This article aims to explain some basic properties of sinusoidal waveforms, and three phase systems. It is important to have a thorough understanding of these concepts before diving into the more advanced topics of motor drive design, or electrics in general. I will assume that the reader is familiar with the use of complex numbers, ohm's law, basic circuit analysis and calculus.

## The square root of three

Only valid for symmetrical three phase systems, the square root of three is commonly seen as a relation between line and phase quantities.

\begin{equation}

V_L = \sqrt{3} V_f

\end{equation}

The power in a three phase system is three times the power in a single phase system:

\begin{equation}

P = 3 \cdot V_f \cdot I_f

\label{eq:3-phase-power-phase-q}

\end{equation}

However, when line quantities are considered, the equation is:

\begin{equation}

P = V_L \cdot I_L \cdot \sqrt{3}

\end{equation}

The following derivations is an attempt to explain how this seemingly magic relation is obtained.

For a Y-connected system, the line voltage is the voltage difference between two phases. I.e. the difference considering the phase difference of \(120^{\circ}\).

\begin{equation}

V_L = V_f \enclose{phasorangle}{0^{\circ}} - V_f \enclose{phasorangle}{-120^{\circ}} = V_f \enclose{phasorangle}{30^{\circ}}

\end{equation}

For a phase voltage of one volts, the result in rectangular form is given as:

\begin{equation}

1 \enclose{phasorangle}{0^{\circ}} + 1 \enclose{phasorangle}{-120^{\circ}} = 1 \enclose{phasorangle}{30^{\circ}} = \frac{3}{2} + j \frac{\sqrt{3}}{2}

\end{equation}

The sum of the two phase voltage vectors may also be expressed as:

\begin{equation}

V_L = 2 \cdot V_f \cdot \sin(120^{\circ})

\end{equation}

Furthermore the trigonometric function evaluates to:

\begin{equation}

\sin(120^{\circ}) = \frac{\sqrt{3}}{2}

\end{equation}

Hence the resulting relation of phase to line voltage is given as:

\begin{equation}

V_L = 2 \cdot V_f \cdot \frac{\sqrt{3}}{2} = V_f \cdot \sqrt{3}

\end{equation}

Still considering a Y-connected system, the line current equals the phase current:

\begin{equation}

I_{L} = I_{f}

\end{equation}

Using the expressions for line quantities in the power equation \eqref{eq:3-phase-power-phase-q}, we obtain:

\begin{equation}

P = 3 \cdot V_f \cdot I_f = 3 \cdot \frac{V_L}{\sqrt{3}} \cdot I_L = \sqrt{3} \cdot V_L \cdot I_L

\end{equation}

Because:

\begin{equation}

\frac{3}{\sqrt{3}} = \sqrt{3}

\end{equation}

It is important to realize that for a \(\Delta\)-connected system a similar derivation to the above will result in exactly the same equation for power. The phase voltage will equal the line voltage, but the current will adhere to the square root of three relation.

Additionally the displacement power factor \(\cos(\phi)\) is added to obtain the well known general power equation for balanced three phase systems:

\begin{equation}

P = V_L \cdot I_L \cdot \sqrt{3} \cdot \cos(\phi)

\end{equation}

## RMS, and the square root of two

Another seemingly magic number is the square root of two, relating the RMS value of a sine wave to it's amplitude. Using voltage as a example, we have that:

\begin{equation}

\hat{V} = V_{RMS} \cdot \sqrt{2}

\end{equation}

Where \(\hat{V}\) is the amplitude voltage, i.e. peak value of the sine wave.

Again using voltage as a example, the RMS value is defined as:

\begin{equation}

V_{RMS} = \sqrt{\frac{1}{T} \int_{t_0}^{t} v(t)^2 \mathrm{d}t }

\end{equation}

The general equation for a sine wave is given as:

\begin{equation}

v(t) = \hat{V} \cdot \sin(\omega t + \theta)

\end{equation}

Where \(\hat{V}\) is the amplitude of the voltage. Ignoring the phase shift \(\theta\), we have the following expression for RMS voltage:

\begin{equation}

V_{RMS} = \sqrt{\frac{1}{T} \int_{t_0}^{t} \hat{V}^2 \cdot \sin^2(\omega t) \mathrm{d}t }

