test.tex

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\begin{document}

\ifthenelse{\boolean{slideshow}}{\mode<presentation>}{}

\subtitle{212MP \& 5018CEM Electromagnetism}
\title{\shadowtext{ Week 10: Driven RLC Circuits (AC circuits)}}


\author[Alex Pedcenko]{ Alex Pedcenko}
\institute[CEM]{
  School of Computing, Electronics and Mathematics \\
  
 }
\date{{ ~Week 10 ~~\today}}


\normalsize
\begin{frame}[plain]

  \titlepage
\end{frame}






\begin{frame}{\shadowtext{Recap: RLC Circuits}}
\vspace{2mm}\p
\begin{columns}[T]
\begin{column}{0.45\linewidth}
\vspace{-7mm}
\begin{figure}
\centering
\includegraphics[width=0.8\linewidth]{./RLC1}
\end{figure}\p
\end{column}

\begin{column}{0.5\linewidth}
\begin{itemize}\itemsep0.35em
  \item[1a)] initially switch S1 is at "a", so capacitor is uncharged ($U_E=0$), but current $I(0)=I_0$  is at maximum ($U_B=\frac{1}{2}LI_0^2)$\p
  \item[1b)] then switch is switched to "b" and inductor is "discharging" (start to charge capacitor), current decaying.
 \end{itemize}
\end{column}
\end{columns}
\p


\begin{columns}[T]
\begin{column}{0.45\linewidth}
\vspace{-3mm}
\begin{figure}
\centering
\includegraphics[width=0.8\linewidth]{./RLC2}

\end{figure}
\end{column}\p

\begin{column}{0.5\linewidth}
\vspace{3mm}

\begin{itemize}\itemsep0.35em
 \item[2a)] Initially capacitor is charged ($U_E=\frac{1}{2}\frac{Q^2}{C}$) and current is zero\p
  \item[2b)] Then the switch is brought to state "b" and capacitor start to discharge producing current $I(t)$ in the circuit.
\end{itemize}
\end{column}
\end{columns}



\end{frame}





\begin{frame}{\shadowtext{Undriven RLC Circuits}}
\vspace{2mm}
\begin{columns}[T]
\begin{column}{0.2\linewidth}
\vspace{-7mm}
\begin{figure}
\centering
\includegraphics[width=1\linewidth]{./RLCa}
\caption{Undriven RLC }
\label{fig:undrivenRLC}
\end{figure}



\end{column}\p
\begin{column}{0.75\linewidth}

\begin{itemize}\itemsep0.35em
	\item So, in RLC circuits we looked at before (including RC, RL, LC, RLC varieties)  the DC battery with e.m.f. $\varepsilon$ was providing "initial conditions" only: \p e.g. initial charge $Q_o$ on the capacitor plates or initial current $I_o$  through the inductor,\p
	\begin{itemize}\itemsep0.5em
	  \item i.e. $V_C(0)=\varepsilon=\frac{Q_0}{C}$ \p  or $I(0)=I_o=\frac{\varepsilon}{R}$ \p
	  \item which is like providing initial displacement for a mass attached to a spring ($\Delta x(0)\rightarrow Q_o$) or initial velocity ($v(0)\rightarrow I_o$) of that mass. \p (i.e. initial potential or kinetic energy). \p
	\end{itemize}

\end{itemize}
\end{column}
\end{columns}
\vspace{-2mm}
\begin{itemize}\itemsep0.35em
\item Then we "released our mass" (i.e. removed battery or closed the switch causing capacitor to discharge) and observed one of few possible transient processes: \p
\begin{itemize}\itemsep0.5em
\item exp. decay of current ($R^2\ge 4L/C$ over/critically damped RLC\&RC\&RL), free oscillations ($R=0$: LC), or  decaying oscillations ($R^2<4L/C$: underdamped RLC)
\end{itemize}

\end{itemize}

\end{frame}




\begin{frame}{\shadowtext{Driven RLC Circuits}}
	\vspace{2mm}
	
	\begin{columns}[T]
		\begin{column}{0.3\linewidth}\vspace{-5mm}
			\begin{figure}
				\centering
				\includegraphics[width=1.1\linewidth]{./drivenRLC}
				\caption{Driven RLC }
				\label{fig:drivenRLC}
			\end{figure}
			\p
			
