# Cells and Membrane Potentials - Equivalent Circuits

The cellular membrane separates the interior of the cell or an organelle from the exterior. This membrane forms a barrier for charged particles like ions. The cell is able to transport ions in and out of the cell in a very controlled manner. If the charge on one side of the cell membrane becomes larger than on the other an electric potential difference is present between theses two sides.

To establish an electric potential, energy has to be invested, thus the transport is active meaning energy sources like ATP or light are used. Once the electric potential is established, it can drive other processes, an important example is ATP Synthase. However, even in the absence of an actively implemented electric potential difference the so called Resting Potential is apparent.

The flux of charged particles through the membrane via a channel leads to a current $J$. On one hand the lipid bilayer provides a barrier, influx or efflux is only possible through channels this slows the current of ions down, so there is a resistance $R$ associated with the channel. On the other hand the membrane separates two sides of unequal charges, which essentially resembles a capacitor with associated capacitance $C$. Let us call the total potential difference between inside and outside of the membrane $V_m$

\begin{align} V_m=V_i - V_o \end{align}

For given ion concentration inside and outside of the cell, we can calculate the chemical potential due to osmosis. Often it is convenient to work with a potential in units of $mV$, thus we will use the expression

\begin{align} \Delta p = \frac{\Delta \mu}{F} = \frac{2.3 RT}{z F} log_{10} \frac{[X]^{OUT}}{[X]^{IN}}= 58mV/z \, log_{10} \frac{[X]^{OUT}}{[X]^{IN}} \end{align} which is sometimes called protonmotive force ($z$ is the charge per ion transported). For example a concentration ration of $1/20$ gives \begin{align} \Delta p = 58mV/z \, log_{10} \frac{[20]}{[400]} \approx -75mV \end{align}

In addition to the protonmotive force there can also be an electric potential, because the sum of charges in the exterior is larger than in the interior (or vice versa). We will call the electric potential $-\Delta \psi$, then the total potential $V_m$ is given by the sum of protonmotive force and electric potential \begin{align} V_m = -\Delta \psi + \frac{2.3 RT}{z F} log_{10} \frac{[X]^{OUT}}{[X]^{IN}} \end{align} where the sign of $\Delta \psi$ depends on the direction of the gradient of the electric potential. In equilibrium when $V_m=0$ the so called Nernst equation is valid \begin{align} E_X=\Delta \psi = \frac{2.3 RT}{z F} log_{10} \frac{[X]_{eq}^{OUT}}{[X]_{eq}^{IN}} \end{align}

This allows us to calculate the potentials for various ion-channels. We can than model the ion flux in analogy to an electronic circuit.

The conductance is given by 1/resistence

\begin{align} \gamma_X=\frac{1}{g_X} \end{align}

so we find the current $J$ as

\begin{align} J= \frac{V_m}{\gamma_X} = \frac{-\Delta \psi}{\gamma_X} + \frac{2.3 RT}{z F \gamma_X} log_{10} \frac{[X]^{OUT}}{[X]^{IN}} \end{align}