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%\title{LCR Circuit to Mimic Surface Plasmon Resonance in Nanostrucures}
%\author{Shivangi Dubey, Kuldeep Kumar\footnote{e-mail: 
%		kuldeep@sgtbkhalsa.du.ac.in}, P.Arun\\ \\
%	Material Science Research Lab, S.G.T.B. Khalsa College,\\ 
%	University of Delhi, Delhi-110 007, India\\
%}

\begin{document}
%	\maketitle
\section*{Appendix A}
The ${\rm 2^{nd}}$ order ODE describing change variation of current with time in a LCR circuit excited by an AC source (fig~1c of manuscript) is given as 
\begin{eqnarray}
L_m\frac{d^2i}{dt^2}+R_m\frac{di}{dt}+\frac{i}{C_m}=\omega V_ocos\omega_{in} t\nonumber
\end{eqnarray}
For solving the ODE, re-arrange
\begin{eqnarray}
\frac{d^2i}{dt^2}+\frac{R_m}{L_m}\frac{di}{dt}+\frac{1}{L_mC_m}i=\frac{\omega_{in}V_m}{L_m}\cos\omega_{in}t\nonumber
\end{eqnarray}
This equation is non-homogeneous and will have two solutions, one the homogeneous part and second a particular part which would depend on the nature of function on the right hand side, The homogeneous solution of the above is found by solving the characteristic equation
\begin{eqnarray}
\lambda^2+\frac{R_m}{L_m}\lambda+\frac{1}{L_mC_m}=0\nonumber
\end{eqnarray}
and is given as
\begin{eqnarray}
i_h(t)=e^{-\frac{R_m}{2L_m}t}(C_1sin\omega_ot+C_2cos\omega_ot)\nonumber
\end{eqnarray}
where
\begin{eqnarray}
\omega_o=\sqrt{\frac{1}{L_mC_m}-\frac{R_m^2}{4L_m^2}}\nonumber
\end{eqnarray}
The non-homogeneous/ particular solution can be obtained by using the guess solution as
\begin{eqnarray}
i_{nh}=A\sin\omega_{in}t+B\cos\omega_{in}t\nonumber
\end{eqnarray}
and substituting back into the original equation. We get a simultaneous equation of A and B as
	%\begin{eqnarray}
	%-\omega_{in}^2A\sin\omega_{in}t-\omega_{in}^2B\cos\omega_{in}t+\frac{R_m\omega_{in}}{L_m}A\cos\omega_{in}t-\frac{\omega_{in}R_m}{L_m}B\sin\omega_{in}t
	%+\frac{A}{L_mC_m}\sin\omega_{in}t+ \frac{B}{L_mR_m}\cos\omega_{in}t=\frac{\omega_{in}V_m}{L_m}\cos\omega_{in}t
	%\end{eqnarray}
\begin{eqnarray}
\left(\frac{1}{L_mC_m}-\omega_{in}^2\right)B+\frac{R_m\omega_{in}}{L_m}A &=& \frac{\omega_{in}V_m}{L_m}\nonumber\\
-\frac{\omega_{in}R_m}{L_m}B+\left(\frac{1}{L_mC_m}-\omega_{in}^2\right)A &=&0\nonumber
\end{eqnarray}
Solving for A and B, we have
\begin{eqnarray}
A=\frac{V_m}{R_m}\frac{1}{\left[1+\frac{L_m^2}{\omega_{in}^2R_m^2}\left(\frac{1}{L_mC_m}-\omega_{in}^2 \right)^2\right]}\nonumber
\end{eqnarray}
\begin{eqnarray}
B=\frac{V_m}{R_m}\frac{\frac{L_m}{\omega_{in}R_m}\left(\frac{1}{L_mC_m}-\omega_{in}^2\right)}{\left[1+\frac{L_m^2}{\omega_{in}^2R_m^2}\left(\frac{1}{L_mC_m}-\omega_{in}^2\right)^2\right]}\nonumber
\end{eqnarray}
Hence, the non-homogeneous solution is 
\begin{eqnarray}
%i_{nh}=\frac{V_m}{R_m}\times \frac{1}{1+x^2}\left[sin\omega_{in}t+x.cos\omega_{in}t\right]\nonumber\\
i_{nh}=\left[\frac{V_m}{R_m}\times \frac{1}{\sqrt{1+x^2}}\right]sin(\omega_{in}t+\phi)\nonumber
\end{eqnarray}
where
\begin{eqnarray}
x&=&\frac{L_m}{\omega_{in}R_m}\left(\frac{1}{L_mC_m}-\omega_{in}^2\right) \nonumber\\
\phi &=& tan^{-1}(x)\nonumber
\end{eqnarray}
The complete current expression can be written as
\begin{eqnarray}
i(t)=e^{-\frac{R_m}{2L_m}t}(C_1sin\omega_ot+C_2cos\omega_ot)+\frac{V_m}{R_m}\left(\frac{1}{\sqrt{1+x^2}}\right)sin(\omega_{in}t+\phi)\nonumber
\end{eqnarray}
or
\begin{eqnarray}
i(t)=e^{-\frac{R_m}{2L_m}t}(C_1sin\omega_ot+C_2cos\omega_ot)+\frac{V_m}{R_m}\left\{\frac{1}{\sqrt{1+\left[\frac{L_m}{\omega_{in}R_m}\left(\frac{1}{L_mC_m}-\omega_{in}^2\right)\right]^2}}\right\}sin(\omega_{in}t+\phi)\nonumber
\end{eqnarray}
Since the first term attentuates rapidly (due to the exponential term), the steady state solution would be represented as
\begin{eqnarray}
i_{steady-state}(t)=\frac{V_m}{R_m}\left\{\frac{1}{\sqrt{1+\left[\frac{L_m}{\omega_{in}R_m}\left(\frac{1}{L_mC_m}-\omega_{in}^2\right)\right]^2}}\right\}sin(\omega_{in}t+\phi)\nonumber
\end{eqnarray}
The voltage drop across the resistance ${\rm R_m}$ would hence be given as
\begin{eqnarray}
v_{steady-state}(t)=\left\{\frac{V_m}{\sqrt{1+\left[\frac{L_m}{\omega_{in}R_m}\left(\frac{1}{L_mC_m}-\omega_{in}^2\right)\right]^2}}\right\}sin(\omega_{in}t+\phi)\nonumber
\end{eqnarray}
\newpage
\section*{Appendix B}
To obtain an expression for currents in the two loops of the equivalent circuit shown in fig~3a, we essentially have to solve the coupled ordinary differential equation using the operator method
%~\cite{kreyzig}.