\end{equation}

We also have that:

\begin{equation}

\sin^2(\omega t) = \frac{1 - \cos(2 \omega t)}{2}

\end{equation}

The amplitude voltage squared \(\hat{V}^2\) is a positive constant, and is moved it front of the expression. The exponent is canceled by the square root. The trigonometric identity is applied, and we obtain:

\begin{equation}

V_{RMS} = \hat{V}\sqrt{\frac{1}{T} \int_{t_0}^{t} \frac{1 - \cos(2 \omega t)}{2} \mathrm{d}t }

\end{equation}

Evaluating the integral, we obtain:

\begin{equation}

V_{RMS} = \hat{V}\sqrt{\frac{1}{T} \left[ \frac{t}{2} - \frac{\sin(2 \omega t)}{4 \omega} \right]_{t_0}^t }

\end{equation}

By definition, the RMS value is taken over a whole number of complete cycles. Hence the \(\sin(2 \omega t)\) term will cancel out.

\begin{equation}

V_{RMS} = \hat{V}\sqrt{\frac{1}{T} \left[ \frac{t}{2} \right]_{t_0}^t }

\end{equation}

The period is given as:

\begin{equation}

T = t - t_0

\end{equation}

Hence:

\begin{equation}

V_{RMS} = \hat{V}\sqrt{\frac{1}{t - t_0} \left[ \frac{t}{2} \right]_{t_0}^t } = \hat{V}\sqrt{\frac{1}{t - t_0} \left( \frac{t}{2} - \frac{t_0}{2} \right) } = \hat{V} \frac{1}{\sqrt{2}}

\end{equation}

Q.E.D

## Average rectified value

The conceptual difference between RMS value, and average rectified value is something that can be intuitively difficult to comprehend. The average of a sine wave around it's symmetry line is always zero, because the positive and negative values are equal. The average rectified value however, is the average of the sine wave if both cycles are positive.

\begin{equation}

V_{ARV} = \frac{1}{T} \int_{t_0}^{t} |v(t)| \mathrm{d}t

\end{equation}

For a sinusoidal waveform, the relation between amplitude, and average rectified value is given as:

\begin{equation}

\hat{V} = V_{ARV} \cdot \frac{\pi}{2}

\end{equation}

The form factor is used to express the difference between the RMS, and the average rectified value:

\begin{equation}

k_f = \frac{V_{RMS}}{V_{ARV}}

\end{equation}

Hence, for the sinusoidal example, the form factor is:

\begin{equation}

\frac{\pi}{2 \sqrt{2}} \approx 1.1107

\end{equation}

I.e. for a regular line RMS voltage of \(230 \;V\), the ARV is \(207.2 \;V\).

### Intuition

So why do we use these two different numbers, and what do they represent?

The RMS is explained as the DC voltage that will dissipate the same amount of power as the AC voltage from which it is calculated. I.e. the use of RMS simplifies the calculation of power in AC circuits.

The explanation is that the power dissipated by a resistor is proportional to the square of the voltage. Double the voltage, and the power is increased by a factor of four:

\begin{equation}

P = \frac{V_{RMS}^2}{R} = \frac{ \left( \cfrac{\hat{V}}{\sqrt{2}} \right)^2}{R} = \frac{\hat{V}^2}{2\cdot R}

\end{equation}

Alternately:

\begin{equation}

P = I_{RMS}^2 \cdot R = \left( \cfrac{\hat{I}}{\sqrt{2}} \right)^2 \cdot R = \frac{\hat{I}^2 R}{2}

\end{equation}

The average on the other hand is useful when dealing with rectifiers, or other non-linear circuits. It is also used when calculating the magnetic flux produced by a coil.

\begin{equation}

\Phi = V_{ARV} \cdot t [Wb]

\end{equation}

If this last statement confuses you, the following might help. Maxwell's fourth equation (Ampère's law) states that the magnetic field integrated around a closed loop is proportional to the current enclosed by the loop. I.e. magnetic field is proportional to current:

\begin{equation}

\oint \vec{B} \cdot \vec{\mathrm{d}S} = \mu_0 i + \frac{1}{c^2} \frac{\partial}{\partial t} \int \vec{E} \cdot \vec{\mathrm{d}A}

\end{equation}

Where \(i\) is the current, \(c\) is the speed of light, \(\mu_0\) is the permeability of vacuum, \(\vec{B}\) is the magnetic field, and \(\vec{E}\) is the electric field. The second term in the right hand side of the equation is Maxwell's addition to Ampère's law. This term is not always needed, ignoring it for simplicity, and also kowing that \(H = \frac{B}{\mu_0} \) yields the following result:

\begin{equation}

\oint \vec{H} \cdot \vec{\mathrm{d}S} = I

\end{equation}

Additionally Faraday's law of induction states that the line integral of the electric field around a closed loop, is equal to the negative of the change in magnetic flux per unit time inside the loop.

\begin{equation}

\oint \vec{E} \cdot \vec{\mathrm{d}S} = - \frac{\mathrm{d}\Phi}{\mathrm{d}t}

\label{eq:faradays-law}

\end{equation}

Unlike the power calculations, there is no squared weight to the current in the case of flux calculations.