		\end{column}
		\begin{column}{0.70\linewidth}
			\textbf{~~~Undriven (unforced)}
			\centered{$L\frac{\d I}{\d t}+IR+\frac{Q}{C}=0$}\p
			\textbf{~~~Driven (forced)}
			\centered{$L\frac{\d I}{\d t}+IR+\frac{Q}{C}=\overbrace{\varepsilon(t)}^{\text{external forcing}}$}\p
			%Voltage source is now sinusoidal with angular frequency $\omega$ and amplitude $V_0$:
		\end{column}
	\end{columns}
	\begin{columns}[T]
		\begin{column}{0.5\linewidth}\vspace{3mm}
			
			%\begin{itemize}\itemsep0.5em
				%\item $\omega=2\pi f$ , rad/s\p
				%\item Period: $T=1/f$, seconds\p
				%\item $f=\omega/2\pi$, Hz \p ~~Amplitude $V_0$, Volts\p
			%\end{itemize}
			
		\end{column}
		\begin{column}{0.5\linewidth}
			
			%\vspace{-8mm}\centered{ \includegraphics[width=0.65\linewidth]{./AC_voltage}}
		\end{column}
	\end{columns}
\end{frame}






\begin{frame}{\shadowtext{Driven RLC Circuits}}
\vspace{2mm}

\begin{columns}[T]
\begin{column}{0.3\linewidth}\vspace{-5mm}
\begin{figure}
\centering
\includegraphics[width=1.1\linewidth]{./drivenRLC}
\caption{Driven RLC }
\label{fig:drivenRLC}
\end{figure}


\end{column}
\begin{column}{0.70\linewidth}
\textbf{~~~Undriven (unforced)}
\centered{$L\frac{\d I}{\d t}+IR+\frac{Q}{C}=0$}
\textbf{~~~Driven (forced)}
\centered{$L\frac{\d I}{\d t}+IR+\frac{Q}{C}=\overbrace{V_0\sin(\omega t+\varphi)}^{\text{external forcing}}$}\p 
Voltage source is now sinusoidal with angular frequency $\omega$ and amplitude $V_0$:
\end{column}
\end{columns}
\begin{columns}[T]
\begin{column}{0.5\linewidth}\vspace{3mm}

\begin{itemize}\itemsep0.5em
  \item $\omega=2\pi f$ , rad/s;\p~~~Period: $T=1/f$, s\p
  \item $f=\omega/2\pi$, Hz; \p ~~~~~Amplitude $V_0$, Volts\p
\end{itemize}

\end{column}
\begin{column}{0.5\linewidth}

\vspace{-6mm}\centered{ \includegraphics[width=0.65\linewidth]{./AC_voltage}}
\end{column}
\end{columns}
\end{frame}



\begin{frame}{\shadowtext{Driven RLC Circuits}}
\vspace{2mm}

\begin{columns}[T]
\begin{column}{0.3\linewidth}\vspace{-5mm}
\begin{figure}
\centering
\includegraphics[width=1.1\linewidth]{./drivenRLC}
\caption{Driven RLC }
\label{fig:drivenRLC}
\end{figure}
\p

\end{column}
\begin{column}{0.70\linewidth}
\begin{itemize}\itemsep0.5em
  \item When a voltage source is connected to an RLC circuit, energy is provided to compensate the energy dissipation in the resistor, and the oscillation will \textbf{no longer damp out}.\p
  \item The oscillations of charge, current and potential difference are called driven or \textbf{forced oscillations}.\p
  \item After the initial "transient time", an AC current will flow in the circuit as a response to the driving voltage. \p The current, written as
  \centered{$I(t)=I_0\sin(\omega t -\varphi)$}\p
  \item will oscillate with the same frequency as the voltage source, with an amplitude $I_0$ and phase $\varphi$ that depends on the driving frequency $\omega$.
\end{itemize}
\end{column}
\end{columns}
\end{frame}

\begin{frame}{\shadowtext{Simple AC Circuits: Purely Resistive (active) Load}}
\vspace{2mm}

\begin{columns}[T]
\begin{column}{0.5\linewidth}\vspace{-5mm}
\begin{figure}
\centering
\includegraphics[width=\linewidth]{./AC_R}
\caption{Purely active load}
\label{fig:AC_R}
\end{figure}
\p

