= Single-channel current-voltage relation
{title2=$i=g(V-E_{\mathrm{rev}})$}
For an ideal open <ion channel> with constant <electrical conductance> $g$, <membrane potential> $V$ and <reversal potential> $E_{\mathrm{rev}}$, define outward <electric current> as positive:
$$
i=g(V-E_{\mathrm{rev}}).
$$
This is an <Ohm's law> model of the open state, not a universal channel law or a rule for opening <probability>. In the synthetic figure, $g=10\,\mathrm{pS}$ and $E_{\mathrm{rev}}=0\,\mathrm{mV}$. Because $1\,\mathrm{pS}\cdot1\,\mathrm{mV}=10^{-3}\,\mathrm{pA}$, the open-state <electric currents> at $-80$ and $-40\,\mathrm{mV}$ are $-0.8$ and $-0.4\,\mathrm{pA}$. Zero <electric current> at the <reversal potential> does not imply a closed <ion channel>. Real <ion channels> can have nonlinear responses, so the straight line is an explicit idealization.
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