# Power in AC systems

As has previously been discussed, the definition of effective value (RMS) is based upon what would make sense in terms of power. Power is important, the very purpose of a electric circuit is very often to perform some kind of power exchange.

Electric power (as with mechanical power, or any other power) is defined as the work done per unit time (i.e. energy per unit time):

\begin{equation}

P = \frac{V \cdot Q}{t} = V \cdot I

\end{equation}

Where \(Q\) is the electric charge in Coulombs.

In direct current (DC) applications the calculation is simple and straight forward.

In systems with alternating voltages (and currents) a interesting phenomena often occurs however. Due to the energy storing capability of capacitors and inductors, currents may appear that are out of phase with the voltage, and thus without any power dissipation. The energy swings back and forth between the components without being dissipated.

A new definition of the product of RMS current and RMS voltage is thus needed, and this is known as *apparent power* (S).

\begin{equation}

S = V_{RMS} \cdot I_{RMS}

\end{equation}

The apparent power is given it's name because it is what we would assume to be the power based on our measurements, but due to some parts of the voltage being out of phase with the current, it is not the actual power. The unit of measurement is Volt-ampere (VA).

Consider the figure to the right. The solid graph represents a voltage, while the dashed graph represents a current. The phase shift in this situation is \(60^{\circ}\), with the current leading (remember that time flows from left to right) the voltage, i.e. it is a capacitive current.

The product of the RMS current and RMS voltage is the apparent power, the product of the actual current and voltage on a given time instant however is the actual resulting power at that time instant. This is illustrated in the next figure, where we observe a large positive power, and a smaller reactive power.

The actual dissipated power in the circuit is given by the *apparent power* multiplied by the cosine of the *phase shift angle* \(\phi\):

\begin{equation}

P = V_{RMS} \cdot I_{RMS} \cdot \cos(\phi)

\end{equation}

Where the *phase shift angle* \(\phi\) is the angle between the sinusoidal voltage and current waveforms.

Finally, we define the remaining power to be the *reactive power* (Q):

\begin{equation}

Q = V_{RMS} \cdot I_{RMS} \cdot \sin(\phi)

\end{equation}

Strictly speeking according to SI convention, the unit of measurement is watts (W), but usually (and in accordance with IEC standards) the unit *volt-ampere reactive (var)* is used* *(note: lowercase letters).

It is common practice to illustrate the two (orthogonal) \(P\), and \(Q\) vectors, along with the resulting \(S\) vector in a right-angled triangle. This is depicted to the right.

A AC component, such as a electric motor is said to have a certain reactive power consumption (largely independent of the active power consumption). This power must be supplied alongside the active power, and affects the voltage balance in the system.

To aid in the understanding of reactive power, the following figures have been included which show how the power changes for various phase shifts. Take particular note at the inductive current \(\varphi = 120^{\circ}\).

### Inductors and capacitors, and how they affect reactive power

The behaviour of the ideal inductor and capacitor is described by the well known differential equations:

\begin{equation}

i_c(t) = C \frac{\mathrm{d}v_c(t)}{\mathrm{d}t}

\end{equation}

\begin{equation}

v_L(t) = L \frac{\mathrm{d}i_L(t)}{\mathrm{d}t}

\end{equation}

If we consider a sinusoidal voltage across the capacitor:

\begin{equation}

v_c(t) = \hat{V} \sin(\omega t)

\end{equation}

The resulting current is given by:

\begin{equation}

i_c(t) = C \frac{\mathrm{d}v_c(t)}{\mathrm{d}t} = C \hat{V} \cos(\omega t) \omega

\end{equation}

As can be seen the current is shifted \(90^{\circ}\) to the right with respect to the voltage, because:

\begin{equation}

\cos(\omega t) = \sin(\omega t + \cfrac{\pi}{2})

\end{equation}

A right shift means that the current leads (appears before) the voltage in time.

