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\begin{document}
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\heading{RLC circuit}
\auname{I.D. 066072570}
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{\bf The problem:}
\Dn
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For the given RLC circle:
\begin{enumerate}
\item find the law of connecting inductances in series and in parallel in general.
What is the total induction of the given circle?
\item The switch is on the right hand side for 3$\tau _{\sec }$ and after is moved to the left side.
\begin{itemize}
\item Find the resonance frequency of the system.
\item find the current through the resistor as a function of time.
\end{itemize}
\end{enumerate}
\includegraphics[scale=0.75]{e_51_1_001_p0.jpg}
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\Dn
{\bf The solution:}
\Dn
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a. inductance in series:
\begin{eqnarray}
I_1 = I_2 \Rightarrow \varepsilon = (L_1 + L_2 )\dot I_1 = L_{eff} \dot I \Rightarrow L_1 + L_2 = L_{eff}
\end{eqnarray}
inductance in parallel:
\begin{eqnarray}
I_1 + I_2 = \frac{\varepsilon }{{L_1 }} + \frac{\varepsilon }{{L_2 }} = \frac{{\varepsilon (L_1 + L_2 )}}{{L_1 L_2 }} \Rightarrow \frac{\varepsilon }{{I_1 + I_2 }} = \frac{{L_1 L_2 }}{{L_1 + L_2 }} = L_{eff} \Rightarrow \frac{1}{{L_{eff} }} = \frac{1}{{L_1 }} + \frac{1}{{L_2 }}
\end{eqnarray}
in our case:
\begin{eqnarray}
L_{eff} = \frac{{(L_1 + L_2 ) \cdot L_3 }}{{L_1 + L_2 + L_3 }}
\end{eqnarray}
after 3$\tau (RC)$ we get:
\begin{eqnarray}
V = \varepsilon (1 - e^{\frac{{ - t}}{{Rc}}} ) = \varepsilon (1 - e^{ - 3} ) \approx 4.75v
\end{eqnarray}
and the equation of the circuit is:
\begin{eqnarray}
\frac{q}{c} + IR + L\dot I = 0 \Rightarrow \frac{q}{c} + R\dot q + L\ddot q = 0
\end{eqnarray}
we shall start analysis with the simplest case- R=0, in this case we have:
\begin{eqnarray}
\ddot q = - \frac{1}{{Lc}}q
\end{eqnarray}
and the solution is:
\begin{eqnarray}
q = Ae^{i\omega _0 t}
\end{eqnarray}
where
\begin{eqnarray}
\omega _0 = \sqrt {\frac{1}{{L_{eff} c}}} = 8\frac{2}{3} \cdot 10^6 H_{rz}
\end{eqnarray}
substitute $q = Ae^{i\omega t}$ in the equation we get:
\begin{eqnarray}
- \omega ^2 + \frac{{iR}}{L}\omega + \omega _0 ^2 = 0 \Rightarrow \omega = \frac{i}{2}\Gamma \pm \sqrt {\omega _0 ^2 - \frac{{\Gamma ^2 }}{4}}
\end{eqnarray}
where
\begin{eqnarray}
\Gamma = \frac{R}{L}
\end{eqnarray}
the solution is
\begin{eqnarray}
q = \re\left[ {Ae^{i(\omega ' + \frac{i}{2}\Gamma )t} } \right]
\end{eqnarray}
where
\begin{eqnarray}
\omega ' = \sqrt {\omega _0 ^2 - \frac{{\Gamma ^2 }}{4}}
\end{eqnarray}
substitute
\begin{eqnarray}
q(t = 0) = c \cdot v(3\tau )
\end{eqnarray}
we get
\begin{eqnarray}
A = 4.75 \cdot 10^{ - 9}
\end{eqnarray}
to find the curent:
\begin{eqnarray}
I = \dot q = RE\left[ {i(\omega ' + \frac{i}{2}\Gamma )Ae^{i(\omega ' + \frac{i}{2}\Gamma )t} } \right] = - Ae^{ - \frac{\Gamma }{2}t} \cdot (\frac{\Gamma }{2}\cos (\omega 't) + \omega '\sin (\omega 't))
\end{eqnarray}
eventually substitute the given parameters we get:
\begin{eqnarray}
I = - 4.75 \cdot 10^{ - 9} e^{ - 37509t} \left[ {37509 \cdot \cos (8\frac{2}{3} \cdot 10^6 t) + 8\frac{2}{3} \cdot 10^6 \sin (8\frac{2}{3} \cdot 10^6 t)} \right]_{Amper}
\end{eqnarray}
The first plot presents a voltage on the resistor $R$ and on the second one the current through it as a function of time.
\includegraphics[scale=0.75]{e_51_1_001_s_p0.jpg}
\includegraphics[scale=0.75]{e_51_1_001_s_p1.jpg}
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