= Coupled resonant circuit model
{title2=$M=k\sqrt{L_1L_2}$}
Let two series loops have <inductances> $L_1,L_2>0$, <capacitances> $C_1,C_2>0$, total <electrical resistances> $R_1,R_2>0$, and reciprocal <mutual inductance> $M=k\sqrt{L_1L_2}$ with $0\leq k<1$. For a steady <sinusoid> of <angular frequency> $\omega>0$, choose complex peak <amplitudes> with time convention $e^{\mathrm i\omega t}$. An ideal source of peak <electric potential difference> $V$ drives loop 1; loop 2 has no independent source. Then
$$
Z_j=R_j+\mathrm i\left(\omega L_j-\frac1{\omega C_j}\right),\qquad
\begin{pmatrix}Z_1&\mathrm i\omega M\\\mathrm i\omega M&Z_2\end{pmatrix}
\begin{pmatrix}I_1\\I_2\end{pmatrix}
=\begin{pmatrix}V\\0\end{pmatrix},\qquad
\frac{I_2}{V}=\frac{-\mathrm i\omega M}{Z_1Z_2+\omega^2M^2}.
$$
Eliminating $I_1$ proves the response formula. Reversing one coil changes the sign of $M$ and the relative <wave phase>, but leaves this <amplitude> response unchanged. The measured sense ratio is $|R_sI_2/V|$, with $R_s$ included in $R_2$, not added a second time.
\Image[/wireless-telegraphy-coupled-resonance-model.png]
{title=Calculated coupled resonance and the distinction between a damped train and a maintained sine}
{description=Original predictions for two identical linear series loops with $L=5\,\mathrm{mH}$, $C=2\,\mathrm{nF}$ and total $R=100\,\Omega$ per loop; the secondary includes $R_s=50\,\Omega$. Top: steady sense-voltage ratios at three coupling coefficients. Bottom: one isolated loop's free damped train, exponential bounds and a maintained sine reference. Neither panel contains historical or laboratory measurements.}
For identical loops at their uncoupled <resonant frequency> $\omega_0=1/\sqrt{LC}$,
$$
\left|\frac{I_2}{V}\right|_{\omega_0}
=\frac{\omega_0M}{R^2+\omega_0^2M^2}.
$$
Writing $x=\omega_0M/R$ makes this $x/[R(1+x^2)]$. Its <derivative> has the sign of $1-x^2$, so response at this fixed <frequency> is largest when $x=1$, equivalently $k=1/Q$. Too little coupling transfers little excitation; excessive coupling reduces the response at the original centre <frequency>. This is a fixed-<frequency> optimum, not a universal optimum over every possible receiver and drive <frequency>.
For the chosen values, $k=1/Q=0.063246$ gives $M=0.31623\,\mathrm{mH}$. At source peak $V=0.100\,\mathrm V$, the secondary peak <electric current> is $0.500\,\mathrm{mA}$ and the sense peak <electric potential difference> is $25.0\,\mathrm{mV}$. The primary peak <electric current> is also $0.500\,\mathrm{mA}$, so source mean <power> is $25.0\,\mathrm{\mu W}$: each loop dissipates $12.5\,\mathrm{\mu W}$, checking <conservation of energy>. Here $V$ is the equivalent ideal source behind $R_1$; a generator's terminal reading must be converted using its measured <Thévenin resistance>.
In the lossless identical-loop limit, the <normal modes> have equal or opposite <electric currents>, making their effective <inductances> $L+M$ or $L-M$. Thus
$$
f_{\rm low}=\frac{f_0}{\sqrt{1+k}},\qquad
f_{\rm high}=\frac{f_0}{\sqrt{1-k}}.
$$
At $k=0.20$ these are $45.944$ and $56.270\,\mathrm{kHz}$. Finite <electrical resistance> broadens and shifts the driven peaks, so the exact response formula, rather than these lossless values, generates the plot. Changing a <capacitor> moves the uncoupled tuning through $f_0\propto C^{-1/2}$: increasing it by $21\%$ lowers $f_0$ by the factor $1/1.1$. The model explains why tuning and coupling must be adjusted together; it supplies no transatlantic propagation or detector-threshold prediction.
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