\documentclass{article}
\begin{document}
\author{Yosun Chang}
\title{AC Circuits}
\maketitle

An AC circuit is characterized by an \emph{alternating current}, which is created from an \emph{alternating voltage} source. Thus, one can formulate the basic 
equations of AC circuits via using either current or voltage. Here, an alternating current (i.e., oscillating in time) will be used,  
\begin{equation}
I(t)=I_0 \sin(\omega t) \label{I}
\end{equation}
where $I_0$ refers to the peak current or the amplitude of the current sinusoid.

The average current is 0, since $\langle I(t)\rangle=\frac{1}{2\pi}\int_0^{2\pi} \sin(\omega t) d(\omega t)=0$. The average current squared is 
$\langle I^2 \rangle=\frac{1}{2\pi}\int_0^{2\pi}I_0^2 (\sin(\omega t))^2 d(\omega t)=\frac{I_0^2}{2}$. Thus, the root mean square current is $I_{rms}=\sqrt{\langle I_0^2\rangle}=\frac{I_0}{\sqrt{2}}$

\section{R Only}
Plug in (\ref{I}) into Ohm's Law and you get $V=IR=I_0 R \sin(\omega t)=V_0 \sin(\omega t)$. Thus, for a resistor, the current and voltage are proportional to each other; in other words, the current and voltage are \emph{in phase}.

\begin{center}
\fbox{$V(t)=V_0 \sin(\omega t)$ and $I(t)=I_0 \sin(\omega t)$}
\end{center}

Energy flows from the source to be dissipated by the resistor. The average power (energy dissipated)  is $\langle P \rangle=\langle I(t)V(t) \rangle=I_0 V_0 \langle \sin^2(\omega t)\rangle=\frac{I_0 V_0}{2}=\frac{I_0^2 R}{2}=I_{rms}^2 R=\frac{V_{rms}^2}{2}$

\section{L Only}

Plugging in the time derivative of (\ref{I}) into $V=L \dot{I}$, one gets $V(t)=L \omega I_0 \cos(\omega t)=V_0 \sin(\omega t + \frac{\pi}{2})$. One sees that the current lags behind the voltage by $\pi/2$. 

\begin{center}
\fbox{$V(t)=V_0 \cos(\omega t)$ and $I(t)=I_0 \sin(\omega t)$}
\end{center}

In the above, the following relation was used: $V_0=I_0 \omega L$. In the spirit of the familiar Ohm's Law, let's define a quantity with the same units of resistance. But, let's call it the \emph{inductive reactance} or impedance. 

\begin{center}
\fbox{$V_0=I X_{L0} \Rightarrow X_{L0}=\omega L$}
\end{center}

The average power is 0 because the integral of the product of $I(t)$ and $V(t)$ averages to 0. This means that energy is sent from the voltage source to the magnetic field of the inductor and back to the voltage source--with no net dissipation. 

\section{C Only}

The voltage across a capacitor is defined as $Q=CV \Rightarrow V=\frac{Q}{C}$. But, $I=\frac{dQ}{dt}\Rightarrow Q=\int I_0 \sin(\omega t)dt =-\frac{I_0}{\omega}\cos(\omega t) $, where (\ref{I}) is used for current. Thus, $V(t)=\frac{Q(t)}{C}=\frac{-I_0\cos(\omega t)}{\omega C}=-V_0 \sin(\omega t - \frac{\pi}{2})$. The current leads the voltage by $\pi/2$. 

\begin{center}
\fbox{$V(t)=-V_0 \cos(\omega t)$ and $I(t)=I_0 \sin(\omega t)$}
\end{center}

In the above, the relation $V_0=I_0 \frac{1}{\omega C}$ is used. This suggests that a \emph{capacitative reactance} be defined as $X_C=\frac{1}{\omega C}$.

\begin{center}
\fbox{$V_0=I X_{C0} \Rightarrow X_{C0}=\frac{1}{\omega C}$}
\end{center}

The average power is 0 because the integral of the product of $I(t)$ and $V(t)$ averages to 0. This means that energy is sent from the voltage source to the electric field between the plates of the capacitor and back to the voltage source--with no net dissipation. 

It turns out that only resistors dissipate energy in AC circuits.


\section{LRC Circuit Analysis via Phasors}
If you get vectors, you get phasors. The only difference is that phasors, as used here, involves rotating vectors that are either (anti)parallel or at right angles to each other. 


\end{document}
