 # How Apparent is the Apparent Power?

When studying AC circuits, we always learn how to calculate the average power of an AC signal generated by a source or consumed by a load.

Here, I will present a diagram that illustrates the voltage and current waveforms on a load with the mathematical expressions of them been provided. Here, the blue line represents the voltage waveform, and the red line represents the current waveform on the load.

## Average Power

The average power consumed by the load, P, is given by:

\begin{align} P=\frac{1}{2}V_{m}I_{m}cos\left ( \phi_{1}-\phi_{2} \right )\end{align}

\begin{align*} =\frac{1}{2}V_{m}I_{m}cos\left ( \theta \right ) \end{align*}

\begin{align*} =V_{rms}I_{rms}cos\left ( \theta \right ) \end{align*}

\begin{align*} =\left | \mathbf{S} \right |cos\left ( \theta \right ) \end{align*}

\begin{align*} =\left | \mathbf{S} \right | \times P.F. \end{align*}

## Power Factor

In equation (1), the last term P.F. represents the Power Factor, which is defined as:

\begin{align} P. F. = cos\left ( \theta \right ) \end{align}

The Power Factor indicates the phase relationship between the voltage and current waveforms of the load.

• When the load is purely resistive, the current is in phase with the voltage, and there is no phase difference. In this case, the load receives a pure real power.
• However, if the load contains reactive components, such as inductance or capacitance, there will be a phase difference between the voltage and current, resulting in a P.F. less than 1.

Alright! Let's stop here. I didn't plan on talking about these formulas or the meanings of real power and reactive power, etc.

## Complex Power

In (1), we see that there is a term |S|, where S represents Complex Power:

\begin{align*} \mathbf{S}=\frac{1}{2}\mathbf{VI}^{*} \end{align*}

with boldface V and I represented as phasors denoting the sinusoidal voltage and current waves. Complex power, S, is measured in VA (volt-amperes) since it is the product of voltage and current.

The phasors used above are in peak representation. {alertInfo}

We won't discuss why S is the product of the phasor V and the conjugate of I here. You can refer to textbooks for further explanation. I also won't discuss what phasors are or their physical meanings. {alertInfo}

### Active Power

\begin{align*}P=\frac{1}{2}Re\left \{ \mathbf{VI}^{*} \right \}  \end{align*}
P is the real part of S, which is known as real power or active power. It is measured in watts (W), though it is still considered as VA, using W or Watt indicates that we are specifically referring to real power.

### Reactive Power

\begin{align*} Q=\frac{1}{2}Im\left \{ \mathbf{VI}^{*} \right \}  \end{align*}
Q is the imaginary part of S, which is known as fictitious power or reactive power. It is measured in VAR (volt-amperes reactive).

### Apparent Power

The absolute value of complex power, |S|, is referred to as apparent power, and the term "apparent" means obvious or evident.

Now the question arises: why is |S| so apparent? Why can we easily see it? Why isn't S as "apparent" as |S|? Does taking the absolute value make it more impressive?

Well, yes, it is impressive!

## Measurement

Let's go back to the Stone Age when technology was not so advanced, and humans first learned to make fire by rubbing sticks together... (Wait, what?!)

Alright! It's like engineers only have basic multimeters at their disposal, which can either measure the voltage across a load in parallel or the current passing through a load in series. Even if they have two multimeters working simultaneously, the measured voltage and current are just numerical values. They can't determine the phase relationship between V and I, right? So, engineers can only see the magnitudes of voltage and current but don't know their relative relationship.

If the multimeter measures the peak voltage Vm and peak current Im or even the RMS values, the engineer would say,

Oh, apparently, the power is given by:

\begin{align} P_{apparent}=\frac{1}{2}V_{m}I_{m}=\left | S \right | \end{align}

or

\begin{align} P_{apparent}=V_{rms}I_{rms} =\left | S \right |  \end{align}

In other words, the engineer can determine the apparent power by using only the magnitude of the voltage and current from measurements.

Regarding the truthfulness of this matter, let this awesome kid answer it! My bullshit ends here... Hahaha!

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simen

An enthusiastic engineer with a passion for learning. After completing my academic journey, I worked as an engineer in Hsinchu Science Park. Later, I ventured into academia to teach at a university. However, I have now returned to the industry as an engineer, again.

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