Phasor Representation of Sinusoids

Consider a sinusoidal voltage $v$, written as $v(t)=V_{peak}\cos (\omega t+\theta )$.

$V_{rms}$ is the RMS of $v$, such that

$$\begin{array}{r}{V}_{\text{rms}}=\sqrt{\frac{1}{T}{\int }_{t_1}^{t_1+T}{v}^2(t)d(t)dt}\\ I_{\text{rms}}=\sqrt{\frac{1}{T}{\int }_{t_1}^{t_1+T}{i}^2(t)d(t)dt}\end{array}$$

The peak and RMS values are related as $V_{\text{rms}}=\frac{V_{peak}}{\sqrt{2}},I_{\text{rms}}=\frac{I_{peak}}{\sqrt{2}}$.

Thus, we can say that alternatively say that $v=\sqrt{2}{V}_{\text{rms}}\cos (\omega t+\theta )$, which we call time-function representation. This can converted to phasor:

$$v(t)=\sqrt{2}{V}_{\text{rms}}\cos (\omega t+\theta )⟶\vec{V}=V_{\text{rms}}{e}^{j\theta }=V_{\text{rms}}\angle \theta$$

The idea of this is that if a vector $\vec{V}$ of magnitude $V_{peak}$ and phase angle $\theta$ with respect to the real axis at $t=0$, rotates in the positive direction (ccw) at an angular speed of $\omega$ radians per second, its real component is $V_{\text{peak}}\cos (\omega t+\theta )$.

Therefore, the value of sinusoidal voltage $v$ is equal to $\sqrt{2}$ times the component of $\vec{V}$ along the real axis. Thus, all the information about $v$ can be represented by $\vec{V}$. A phasor is a vector that rotates counterclockwise.

Conversion

We can convert between phasor and time-function representation with:

$$\begin{array}{rl}v(t)& =\text{Re}\{\sqrt{2}{\vec{V}}_{\text{rms}}{e}^{j\omega t}\}\\ & =\text{Re}\{\sqrt{2}(V_{\text{rms}}{e}^{j\theta })e^{j\omega t}\}\\ & =\text{Re}\{\sqrt{2}{V}_{\text{rms}}{e}^{j(\omega t+\theta )}\}\\ & =\sqrt{2}{V}_{\text{rms}}\cos (\omega t+\theta )\end{array}$$