A similar approach is taken at deriving the inductor voltage, this time by assuming a current:

\begin{equation}

i_L(t) = \hat{I} \sin(\omega t)

\end{equation}

The resulting voltage is given by:

\begin{equation}

v_L(t) = L \frac{\mathrm{d}i_L(t)}{\mathrm{d}t} = L \hat{I} \cos(\omega t) \omega

\end{equation}

Again we see the \(90^{\circ}\) right shift, but this time it is the voltage that is shifted to the right of the current. I.e. the voltage appears before the current, or the current lags the voltage.

To summarize:

- Capacitor: current leads the voltage
- Inductor: current lags the voltage

The reactive power produced (or consumed) by a capacitor is given by:

\begin{equation}

Q_C = \frac{V^2}{\cfrac{1}{j \omega C}} = V^2 \cdot j(\cdot 2 \cdot \pi \cdot f \cdot C)

\end{equation}

Similarly for the inductor the reactive power is given as:

\begin{equation}

Q_L = \frac{V^2}{j \omega L} = \frac{V^2}{j (2 \cdot \pi \cdot f \cdot L)}

\end{equation}

The symbol \(j\) is the imaginary unit, and:

\begin{equation}

\frac{1}{j} = -j

\end{equation}

Thus the inductive and capacitive reactive powers have opposite signs.

If you find these concepts difficult, here is a different way to look at it. Consider the figure to the right where a switch is used to connect the capacitor to the voltage source at \(t = t_0\). The capacitor is initially discharged, and because there will always be a certain source impedance it takes some time to charge the capacitor to the voltage \(V_{DC}\). The initial voltage will be zero, while the initial current will be large. Hence the current comes before (leads) the voltage.

In a similar manner, when a inductor is initially connected to a DC voltage, the inductor will immediately induce a magnetic field opposing the current. Thus the voltage across the inductor will rise instantly. The current will ramp up slowly, at a rate determined by the inductance. Hence the voltage comes before (leads) the current.

### Reactive power in power electronics

In a voltage source inverter we are able to control the output voltage to a large degree, within the limitations of the DC-link voltage magnitude.

The figure to the right attempts to illustrate the connection between a inverter and the power grid.

The power equations are given as:

\begin{equation}

P_{out} = \frac{V E}{X_s} \sin (\delta)

\end{equation}

\begin{equation}

Q_{out} = \frac{V E}{X_s} \cos (\delta) - \frac{V^2}{X_s}

\end{equation}

Where \(E\) is the inverter voltage output magnitude, \(V\) is the grid voltage magnitude, and \(\delta\) is the angle between grid and inverter voltage.

In practice we have two control variables, the inverter output voltage magnitude \(E\), and the voltage angle \(\delta\). The voltage magnitude is used to control the production (or consumption) of reactive power, while the voltage angle is used to control active power. They are both easily controlled by modern modulation techniques such as space vector pulse width modulation.

If the voltage angle \(\delta = 90^{\circ}\), \( \sin(\delta) = 1\), and we have maximum power output for the given voltage. If on the other hand \(\delta = 0^{\circ}\), we have zero active power, and maximum reactive power.

The figure to the right gives a somewhat more detailed picture of the operation of the inverter. The grid is modelled as a series connection of a resistor, a inductor and a voltage source (this could also be the model of a motor). The DC-bus of the inverter is shown with a parallel connection of a capacitor and a voltage source. The reader should not be confused into believing that the capacitor is needed for supplying reactive power however. It's purpose is to maintain a fixed DC-link voltage, and ideally at a level higher than the peak of the grid voltage. When the voltage is constant it does not make sense to talk about reactive power, although the same currents that are regarded as reactive on the AC side will also flow on the DC side (and regenerative currents will charge the capacitor).

## The measurement of power in (three) phase systems

Three phase power measurement methods are often classified by the number of single phase watt meters that are used in the measurement.

The single watt meter method is only applicable to symmetrical loads, while the two and three watt meter methods are applicable to all kinds of loads. The difference between the two, and three watt meter method is that the latter supports unsymmetrical systems with neutral conductors, while the former requires a isolated neutral point.

### Blondel's theorem

Blondel's theorem, from electrical engineer André Blondel, states that for a system of \(N\) conductors, the required number of watt meters is \(N-1\).