Sodium Channel Current in Membrane Circuit

Based on your handwritten diagram, I interpret this as a circuit representing a biological membrane model (likely a Hodgkin-Huxley type sub-circuit). The diagram shows a parallel arrangement consisting of a capacitor $C_m$ on the left and a branch on the right containing an electromotive force (EMF) $E_0$, a potential point $V_{na}$, and a conductance $g_{na}$ (represented as a resistor). The top wire is grounded ($0\text{V}$).

Answer

The current $i$ flowing through the sodium channel branch is determined by Ohm's Law for a circuit with an EMF, calculated as the product of the conductance and the driving force ($V_{na} - E_0$). Assuming the bottom wire connects to the intracellular potential $V_m$, the relationship is $i = g_{na}(V_m - E_0)$.

Explanation

1. Analysis of the Diagram The image displays a circuit with two main parallel branches. The left branch contains a membrane capacitor ($C_m$). The right branch is the conductive pathway for a specific ion (labeled $na$ for Sodium). It contains a battery $E_0$ (the Nernst equilibrium potential) and a variable resistor/conductor $g_{na}$. The top of the circuit is connected to a ground symbol, implying a reference potential of $0\text{V}$.

2. Identifying the Knowns and Variables

   • Known: Conductance $g_{na}$, Equilibrium potential $E_0$, Ground $= 0\text{V}$.
   • Find: The expression for the current $i$ in the right branch.
   • Formula: $i = g \cdot \Delta V$ (Ohm's Law in terms of conductance).
   • Note: In electrophysiology, $g = 1/R$.

3. Determining the Driving Force The "Driving Force" is the difference between the actual membrane potential ($V_m$) and the equilibrium potential for that ion ($E_0$). If we assume the potential at nodes $B$ and $C$ is the membrane potential $V_m$ relative to the grounded extracellular side: $V_{driving} = V_m - E_0$ This represents the net electrical pressure pushing ions through the channel.

4. Applying Ohm's Law (Substitution) ⚠️ This step is required on exams: High school physics often uses $V = IR$, but in biology, we use $I = gV$. Substituting our driving force into the conductance formula: $i = g_{na} \cdot (V_m - E_0)$ This equation describes the current $i$ indicated by the arrow in your diagram.

5. Unit Check and Dimensional Analysis

   • $g$ (Conductance) is measured in Siemens ($S$), which is $A/V$.
   • $(V_m - E_0)$ is measured in Volts ($V$).
   • $S \times V = (A/V) \times V = A$ (Amperes). The units are consistent with current.

Final Answer

The current $i$ flowing through the conductance $g_{na}$ is: $\boxed{i = g_{na}(V_m - E_0)}$

Common Mistakes

• Sign Errors: Students often flip $V_m$ and $E_0$. Remember that if the membrane potential is more positive than the equilibrium potential, the driving force is positive, leading to an outward current (depending on ion charge).
• Confusing $g$ and $R$: Ensure you are multiplying by conductance ($g$). If the diagram provided resistance ($R$), the formula would be $i = \frac{V_m - E_0}{R}$.