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%% DESIGN CIRCUITS
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\usepackage[siunitx, american, smartlabels, cute inductors, europeanvoltages]{circuitikz}
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\title{\blue Introduction to Electrical Engineering 2 \\
\blueb Assignment Solution $1$}
\author{Igor Shevtsov}
\date{\today}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\maketitle
\section{Question 1 Solution}{%
\subsection{Section 1}{Following equations describe current-voltage relations between some%
\footnote{Resistor, Capacitor and Inductor} circuit elements:%
\begin{align}
v_R&=iR \label{eq:1} \\
i_C&=C\dot{v}_C \label{eq:2} \\
v_L&=L\dot{i}_L \label{eq:3}
\end{align}
Applying nodal method at the $v_L$ node of circuit on Figure~\ref{fig:q1fig}
\begin{align}
I_0=\dfrac{v_L}{R}+i_L \label{eq:4}
\end{align}
Using \eqref{eq:3} one could obtain a differential equation
\begin{align}
\Aboxed{\dot{i_L}(t)+\tfrac{R}{L}i_L(t)&=\tfrac{R}{L}I_0 \quad,\, t>0} \label{eq:5}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\ctikzset {bipoles/length=1.2cm}
\begin{figure}[!htb]
\centering
\begin{circuitikz}[scale =1.2]\draw
(0,0) to [current source = $I_0$] (0,2) -- (2,2)
node[anchor=south] {$v_L$}
to [R, l_= $R$, i>^=$i_R(t)$, *-] (2,0)
(2,2) -- (4,2) to [L=$L$, i>^=$i_L(t)$, v=$v_L(t)$] (4,0) -- (0,0)
;\end{circuitikz}
\caption{\green Electric Circuit}
\label{fig:q1fig}
\end{figure}
}\label{sub:sub1}
%%% END UBSECTION 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Section 2}{
Assuming circuit's steady state, yields constant\footnote{Resistor Shortened by Inductor} current, therefore
\begin{align}
\textsl{ZSR:\quad} \Aboxed{i_L&=I_0 \label{eq:6}}
\end{align}
\begin{proof}
Solving ZIR of the equation~\eqref{eq:5} :
\begin{align}
i_{L,ZIR}(t)=A\exp({\dfrac{-t}{\tau}})\label{eq:7}\\
\lim_{t \to \infty}i_{L}(t)=\lim_{t \to \infty}\left(i_{L, ZIR}+i_{L,ZSR}\right)=i_{L, ZSR}=I_0 \nonumber
\end{align}
\end{proof}
}\label{sub:sub2}%
%%% END SUBSECTION 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Section 3}{
Summarizing equation~\eqref{eq:6} with \eqref{eq:7} and activating given initial condition $i_L(t=0)=2~\si{\A}$ results in final solution $i_L(t)$, where $\tau=\frac{L}{R}$:
\begin{align*}
\Aboxed{i_{L}(t)=I_0+\left(2- I_0\right)exp\left({\dfrac{-t}{\tau}}\right) \quad, t>0}
\end{align*}
\begin{figure}[!htb]
\centering
\includegraphics[scale=.3]{fig2.eps}
\caption{\green Inductor Current with $I_0=1A$}
\label{fig:q1fig2}
\end{figure}
}\label{sub:sub3}
%%% END SUBSECTION 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
}\label{sec:q1sec}
%%% END SECTION 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Question 2 Solution}{
\subsection{Section 1}{
Consider the circuit on Figure~\ref{fig:q2fig1b}. Now\footnote{Inductor is Short} applying current divider technique one could easily calculate the following currents:\par
\begin{align*}
i_1 = \dfrac{40}{500+2k||6k}\cdot\dfrac{2k}{500+2k||6k}\\ \\
