AC Theory

Note

Only sinusoidal AC supply are considered in s1.

Say $v$ is alternating as in $v=V_{m}\sin(\omega{t}+\phi)$.

Why AC instead of DC?

• Production of AC is less expensive
• AC devices are efficient and require less maintenance

Peak value

Maximum instantaneous value. $V_m$ in the example. $$

Peak-to-peak value

Maximum variation between maximum positive and negative instantaneous values. For a sinusoidal waveform, this is twice the peak value. $2V_m$ in the example. $$

Mean value

$$v_{\text{mean}}= \frac{1}{T} \int_{T_0}^{T_0+T}{v(t)\text{d}t}$$

Here:

• $T_0$ is the starting time of a cycle
• $T$ is the periodic time

For any symmetric waveform, mean value is $0$. $$

Average value

Mean value of the rectified version of a waveform.

For symmetric waveforms, half-cycle mean value is taken as the average value.

$$v_{\text{average}}= \frac{2}{T} \int_{T_0}^{T_0+\frac{T}{2}}{v(t)\,\text{d}t}$$

For sinusoidal waveforms, from the example:

$$v_{\text{average}} = \frac{2}{T} \int_{T_0}^{T_0+\frac{T}{2}}{V_{m}\sin(\omega{t}+\phi)\,\text{d}t}$$ $$= \frac{2}{\pi}V_m = 0.637V_m$$

rms value

Aka. effective value. rms value is always used to express the magnitude of a time varying quantity.

$$v_{\text{rms}}= \sqrt{ \frac{1}{T} \int_{T_0}^{T_0+T}{v(t)^2\,\text{d}t} }$$

For sinusoidal waveforms:

$$v_{\text{rms}}= V_m \sqrt{ \frac{1}{T} \int_{T_0}^{T_0+T}{\sin^2{(\omega{t}+\phi)}\,\text{d}t} } = \frac{V_m}{\sqrt{2}}$$

Note

$i_{\text{rms}}$ is the equivalent current that dissipates same amount of power across a resistor $R$ in time $T$ as $i(t)$. Similar for voltage.

Instantaneous power

$$P=vi=i^2R$$

Form factor

$$\text{Form factor} = \frac{\text{rms value}}{\text{average value}} = {\frac{V_m}{\sqrt{2}}}\times{\frac{\pi}{2{V_m}}} =1.111$$

Peak factor

$$\text{Peak factor} =\frac{\text{peak value}}{\text{rms value}} ={V_m}\times{\frac{\sqrt{2}}{V_m}} =1.412$$