= Equal-loop RL induction step
{title2=$\tau_\pm=(L\pm M)/R$}
For an exactly soluble example, set $L_1=L_2=L$, $R_1=R_2=R$, and $0<M<L$. Start with zero <electric currents>, and apply a constant <electric potential difference> $V_0>0$ at $t=0$. The sums $s=i_1+i_2$ and $d=i_1-i_2$ obey separate <first-order linear differential equations>:
$$
(L+M)\dot s+Rs=V_0,\qquad
(L-M)\dot d+Rd=V_0.
$$
Their <time constants> are $\tau_+=(L+M)/R$ and $\tau_-=(L-M)/R$. Thus, for $t\geq0$,
$$
i_1(t)=\frac{V_0}{2R}\left(2-e^{-t/\tau_+}-e^{-t/\tau_-}\right),\qquad
i_2(t)=\frac{V_0}{2R}\left(e^{-t/\tau_-}-e^{-t/\tau_+}\right).
$$
Since $\tau_-<\tau_+$, \b[$i_2(t)<0$ for $t>0$, but $i_2(t)\to0$ as $t\to\infty$]. The primary approaches $V_0/R$. The secondary response opposes the rise of linked <magnetic flux>, in accordance with <Lenz's law>. Removing <mutual inductance> by setting $M=0$ makes the two exponentials equal and gives zero secondary <electric current>.
For stipulated $L=0.1\,\mathrm H$, $M=0.06\,\mathrm H$, $R=10\,\Omega$ and $V_0=1\,\mathrm V$, the two <time constants> are $16\,\mathrm{ms}$ and $4\,\mathrm{ms}$. Setting $\dot i_2=0$ gives its largest negative magnitude at
$$
t_* =\frac{\tau_+\tau_-}{\tau_+-\tau_-}\log\frac{\tau_+}{\tau_-}
\simeq7.394\,\mathrm{ms},\qquad
i_2(t_*)\simeq-23.62\,\mathrm{mA}.
$$
The continuing primary <electric current> eventually reaches $100\,\mathrm{mA}$ while the secondary returns to zero. These numbers belong to the stated model, not the 1831 apparatus.
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