i_2 = \dfrac{40}{500+2k||6k}\cdot\dfrac{6k}{500+2k||6k}\\ \\
\Aboxed{i_1(0^-)=20\,\si{mA} \qquad i_2(0^-)=60\,\si{mA}}
\end{align*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\ctikzset {bipoles/length=.8cm}
\begin{figure}[!htb]
\centering
\begin{subfigure}{.5\textwidth}
\begin{circuitikz}[scale =.6]\draw
(0,0) to [voltage source = $40\,\si{\volt}$] (0,3)
to [R = $500~\si{\ohm}$] (2,3)
to [opening switch, l_= {t>0}] (3,3) -- (4,3)
node[anchor=south]{$v_x$}
to [R = $6~\si{\kohm}$, i>_=$i_1$, *-] (7,3)
to [L, l_= $400~\si{m\henry}$] (7,0) -- (0,0)
(4,3) to [R, l_= $2~\si{\kohm}$, i>_=$i_2$] (4,0)
;\end{circuitikz}
\caption{\green \,Superposed Circuit}
\label{fig:q2fig1a}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
\begin{circuitikz}[scale =.6]\draw
(0,0) to [voltage source = $40\,\si{\volt}$] (0,3)
to [R = $500~\si{\ohm}$] (2,3) -- (4,3)
to [R = $6~\si{\kohm}$, i>_=$i_1$] (6,3)
to [L, l= $400~\si{m\henry}$] (6,0) -- (0,0)
(3,3) node[anchor=south]{$v_x$}
to [R, l_= $2~\si{\kohm}$, i>_=$i_2$,*-] (3,0)
;\end{circuitikz}
\caption{\green \,Past Time Circuit}
\label{fig:q2fig1b}
\end{subfigure}
\begin{subfigure}{.4\textwidth}
\begin{circuitikz}[scale =.6]\draw
(0,0) to [R = $8\,\si{\kohm}$, i_<=$i_2$,-*] (0,3) -- (2,3)
node[anchor=south]{$v_L$}
to [L = $400\,\si{m\henry}$, i>^=$i_1$,*-] (2,0) -- (0,0)
;\end{circuitikz}
\caption{\green \,Future Time Circuit}
\label{fig:q2fig1c}
\end{subfigure}
\caption{\green Question 2 Circuit}
\label{fig:q2fig1}
\end{figure}
}
%%% END SUBSECTION 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Section 2}{
Assuming continuity in the inductor current $i_1(0^-)=i_1(0^+)=20\,\si{\mA}$ on Figure~\ref{fig:q2fig1c}, where $i_1=-i_2$, thus, setting $i_1(0^+)=-i_2(0^+)=20\,\si{\mA}$ we get the currents
\begin{align*}
\Aboxed{i_1(0^+)=20\,\si{mA} \qquad i_2(0^+)=-20\,\si{\mA}}
\end{align*}
}
%%% END SUBSECTION 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Section 3}{It's obvious, that circuits \ref{fig:q2fig1b} and \ref{fig:q2fig1c} are equivalent to the one on Figure~\ref{fig:q1fig} (Page~\pageref{fig:q1fig}) with $\tau=50~\si{\ms}$ and $I_0=V_0/R_{eq}=20~\si{\mA}$:
\begin{align}
\Aboxed{\dot{i_L}(t)+\tfrac{R}{L}i_L(t)&=\tfrac{R}{L}I_0 \quad,\, t>0}
\end{align}
}
%%% END SUBSECTION 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Section 4}{Adopting ZIR\footnote{Since, there is no source in the circuit} solution~\eqref{eq:7}
\begin{align*}
\Aboxed{i_1(t)&=i_1(0^+)\exp\left({\dfrac{-t}{\tau}}\right) \quad, t>0} \\
\Aboxed{i_2(t)&=-i_1(t)}
\end{align*}
}
%%% END SUBSECTION 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
}\label{sec:q2sec}
%%% END SECTION 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Question 3 Solution}{
\subsection{Section 1}{
The voltage drop on the capacitor\footnote{Open-Circuited Capacitor } on Figure~\ref{fig:q3fig1b} as follows
\begin{align}
\Aboxed{v_C(0^-)&=15~\si{\mA} \cdot 2.4~\si{\kohm}=36~\si{\volt} \label{eq:8}}
\end{align}
\ctikzset {bipoles/length=.8cm}
