure capacity.
First, let's clear up the common misconception that "reactive power is wasted power." While it's true it doesn't do work directly, it is absolutely essential for motors to create magnetic fields and for transformers to operate. The goal isn't to eliminate it; the essence of power factor improvement is to suppress the "excessive flow of reactive power." Next, a common mistake when experimenting with simulators is confusing the power factor angle with the phase difference. In this tool, the "power factor angle" is defined as the phase difference (φ) of the current relative to the voltage. This means φ>0 when the current lags (inductive load) and φ<0 when it leads (capacitive load). This angle matches the one shown in the power triangle, so be sure to grasp this point firmly.
In practical applications, be careful not to treat the apparent power S unit [VA] and the active power P unit [W] as the same thing. For example, if you connect a load with a power factor of 0.6 (300W) to a 500VA UPS, you might be able to use it right up to its 500VA limit. However, if you connect a load with a power factor of 0.9 (450W), the active power is higher, but the apparent power is 500VA (=450W/0.9), so this also uses the capacity to its limit. As you can see, equipment capacity is often specified in apparent power, so you must always consider the power factor in your design.
A three-phase induction motor draws 30 A at 380 V with power factor 0.88 lagging at 50 Hz. Real power P = 380 × 30 × 0.88 = 10,032 W (10.0 kW). Reactive power Q = 10,032 × tan(arccos 0.88) = 6,124 VAR (6.1 kVAR). Apparent power S = √(10,032² + 6,124²) = 11,795 VA (11.8 kVA). The phasor diagram shows current lagging voltage by 28.4°, with the S vector at the hypotenuse of the right triangle formed by P and Q.