# 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.

$$\omega = 2 \pi f$$

## 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

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