\end{column}
\begin{column}{0.5\linewidth}
\begin{itemize}\itemsep0.5em
  \item Voltage source provide sinusoidal e.m.f 
  
  \centered{$V(t)=V_0\sin\omega t$} \p
  
  Kirchhoff's 2nd law:~~~~ \centered{$V(t)-I(t)R=0$ \p}
  
  \centered{$\therefore~~~I(t)=\frac{V_0 \sin\omega t}{R}=I_0\sin\omega t$}\p
  
  

\end{itemize}
\end{column}
\end{columns}

\centered{Amplitudes:~~~~~~{$V_0$~~~~~~and~~~~~$I_0=\frac{V_0}{R}$}}


\end{frame}

\begin{frame}{\shadowtext{Phasor Diagram for Active Load}}
\vspace{2mm}

\begin{columns}[T]
\begin{column}{0.3\linewidth}
\vspace{-5mm}

~~\includegraphics[width=\linewidth]{./Active_VI}


\end{column}\p
\begin{column}{0.7\linewidth}
\begin{itemize}\itemsep0.5em

  \item  $I(t)$ and  $V(t)$ \textbf{oscillate in phase}:
  
  \centered{$V(t)=V_0 \sin\omega t$~~~~$I(t)=I_0\sin\omega t$}

\end{itemize}
\end{column}
\end{columns}
\p

\begin{columns}[T]
\begin{column}{0.3\linewidth}\vspace{-3mm}
\begin{center}
\includegraphics[width=1\linewidth]{./phasor_R}
\end{center}
\end{column}\p
\begin{column}{0.7\linewidth}
\vspace{-15mm}
\begin{itemize}\itemsep0.5em
\item The behavior of $I(t)$ and $V(t)$ can also be represented with a \textbf{phasor diagram}.\p
\item A phasor is a rotating vector having the following properties:\p
\begin{itemize}\itemsep0.5em
\item \textbf{length}: the length corresponds to the amplitude.\p
\item angular speed: the vector rotates counter-clockwise with an angular speed $\omega$.\p
\item projection: the projection of the vector along the vertical axis corresponds to the value of the alternating quantity at time $t$.

\end{itemize}
\end{itemize}

\end{column}
\end{columns}


\end{frame}

\begin{frame}{\shadowtext{RMS or "effective" value of $I(t)$ or $V(t)$}}
\vspace{2mm}
\begin{itemize}\itemsep0.5em
\item AC Voltage can do work: i.e. you connect your electric kettle to a wall-socket, it will boil water. \p 
\item However \textbf{mean value} of sinusoidal current which flows in the heating element of that kettle is 0: \p
\centered{$\langle I \rangle=\frac{1}{T}\int_0^T I(t)~d t\p =\frac{1}{T}\int_0^T I_0\sin\omega t~ dt=0$}\p
\item But electric power isn't sensitive to the instantaneous direction of current:
\centered{$P=I(t)^2 R=I_0^2R\sin^2\omega t$~~~~or~~~~$P=V(t)^2/R$}\p
\item Therefore we need mean of the $I^2(t)$ or $V^2(t)$, \p i.e. "effective" value (\textbf{root mean square} values) of  current or  voltage:\p

\hspace{-10mm}{$\langle I^2(t) \rangle =\frac{I_0^2}{T}\int_0^T \sin^2\omega t~d t\p=\frac{I_0^2}{T}\int_0^T\frac{1-\cos 2\omega t}{2}~d t\p=\frac{I_0^2}{2T}\bigg( T+\cancelto{0}{\biggl[\frac{\sin 2\omega t}{2\omega}\biggr]_0^T} \bigg)\p = \frac{I_0^2}{2}$}\p
 
\includegraphics[width=0.2\linewidth]{cos2x} \hspace{2cm} \p {\fbox{$I_{RMS}=\sqrt{\langle I\rangle}=I_0/\sqrt{2}$}~~~~~~$V_{RMS}=V_0/\sqrt{2}$}
 