\begin{eqnarray}
v(t) &=& \frac{1}{C_m}\int i_1dt+L_m\frac{di_1}{dt}+M\frac{d}{dt}(i_1-i_2)+R_mi_1\nonumber\\
0 &=& \frac{1}{C_h}\int i_2dt+L_h\frac{di_2}{dt}+M\frac{d}{dt}(i_2-i_1)+R_hi_2\nonumber
\end{eqnarray}
Removing integral sign, we have
\begin{eqnarray}
\omega V_o cos\omega t &=& \frac{i_1}{C_m}+L_m\frac{d^2i_1}{dt^2}+M\frac{d^2}{dt^2}(i_1-i_2)+R_m\frac{di_1}{dt}\nonumber\\
0 &=& \frac{i_2}{C_h}+L_h\frac{d^2i_2}{dt^2}+M\frac{d^2}{dt^2}(i_2-i_1)+R_h\frac{di_2}{dt}\nonumber
\end{eqnarray}
Solving the coupled differential equation for ${\rm i_1}$, we have (for simplification of calculations, we have made a reasonable assumption that ${\rm R_m\sim R_h\sim 0}$ or negligiable compared to the remaining terms. In simulations, we have used ${\rm R_m\sim R_h=0.5\Omega}$.)
\begin{eqnarray}
i_1=\frac{\frac{\omega V_o}{L_m}\left[\frac{1}{L_hC_h}-\left(1+\frac{M}{L_h}\right)\omega^2\right]cos\omega t}{\left[1+M\left(\frac{1}{L_m}+\frac{1}{L_h}\right)\right]\omega^4-\left[\frac{1}{L_hC_h}\left(1+\frac{M}{L_m}\right)+\frac{1}{L_mC_m}\left(1+\frac{M}{L_h}\right)\right]\omega^2+\frac{1}{L_hL_mC_hC_m}}\label{newuse}
\end{eqnarray}
From this equation, the frequency at which resonance occurs (${\rm \omega_p}$) due to culmunated effect of metal nano-particle and surrounding dielectric can be found from
\begin{eqnarray}
\omega_p=\sqrt{\frac{\left[\frac{1}{L_hC_h}\left(1+\frac{M}{L_m}\right)+\frac{1}{L_mC_m}\left(1+\frac{M}{L_h}\right)\right]}{\left[1+M\left(\frac{1}{L_m}+\frac{1}{L_h}\right)\right]}}\label{pp}
\end{eqnarray}
Consider both metal and dielectric materials are non-magnetic, similar permeability would result in ${\rm L_m \approx L_h=L}$. Hence, above equation would reduce to
\begin{eqnarray}
\omega_p=\sqrt{\frac{\left[\frac{1}{L}\left(1+k\right)\left(\frac{1}{C_h}+\frac{1}{C_m}\right)\right]}{\left(1+2k\right)}}\nonumber
\end{eqnarray}
Since the dielectric surrounding medium is EM radiation inactive compared to the metal nano-particle, ${\rm C_h \ll C_m}$. Hence, we write
\begin{eqnarray}
\omega_p &\approx& \sqrt{\frac{\left[\frac{1}{LC_h}\left(1+k\right)\right]}{\left(1+2k\right)}}\nonumber\\
\omega_p &\approx& \frac{1}{\sqrt{LC_h}}\left(\sqrt{\frac{1+k}{1+2k}}\right)\nonumber
\end{eqnarray}	
Since we talk of SPR peak position in terms of wave-lengths
\begin{eqnarray}
\lambda_p &\propto& \sqrt{LC_h}\left(\sqrt{\frac{1+2k}{1+k}}\right)\nonumber
\end{eqnarray}	
	
\end{document}
