Derivation and phasor diagram of complex power

Complex power S is the phasor product of the voltage and the conjugate of the current. Its real part is the average (active) power and its imaginary part is the reactive power.

Derivation

Let the instantaneous voltage and current be

$$v = V_m\cos(\omega t + \theta_v), \qquad i = I_m\cos(\omega t + \theta_i)$$

The instantaneous power is

$$\begin{aligned} p = vi &= \frac{V_mI_m}{2}\left[\cos(\theta_v - \theta_i) + \cos(2\omega t + \theta_v + \theta_i)\right]\\ &= |V||I|\cos\phi\,[1 + \cos 2(\omega t + \theta_v)] + |V||I|\sin\phi\,\sin 2(\omega t + \theta_v) \end{aligned}$$

Here $|V| = V_m/\sqrt2$, $|I| = I_m/\sqrt2$ and $\phi = \theta_v - \theta_i$.

• The first term has average $P = |V||I|\cos\phi$. This is the active power.
• The second term has zero average and peak $Q = |V||I|\sin\phi$. This is the reactive power.

In phasor form, $V = |V|\angle\theta_v$ and $I = |I|\angle\theta_i$. Then

$$\begin{aligned} S = VI^* &= |V|\angle\theta_v\cdot|I|\angle-\theta_i = |V||I|\angle(\theta_v - \theta_i)\\ &= |V||I|\cos\phi + j|V||I|\sin\phi = P + jQ \end{aligned}$$

For a load $Z = R + jX$: $S = |I|^2Z = |I|^2R + j|I|^2X$.

Phasor diagram and power triangle

 V reference (lagging load)       Power triangle
  ---------------> V                  /|
  \  phi                         S  /  | Q = |V||I|sin(phi)
    \                             /    |
      \  I (lags V)             / phi  |
        v                      +-------+
                                P = |V||I|cos(phi)

Multiplying V by the conjugate of I turns the V–I phasor diagram into a right-angled triangle. The hypotenuse is $|S|$, the base is P, the height is Q, and the angle is $\phi$.

Significance of the components

• Q > 0 means a lagging (inductive) load and Q < 0 a leading (capacitive) load.
• A low power factor means more current for the same P. This causes more $I^2R$ loss, more voltage drop and larger equipment. So utilities add capacitors to reduce Q.

Example: $V = 230\angle0^\circ$ V and $I = 10\angle-36.87^\circ$ A give $S = 2300\angle36.87^\circ = 1840 + j1380$ VA, with pf 0.8 lagging.