# Electronics Handbook/Electricity Source

There are 2 sources of Electricity used in provide power to Electronic circuits to operate . DC which stands for Direct Current and AC which stands for Alternating Current

## DC

Direct Current provides constant voltage over time and the flow of electric charge is only in one direction. Mathematically DC can be represented by a function

$v(t) = v$

Direct current is generated from many sources

## AC

Alternate Current provide voltage varies sinusoidally over time with one complete cycle equals to $2\pi$

A sine wave, over one cycle (360°). The dashed line represents the root mean square (RMS) value at about 0.707

An AC voltage v can be described mathematically as a function of time by the following equation:

$v(t)=V_\mathrm{peak}\cdot\sin(\omega t)$,

where

• $\displaystyle V_{\rm peak}$ is the peak voltage (unit: volt),
• $\displaystyle\omega$ is the angular frequency (unit: radians per second)
• The angular frequency is related to the physical frequency, $\displaystyle f$ (unit = hertz), which represents the number of cycles per second , by the equation $\displaystyle\omega = 2\pi f$.
• $\displaystyle t$ is the time (unit: second).

The peak-to-peak value of an AC voltage is defined as the difference between its positive peak and its negative peak. Since the maximum value of $\sin(x)$ is +1 and the minimum value is −1, an AC voltage swings between $+V_{\rm peak}$ and $-V_{\rm peak}$. The peak-to-peak voltage, usually written as $V_{\rm pp}$ or $V_{\rm P-P}$, is therefore $V_{\rm peak} - (-V_{\rm peak}) = 2 V_{\rm peak}$.

### Power and root mean square

The relationship between voltage and the power delivered is

$p(t) = \frac{v^2(t)}{R}$ where $R$ represents a load resistance.

Rather than using instantaneous power, $p(t)$, it is more practical to use a time averaged power (where the averaging is performed over any integer number of cycles). Therefore, AC voltage is often expressed as a root mean square (RMS) value, written as $V_{\rm rms}$, because

$P_{\rm time~averaged} = \frac{{V^2}_{\rm rms}}{R}.$

For a sinusoidal voltage:

$V_\mathrm{rms}=\frac{V_\mathrm{peak}}{\sqrt{2}}.$

The factor $\sqrt{2}$ is called the crest factor, which varies for different waveforms.

$V_\mathrm{rms}=\frac{V_\mathrm{peak}}{\sqrt{3}}.$
$\displaystyle V_\mathrm{rms}=V_\mathrm{peak}.$

### Example

To illustrate these concepts, consider a 230 V AC mains supply used in many countries around the world. It is so called because its root mean square value is 230 V. This means that the time-averaged power delivered is equivalent to the power delivered by a DC voltage of 230 V. To determine the peak voltage (amplitude), we can rearrange the above equation to:

$V_\mathrm{peak}=\sqrt{2}\ V_\mathrm{rms}.$

For our 230 V AC, the peak voltage Vpeak is therefore $\displaystyle 230 V \times\sqrt{2}$, which is about 325 V. The peak-to-peak value $\displaystyle V_{P-P}$ of the 230 V AC is double that, at about 650 V.

Note that some countries use a frequency of 50 Hz, while others use a frequency of 60 Hz. The calculation to convert from RMS voltage to peak voltage is independent of the frequency.

In certain applications, different waveforms are used, such as triangular or square waves. Audio and radio signals carried on electrical wires are also examples of alternating current. In these applications, an important goal is often the recovery of information encoded (or modulated) onto the AC signal.