Electric Power Calculator Back
Electrical Engineering

Electric Power Calculator

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.

Circuit Type
Power Supply
V
A
°
Hz
RLC Component Values
Ω
mH
μF
Results
Active Power P
Reactive Power Q
Apparent Power S
Power Factor PF
Phasor

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: $P_{3\phi}= \sqrt{3}\,V_L I_L \cos\phi$
Theory & Key Formulas

$$P = VI\cos\phi = I^2 R$$

有効電力(W):$V$ は電圧(V)、$I$ は電流(A)、$\cos\phi$ は力率。

$$Q = VI\sin\phi$$

無効電力(var):力率改善のためにコンデンサ補償が行われる。

$$S = VI = \sqrt{P^2 + Q^2}$$

皮相電力(VA):有効電力と無効電力のベクトル和。力率 $\cos\phi = P/S$。

What is AC Power?

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What exactly is the "power factor" I see on the simulator slider? It sounds important, but I'm not sure what it physically represents.
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Basically, the power factor tells you how effectively the current you're drawing is being converted into useful work. In practice, it's the cosine of the phase angle ($\phi$) between the voltage and current waveforms. Try moving the "Power Factor" slider in the simulator from 1.0 down to 0.5. You'll see the current phasor lag behind the voltage, and the "Real Power" bar gets smaller even if voltage and current stay the same. That's wasted capacity!
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Wait, really? So if the voltage and current numbers are high, but the power factor is low, I'm not actually getting much useful power? What's happening to the rest of the energy?
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Exactly! The rest is "Reactive Power" (Q), which sloshes back and forth between the source and reactive components like motors or transformers. It doesn't do useful work but heats up the wires. In the simulator, watch the power triangle. As you lower the power factor, the reactive power side (Q) grows, and the hypotenuse—the "Apparent Power" (S)—stays long, meaning the utility has to supply that large current. A common case is a factory full of induction motors at low load, which has a poor power factor.
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That makes sense for the triangle. But the simulator also has a "Circuit" parameter. What does changing that from "Resistive" to "Inductive" actually do to the math?
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Great question! Changing the "Circuit" type automatically sets a typical power factor and changes the phase angle $\phi$. For a purely resistive load (like a heater), $\phi = 0$, so the power factor is 1. For an inductive load (like a motor), current lags voltage ($\phi \gt 0$). The simulator calculates this using $\phi = \arctan((X_L - X_C)/R)$. When you select "Inductive," it models a positive $\phi$, which you can see as the angular shift in the phasor diagram.

Physical Model & Key Equations

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.

Frequently Asked Questions

The simulator may be in a paused state. Check the "Play/Stop" button at the bottom of the screen and switch it to the playing state (▶). Additionally, if the browser tab is in the background, updates may stop.
This occurs in the resonant state where the inductive reactance XL equals the capacitive reactance XC. At this point, the circuit behaves as a pure resistor, eliminating the phase difference between voltage and current, resulting in a power factor angle of 0° and a power factor of 1.0. Try adjusting the L and C values using the sliders.
It can be correct. When calculating three-phase power from line voltage and line current, it becomes √3 times. When calculated from phase voltage and phase current, it becomes three times. The simulator automatically performs the correct conversion based on the settings, so please check the displayed values.
The sign of reactive power corresponds to the sign of the power factor angle. Q is negative when the current leads the voltage (capacitive load) and positive when it lags (inductive load). In the power triangle, the sign of Q helps identify the type of reactance.

Real-World Applications

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.

Common Misconceptions and Points to Note

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.

How to Use

  1. Enter RMS voltage (V) in the vvNum field—typical range 120–480 V for industrial circuits.
  2. Set RMS current (A) in the iiNum field—motor loads typically draw 5–50 A.
  3. Input power factor (cos φ) in phiNum as a decimal 0–1; lagging induction motors are typically 0.85–0.92.
  4. Set line frequency in freqNum: 50 Hz (Europe, Asia) or 60 Hz (North America).
  5. Observe the real-time phasor diagram and P-Q-S triangle updating as you adjust parameters.

Worked Example

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.

Practical Notes

  1. Power factor below 0.95 incurs utility penalties on industrial bills; add capacitor banks to improve PF toward unity.
  2. Reactive power (VAR) represents magnetic energy storage in motor windings and transformers—it does no useful work but increases system losses.
  3. At 60 Hz (North America), the same 30 A motor at 480 V yields higher kilowatt capacity than 50 Hz circuits; verify frequency before equipment selection.
  4. Use phasor angle φ to diagnose circuit behavior: φ near 0° indicates resistive loads (heating), φ near 90° indicates inductive loads (motors, inductors).