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\textbf{Physics 120  \smallskip - \smallskip David Kleinfeld \smallskip - \smallskip Spring 2019}
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 \section{RLC resonator with a lossy inductor}

The circuit in laboratory exercise 3.1 uses an indictor that has a modest resistance. Let's see how this effects the amplitude of the signal output.

\includegraphics[width=8cm]{LRCr}

The equivalent impedance of the (L+r)$||$C circuit is
\begin{eqnarray}
Z_{eq} &=&  \frac {\frac{r + i \omega L}{i \omega C}}{r + i\omega L + \frac{1}{i \omega C}} \\
           &=&  \frac{r + i \omega L}{i + i \omega C (r + i \omega L)} \nonumber
\end{eqnarray}
Then
\begin{eqnarray}
\frac{V(\omega)}{V_o(\omega)} &=& \frac{ \frac{r + i \omega L}{i + i \omega C (r + i \omega L)}}{\frac{r + i \omega L}{i + i \omega C (r + i \omega L)} +R}  \\
           &=&  \frac{r + i \omega L}{r + i \omega L  + R(1+i\omega C(r + i\omega L))}   \nonumber \\
          &=&  \frac{r + i \omega L}{r + R - R\omega^2 LC + i\omega L  + i\omega rRC}   \nonumber.
\end{eqnarray}
We let
\begin{equation}
\omega_0 = \frac{1}{\sqrt{LC}},
\end{equation}
\begin{equation}
\rho = r/R,
\end{equation}
where the limit of a lossless inductor then corresponds to $\rho \rightarrow 0$, and
\begin{equation}
\tau = L/R.
\end{equation}
In terms of these normalized units,
\begin{equation}
\frac{V(\omega)}{V_o(\omega)}	 =    \frac{\rho + i \omega \tau}{1 + \rho - \left( \frac{\omega}{\omega_0} \right)^2 + i\omega \tau \left(1 +  \frac{\rho}{( \omega_0 \tau)^2} \right) }.
\end{equation}
The magnitude of the output is then
\begin{eqnarray}
\left|\frac{V(\omega)}{V_o(\omega)}\right| &=&  \sqrt{\frac{{\rho^2 + (\omega \tau)^2}} {{\left( 1 + \rho -  \left( \frac{\omega}{\omega_0} \right)^2  \right)^2  + (\omega \tau)^2 \left(1 +  \frac{\rho}{( \omega_0 \tau)^2} \right)^2 }}} \\
		&=& \sqrt{ \frac{{ (\frac{\omega}{\omega_0})^2 + (\frac{\rho}{\omega_0 \tau})^2}} {{\left( \frac{1 -  \left( \frac{\omega}{\omega_0} \right)^2 + \rho}{\omega \tau}  \right)^2  + (\frac{\omega}{\omega_0})^2 \left(1 +  \frac{\rho}{( \omega_0 \tau)^2} \right)^2 }}}  \nonumber
\end{eqnarray}
and this has the expected limit as $\rho \rightarrow 0$, i.e.,
\begin{equation}
\left|\frac{V(\omega)}{V_o(\omega)}\right| \;  \overrightarrow{\rho \rightarrow 0} \;  \frac{{ \frac{\omega}{\omega_0} }} {\sqrt{\left( \frac{1 -  \left( \frac{\omega}{\omega_0} \right)^2}{\omega \tau}  \right)^2  + (\frac{\omega}{\omega_0})^2 }}
\end{equation}

We return to the general expression of the magnitude of the output and note that for $\omega$ = $\omega_0$, the magnitude of the output is
\begin{equation}
\left|\frac{V(\omega)}{V_o(\omega)}\right|  \;  \overrightarrow{\omega = \omega_0} \;  \sqrt{ \frac{{ 1 + \left(\frac{\rho}{\omega_0 \tau}\right)^2}} {{\left( \frac{ \rho}{\omega_0 \tau}  \right)^2  + \left(1 +  \frac{1}{\rho} \left(\frac{\rho}{ \omega_0 \tau} \right)^2 \right)^2 }}} .
\end{equation}
The dependence on $\frac{\rho}{\omega_0 \tau}$ = $r \sqrt {\frac{C}{L}}$ shows that loss in the inductors is minimized at higher frequencies.

 For the components in laboratory 3.1, i.e., $L$ = 10 mH, $R$ = 100 k$\Omega$, $C$ = 0.01 $\mu$F and $r$ measured to be 150 $\Omega$ for one inductor, we find $\rho$ = 0.015, $\omega \tau$ = 0.01, and the decrement in amplitude is large, a factor of 16. This occurs even through $\rho$ is rather small.


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