\end{itemize}


\end{frame}

\begin{frame}{\shadowtext{Purely Inductive (Reactive) Load}}
\vspace{2mm}

\begin{columns}[T]
\begin{column}{0.3\linewidth}\vspace{-5mm}
\begin{figure}
\centering
\includegraphics[width=\linewidth]{./AC_L}
\caption{Purely Inductive load}
\label{fig:AC_R}
\end{figure}
\p

\end{column}
\begin{column}{0.7\linewidth}
\begin{itemize}\itemsep0.5em
  \item Kirchhoff's rule now reads~~~$V(t)-V_L(t)=0$\p
  \centered{$V(t)-\p {L\frac{\d I}{\d t}}=0$}\p
  \item To find current through the inductor we make $\frac{\d I(t)}{\d t}$ the subject:

\end{itemize}
\end{column}
\end{columns}
\begin{itemize}\itemsep0.5em
\item $\frac{\d I(t)}{\d t}=\frac{V(t)}{L}\p = \frac{V_{0L}}{L}\sin\omega t $\p ~~~~where ~~~$V_{0L}=V_0$\p
\item $I(t)=\int \d I \p = \frac{V_0}{L}\int\sin\omega t \d t \p = -\frac{V_0}{\omega L}\cos\omega t \p = \frac{V_0}{\omega L}\sin(\omega t-\varphi)$; \p~~ $\varphi=\frac{\pi}{2}$\p
\item We see that the amplitude value of the current through the inductor is
\centered{$I_0\Rightarrow \frac{V_0}{\omega L} \p = \frac{V_0}{X_L}$ \p ~~~where~~~\fbox{$X_L=\omega L$} ~~~~ is \textbf{inductive reactance}, Ohms, [$\Omega$]}
\end{itemize}

\end{frame}

\begin{frame}{\shadowtext{Phasor Diagram for purely inductive load}}
\vspace{-3mm}

\begin{columns}[T]
\begin{column}{0.5\linewidth}

\begin{figure}
\centering
\includegraphics[width=0.7\linewidth]{./Reactive_VI_plot}
\caption{$I(t)$ and $V(t)$ for inductive load}
\label{fig:Reactive_VI_plot}
\end{figure}


\end{column}
\begin{column}{0.5\linewidth}
\vspace{-3mm}
\begin{figure}
\centering
\includegraphics[width=0.6\linewidth]{./phasor_L}
\caption{Phasor diagram}
\label{fig:phasor_L}
\end{figure}

\end{column}
\end{columns}
\begin{itemize}\itemsep0.5em
\item The current $I(t)$ is out of phase with $V(t)$ by $\varphi=\pi/2$ \p
\item it reaches its maximum value after $V(t)$ does, delayed by one quarter of a cycle.\p
\item  \fbox{\textbf{The current LAGS the voltage by  $\pi/2$ in a purely inductive circuit}}

\end{itemize}
 

\end{frame}


\begin{frame}{\shadowtext{Purely Capacitive Load}}
\vspace{2mm}

\begin{columns}[T]
\begin{column}{0.3\linewidth}\vspace{-5mm}
\begin{figure}
\centering
\includegraphics[width=\linewidth]{./AC_C}
\caption{Purely Capacitive load}
\label{fig:AC_C}
\end{figure}
\p

\end{column}
\begin{column}{0.7\linewidth}
\begin{itemize}\itemsep0.5em
  \item Kirchhoff's rule now reads~~~$V(t)-V_C(t)=0$\p
  \centered{$V(t)-\frac{Q}{C}=0$}\p
  \item To find current we make $Q$ the subject and differentiate: \p \centered{$Q=V(t)C \p = CV_0\sin\omega t ~~~\p~~~ V_{C0}=V_0$}\p 
  
  \item current  is then \p

\end{itemize}
\end{column}
\end{columns}
\vspace{-1mm}
\begin{itemize}\itemsep0.5em
\item $I(t)=\frac{\d Q}{\d t} \p =\omega C V_0\cos\omega t \p = \omega C V_0\sin(\omega t + \frac{\pi}{2})$ \p

