## Single Phase Power Review

Voltages waveforms in a power system are typically sinusoidal. In single phase power systems the voltages in the system can be described as

\begin{aligned} v(t) & =\hat{V}\cos\left(\omega t +\psi \right) \\ v(t) & = \Re \left\{ \sqrt{2} \mathbf{V}e^{j\omega t} \right\} \end{aligned}

where $$\psi$$is the phase angle of the voltage. $$\mathbf{V}$$ is a phasor defined in terms of the rms voltage. Unlike other areas (e.g. signal processing, communications) phasors in power systems are always defined in terms of rms magnitude, not peak magnitude. The rms of a sinusoidal quantity is given by

$V=\frac{\hat{V}}{\sqrt{2}}$

and the phasor notation is

\begin{aligned} \mathbf{V}=Ve^{j\psi} \\ \mathbf{V}=V\angle\psi \end{aligned}

Note that $$\psi_v$$ is often set to zero and voltage phase angle is used as the reference for the rest of the system.

Currents flowing in a linear system will also be sinusoidal. If we let $$\psi_v = 0$$ then:

\begin{aligned} v(t) & =\hat{V}\cos\left(\omega t\right) \\ i(t) & =\hat{I}\cos\left(\omega t\ +\theta\right) \\ \mathbf{V} & =V \\ \mathbf{I} & =I\angle\theta =Ie^{j \theta} \end{aligned}

Where $$\theta$$ is the electrical angle between the voltage and current. In power systems, which are usually inductive, it is common for $$\theta$$ to be negative.

Power flow in a system is the rate of change of energy past the point in the system where voltage and current are measured. Instantaneous power is the product of the voltage and current functions of time:

\begin{aligned} p(t) & =v(t) i(t) \\ p(t) & = 2VI \cos\left( \omega t \right) \cos\left( \omega t + \theta \right) \\ p(t) & = 2VI \left\{\frac{1}{2} \left[ \cos\left( 2 \omega t+\theta \right) +\cos \left( - \theta \right) \right] \right\} \\ p(t) & = VI\cos\theta +VI \cos \left(2 \omega t+\theta \right) \end{aligned}

The graph below plots voltage, current and power waveforms as functions of time. In the graph, the voltage waveform is 10V rms, current is 5A rms and the current lags voltage by 45 degrees.

From the above equation and plots, it can be seen that the power is comprised of two components, a steady value and a component oscillating at twice the supply frequency. It can also be seen that at certain points, the instantaneous power is negative. Negative power indicates that energy flow is in the opposite direction to the defined direction of positive current flow. To obtain the average power flow with time, it is necessary to integrate power over one supply cycle:

\begin{aligned} P&=\frac{1}{2\pi}\int_0^{2\pi}p\left(\omega t\right) d{\omega t} \\ P&=VI\cos\theta \end{aligned}

It should be clear that there is a current flow and instantaneous power flow which is not associated with actual power transfer. This occurs when a circuit contains energy storage components including capacitors and inductors. With these components, energy is transferred back and forth between the system and the component each cycle, with no net power transfer. To quantify the current and energy flow due to these component use the terms Apparent Power or Voltamperes, denoted using $$S$$, Reactive Power or Voltamperes-Reactive $$Q$$, and power factor. Complex Voltamperes are defined using them product of the voltage phasor and complex conjugate of the current phasor:

\begin{aligned} \mathbf{S} & = \mathbf{V}\mathbf{I}^* \\ \mathbf{S} & = VI\cos\theta +jVI\sin\left( - \theta \right) \\ \mathbf{S} & = P+jQ \\ S & = VI \\ S & = \sqrt{P^2+Q^2} \end{aligned}

$$S$$, the magnitude of $$\mathbf{S}$$, is the magnitude of the oscillating component of instantaneous power with units of VA, and Q is proportional to the energy stored in the system, with units of VAR. (Note that Watts are reserved for real power flow). The relationship between Voltamperes and power is given by the power factor:

\begin{aligned} {PF}=\frac{P}{S} \\ {DPF} = \cos\theta \end{aligned}

The cosine of the angle between voltage and current is sometimes called displacement power factor (DPF) or fundamental power factor (FPF). In a system with only fundamental current, the power factor equals the displacement power factor.