# AC Systems

## Properties of a Sinusoid

### General Form

$$ v(t) = \hat{V} \sin ( \omega t + \phi ) $$

Where \( \hat{V} \) is the amplitude, \( \omega \) is the angular frequency, \( \phi \) is the phase.

### Period

Time interval for the wave to repeat itself.

$$ v(t) = v(t+ T) $$

### Frequency

Number of repetitions per unit time (second).

$$ f = \frac{1}{T} $$

### Angular Frequency

Radians per second, used as mathematical convenience.

$$ \omega = 2 \pi f $$

### Phase

## RMS Values

To achieve the same power dissipation, the square of the peak voltage of an AC system must be twice the square of the DC system.

$$ \hat V ^2 = 2 V_{DC} ^2 $$

Leading to the definition of RMS:

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

$$ I_{RMS} = \frac{\hat I}{\sqrt{2}} $$

Where \( \hat V, \hat I \) are the peak voltage and current in an AC system, \( V_{RMS}, I_{RMS} \) are the voltage and current of a DC system required to achieve the same power dissipation.

## Components under AC Excitation

### Resistors

Assuming current takes the form

$$ i(t) = \hat{I} \sin (\omega t) $$

By Ohm's law, voltage will take the form

$$ v(t) = R \hat I \sin ( \omega t ) = \hat V \sin ( \omega t ) $$

There is no phase difference between the current and voltage waveforms.

The peak value of the current and voltage can be related with

$$ \hat V = R \hat I $$

Instantaneous power of the resistor under AC excitation can be found by multiplying the current and voltage waveforms.

$$ p(t) = \frac{\hat V ^2}{R} \frac{1}{2} [1 + \cos (2 \omega t )] $$

Average power can be found by integrating instantaneous power and dividing by the period.

$$ P_{ave} = \frac{\hat V ^2}{2R} $$

### Capacitors

$$ v(t) = \hat V \sin (\omega t - \frac{\pi}{2}) $$

Voltage lags behind current by \( \frac{\pi}{2} \), or \( 90 ^\circ \).

Peak voltage and current related by

$$ \hat V = \frac{1}{\omega C} \hat I $$

Where reactance, \( X_C \)

$$ X_C = \frac{1}{\omega C} $$

Since capacitors charge and discharge over one period, the average power dissipation is zero. Instantaneous power can be found by

$$ p(t) = - \frac{1}{2} \frac{1}{\omega C} \hat I ^2 (\sin (2 \omega t) + \sin (0)) $$

### Inductors

$$ v(t) = \hat V \sin (\omega t + \frac{\pi}{2}) $$

Voltage leads current by \( \frac{\pi}{2} \), or \( 90 ^\circ \).

Peak voltage and current related by

$$ \hat V = \omega L \hat I $$

Where reactance, \( X_L \)

$$ X_L = \omega L $$

Average power dissipation also integrates to zero, instantaneous power is found by

$$ p(t) = - \frac{1}{2} \omega L \hat I ^2 (\sin (2 \omega t) + \sin (0)) $$