\item  Amplitude of the current  is ~~~~~ $I_0=\omega C V_{c0} \p = \frac{V_{CO}}{X_C}$~~~ \p where ~~\fbox{~$X_C=\frac{1}{\omega C}$}

is \textbf{capacitance reactance}. \p 

\item Angle between $V(t)$ and $I(T)$ is then $\varphi=-\pi/2$ 

\end{itemize}

\end{frame}




\begin{frame}{\shadowtext{Phasor Diagram for purely capacitive load}}
\vspace{-3mm}


\begin{figure}
\centering
\includegraphics[width=0.7\linewidth]{./phasor_C}
\caption{$I(t)$ and $V(t)$ for capacitive load}
\label{fig:Reactive_VI_plot}
\end{figure}



\begin{itemize}\itemsep0.5em
\item Notice that at $t=0$, the voltage across the capacitor is zero while the current in the circuit is at a maximum (beginning of charging). \p

\item In fact, $I(t)$ reaches its maximum before $V_C(t)$ by one quarter of a cycle. Thus, we say that \p
\item \fbox{\textbf{The current LEAD the voltage by  $\pi/2$ in a purely capacitive circuit}}
\end{itemize}
 

\end{frame}



\begin{frame}{\shadowtext{The RLC Series Circuit}}
\vspace{2mm}

\begin{columns}[T]
\begin{column}{0.3\linewidth}\vspace{-5mm}
\begin{figure}
\centering
\includegraphics[width=\linewidth]{./drivenRLC}
\caption{RLC Circuit}
\label{fig:AC_RLC}
\end{figure}
\p

\end{column}
\begin{column}{0.7\linewidth}
\begin{itemize}\itemsep0.5em
  \item Kirchhoff's rule now reads~~~$V(t)-V_R(t)-V_L(t)-V_C(t)=0$\p
  \centered{$V(t)-IR-L\frac{\d I}{\d t}-\frac{Q}{C}=0$}\p
  \item or: ~~~~~~~~~~~{$L\frac{\d I}{\d t}+IR+\frac{Q}{C}=V_0\sin\omega t$}\p 
  
  \item or in terms of charge $Q$:
  \centered{$L\frac{\d^2 Q}{\d t^2}+R\frac{\d Q}{\d t}+\frac{Q}{C}=V_0\sin\omega t$}\p

\end{itemize}
\end{column}
\end{columns}

\begin{itemize}\itemsep0.5em
\item Solution to the above inhomogeneous equation is 
$Q(t)=Q_0\cos(\omega t-\varphi)$

\item  Amplitude can be found to be $Q_0=\frac{V_0}{\omega \sqrt{R^2+(X_L-X_C)^2)}}$ \p

\item and phase angle ~~~~~~~ $\varphi=\tan^{-1}\left(\frac{X_L-X_C}{R}\right)$

\end{itemize}

\end{frame}

\begin{frame}{\shadowtext{The RLC Series Circuit}}
\vspace{0mm}
\begin{columns}[T]
\begin{column}{0.3\linewidth}\vspace{-5mm}
\begin{figure}
\centering
\includegraphics[width=\linewidth]{./drivenRLC}
\caption{RLC Circuit}
\label{fig:AC_RLC}
\end{figure}
\p

\end{column}

\begin{column}{0.7\linewidth}

\begin{itemize}\itemsep0em
\item Current is ~~~$I(t)=\frac{\d Q}{\d t}=I_0\sin(\omega t-\varphi)$ \p
\item Where ~~~$I_0=-Q_0\omega \p =-\frac{V_0}{\sqrt{R^2+(X_L-X_C)^2}}$ \p
\item The current has the same amplitude and phase at all points in the series RLC circuit.\p
\item But the instantaneous voltage across each of  $R$, $L$ and $C$ has a different amplitude and phase relationship with the current!\p



\end{itemize}

\end{column}
\end{columns}


\begin{itemize}\itemsep0em
\item Ohm's law still hold: \centered{\vspace{-5mm}$I(t)=\frac{V(t)}{\sqrt{R^2+(X_L-X_C)^2}}\p =\frac{V(t)}{|Z|}$}\p
 
