Single-Phase AC Power Analyzer
Interactive visualization of Voltage, Current, and Instantaneous Power Flow
Reactive Power (Q)
0.500 VAR
Apparent Power (S)
1.000 VA
1. Instantaneous Waveforms
2. Average Power Components (Macroscopic)
Live Calculation Example:
Given fixed amplitudes: V_rms = 1.0 V, I_rms = 1.0 A
Current Phase Angle (θ) = 30°
Active Power (P) calculation:
P = V_rms × I_rms × cos(θ)
P = 1.0 × 1.0 × cos(30°)
P = 0.866 W
Reactive Power (Q) calculation:
Q = V_rms × I_rms × sin(θ)
Q = 1.0 × 1.0 × sin(30°)
Q = 0.500 VAR
Mathematical Derivation of Active and Reactive Power
In a single-phase AC circuit, we define the instantaneous voltage $v(t)$ and current $i(t)$ as sinusoidal functions of time. Let us take the voltage as the reference phasor (phase angle $= 0$). To denote RMS values, we use $V$ and $I$, such that the peak amplitudes are $\sqrt{2}V$ and $\sqrt{2}I$.
$$ v(t) = \sqrt{2} V \sin(\omega t) $$
$$ i(t) = \sqrt{2} I \sin(\omega t - \theta) $$
Where $\theta$ is the phase angle by which the current lags the voltage. If $\theta > 0$, the load is inductive (lagging). If $\theta < 0$, the load is capacitive (leading).
1. Instantaneous Power $p(t)$
The instantaneous power transferred to the load is the product of instantaneous voltage and current:
$$ p(t) = v(t) \cdot i(t) = \Big(\sqrt{2} V \sin(\omega t)\Big) \cdot \Big(\sqrt{2} I \sin(\omega t - \theta)\Big) $$
$$ p(t) = 2 V I \sin(\omega t) \sin(\omega t - \theta) $$
2. Applying Trigonometric Identities
We apply the product-to-sum trigonometric identity: $\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]$.
$$ p(t) = V I \Big[ \cos(\omega t - (\omega t - \theta)) - \cos(\omega t + \omega t - \theta) \Big] $$
$$ p(t) = V I \cos(\theta) - V I \cos(2\omega t - \theta) $$
Next, we expand the second term using the angle difference identity: $\cos(A - B) = \cos A \cos B + \sin A \sin B$.
$$ p(t) = V I \cos(\theta) - V I \Big[ \cos(2\omega t)\cos(\theta) + \sin(2\omega t)\sin(\theta) \Big] $$
3. Isolating Active and Reactive Components
By factoring out the common $\cos(\theta)$ and $\sin(\theta)$ terms, we decompose the instantaneous power equation into two distinct parts:
$$ p(t) = \underbrace{V I \cos(\theta) \Big[ 1 - \cos(2\omega t) \Big]}_{\text{Instantaneous Active Power}} - \underbrace{V I \sin(\theta) \sin(2\omega t)}_{\text{Instantaneous Reactive Power}} $$
We define the macroscopic power quantities from these amplitudes:
- Active Power ($P$): The average power that performs real work (measured in Watts).
$$ P = V I \cos(\theta) $$
- Reactive Power ($Q$): The peak power that continuously oscillates between the source and the load's reactive elements (measured in Volt-Amperes Reactive, VAR).
$$ Q = V I \sin(\theta) $$
Substituting $P$ and $Q$ back into the instantaneous power equation yields the most elegant form of the expression:
$$ p(t) = P \Big( 1 - \cos(2\omega t) \Big) - Q \sin(2\omega t) $$
Observation from the graph above: Notice that the $p(t)$ waveform oscillates at twice the fundamental frequency ($2\omega$). The Inst. Active Power component is always positive and pulses from $0$ to $2P$, yielding an average of $P$. The Inst. Reactive Power component alternates symmetrically around the zero axis, yielding a net average of $0$ over a cycle. Their sum creates the total instantaneous power $p(t)$.