\begin{figure}[!htb]
\begin{subfigure}{\textwidth}
\begin{circuitikz}[scale =.8]\draw
(0,0) to [current source = $15\,\si{\mA}$] (0,3) -- (2,3)
to [R, l= $2.4\,\si{\kohm}$] (2,0) -- (4,0)
to [C,l_=$0.25\,\si{\micro\farad}$] (4,2.5)
to[short, -o](3.3,3.3)
(4,3)node[anchor=south]{t=0}
node[ocirc] (A) at (3,3) {}
node[ocirc] (B) at (5,3) {}
(B) to [open, v=${}$] (A)
(5,3)to[short, o-](6,3)--(6,5)
to [R=$25\,\si{\kohm}$] (8,5)--(8,3)
to [cI,l_=$\alpha \cdot v_{\phi}$] (6,3)
(2,3)to[short,-o](3,3)
(8,3)--(9,3)
to [R, l_= $15~\si{\kohm}$,v^=$v_{\phi}$] (9,0) -- (0,0)
;\end{circuitikz}
\caption{\green \,Superposed Circuit}
\label{fig:q3fig1a}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
\begin{circuitikz}[scale =1]\draw
(0,0) to [current source = $15~\si{\mA}$] (0,2) -- (2,2)
to [R, l_= $2.4~\si{\kohm}$] (2,0)
(2,2) -- (4,2)
node[ocirc] (A) at (4,0) {}
node[ocirc] (B) at (4,2) {}
(B) to [open, v^=${v_c}$] (A)
(0,0)--(4,0)
;\end{circuitikz}
\caption{\green \,Past Time Circuit}
\label{fig:q3fig1b}
\end{subfigure}
\begin{subfigure}{.4\textwidth}
\begin{circuitikz}[scale =.6]\draw
node[ocirc] (A) at (0,0) {}
node[ocirc] (B) at (0,3) {}
(B) to [open, v=${v_t}$] (A)
(0,3) to [short, i=${i_t}$](1,3)--(2.7,3)--(2.7,4.5)
to [R, l_= $25~\si{\kohm}$] (6.5,4.5)--(6.5,2.2)
to [cI, l= $\alpha \cdot v_{\phi}$] (2.7,2.2)--(2.7,3)
(6.5,3)--(8,3) to [R, l_= $15~\si{\kohm}$,v^=$v_{\phi}$] (8,0)--(0,0)
;\end{circuitikz}
\caption{\green \,Future Time Circuit}
\label{fig:q3fig1c}
\end{subfigure}
\caption{\green Question 3 Circuit}
\label{fig:q3fig1}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
}
%%% END SUBSECTION 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Section 2}{
Replacing capacitor with a test voltage, as shown on Figure~\ref{fig:q3fig1c}, allows input resistance calculation $R_{in}=\dfrac{v_t}{i_t}$
\begin{align}
v_t-v_{\phi}&=25k\alpha \cdot v_{\phi} \nonumber \\
v_t=(1+25k\alpha) \cdot v_{\phi}&=15k(1+25k\alpha) \cdot i_t \nonumber \\
\Aboxed{R_{in}(C)&=15k(1+25k\alpha) \label{eq:10}}
\end{align}
}
%%% END SUBSECTION 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Section 3}{Given circuit's time constant $\tau=R \cdot C=25~\si{ms}$ requires resistor $R=100~\si{\kohm}$. Setting into~\eqref{eq:10} leads to the value of $\alpha$-parameter
\begin{align*}
\Aboxed{\alpha \approx 2.26 \cdot 10^{-4}}
\end{align*}
}
%%% END SUBSECTION 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Section 4}{As before\footnote{Due to circuits analogy} using ZIR solution~\eqref{eq:7} adapted for voltages, where $\tau=25~\si{ms}$ and initial condition of the capacitor\footnote{Continuity in voltage drop} $v_C(0^-)=v_C(0^+)=36~\si{\volt}$
\begin{align*}
\begin{cases}
\Aboxed{v_C(t)&=v_C(0^+)\exp\left({\dfrac{-t}{\tau}}\right) \quad, t\geqslant0} \\
\Aboxed{v_C(t)&=v_C(0^-) \hspace{5.1em}, t<0} \\
\end{cases}
\end{align*}
}
%%% END SUBSECTION 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
}\label{sec:q3sec}
%%% END SECTION 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}