 \item where \fbox{$|Z|=\sqrt{R^2+(X_L-X_C)^2}$} is called \textbf{impedance}, Ohms,  of an AC circuit.\p
  \item Reciprocal of impedance $Z$ is \textbf{complex admittance}:  \fbox{$Y=\frac{1}{Z}$}.
\end{itemize}


\end{frame}

\begin{frame}{\shadowtext{Complex Impedance}}
\vspace{2mm}
\begin{itemize}\itemsep0.5em
\item For a "purely ohmic" resistor, $Z = R$. Due to the phase effects,  the contributions of $X_C$ and $X_L$ differ in phase from resistive components by $\pi/2$ radians, a process like vector addition (phasors) is used to develop expressions for impedance. More general is the \textbf{complex impedance} method.\p
%\item The handling of the impedance of an AC circuit with multiple components quickly becomes unmanageable if sines and cosines are used to represent the voltages and currents.
\item A mathematical method which eases the difficulty is the \textbf{use of complex exponential functions}. The basic parts of the strategy are as follows\p
\centered{$e^{j\omega t}=\cos\omega t+j\sin\omega t$}

\item $V=V_0~e^{j\omega t}$ ~~~~~$I=I_0 ~e^{j\omega t-\varphi}$ \p
\item The impedance then can be expressed as complex exponential
\centered{$Z=\frac{V_0}{I_0}~e^{-j\varphi}\p = R+jX = R+j(X_L-X_C)$}\p

\item The impedance of individual elements can be expressed as pure real ($R$) or imaginary numbers: \p $R$~~~$X_L=j\omega L$~~~~~$X_C=-\frac{j}{\omega C}$
\end{itemize}



\end{frame}

\begin{frame}{\shadowtext{Impedance diagrams}}
\vspace{2mm}

\begin{columns}[T]
\begin{column}{0.5\linewidth}

\begin{center}
\includegraphics[width=1\linewidth]{./impedance_diag1}
\end{center}


\end{column}
\begin{column}{0.5\linewidth}

\begin{center}
\includegraphics[width=1\linewidth]{./impedance_diag}\\
Impedance diagram is basically the "Argand diagram" for impedances.
\end{center}


\end{column}
\end{columns}

\end{frame}

\begin{frame}{\shadowtext{Resonance}}
\vspace{2mm}


\begin{columns}[T]
\begin{column}{0.6\linewidth}
\begin{itemize}\itemsep0.5em

\item \textbf{Resonance} happen when $\omega$ of the source is the same as natural (resonant) frequency of the correspondent $LC$ circuit:\p

\item~~~~ $\omega=\omega_0=\sqrt{\frac{1}{LC}}$.\p
\item In this case $|X_L|=|X_C|$ and both reactive voltages compensate each other exactly, so that effective remaining load is purely active ($R$). \p That lead to dramatic drop in impedance, which becomes 
$$Z=\sqrt{R^2+\cancelto{0}{(X_l-X_C)^2}}=R$$ \p
\end{itemize}
\end{column}
\p
\begin{column}{0.35\linewidth}
\vspace{-8mm}
\begin{center}
\includegraphics[width=0.8\linewidth]{./RESONANCE}

Dependence of the current amplitude in RLC circuit on frequency of the source: $\omega=\omega_0$ is resonant frequency. 
\end{center}
\end{column}
\end{columns}
\vspace{2mm}

\begin{itemize}
	\item {In other words, at a \underline{resonance}, \textbf{imaginary parts of both complex impedance and admittance are zero}: $Im(Z)=Im(Y)=0$}
\end{itemize}
\end{frame}


\begin{frame}{\shadowtext{Tutorial/Homework questions}}
\vspace{2mm}
\begin{enumerate}\itemsep0.5em
\item I have to drive a solenoid with AC voltage from wall socket, say $V_{RMS}=220~V$, 50 Hz. Its active resistance is $R=2$ Ohms and inductance is $L=10~mH$. What maximum RMS current value  I can get through it?
\p

\item I need to optimise my circuit from above question. What must I do to maximize the current through the solenoid? Provide quantitative answer. Draw impedance diagram.
\p
\item If I only have a capacitor for 1 mF, how must I adjust frequency of my power supply to maximize the current through the solenoid? 
\p
 \item Solve  Homework questions from  Aula.
\end{enumerate}


\end{frame}


\end{document}