At resting state, membrane is negatively charged
When negative current is introduced, membrane potential becomes more negative - hyperpolarisation
When positive current is introduced, membrane potential becomes more positive - depolarisation
If current injected is strong enough, and a membrane potential reaches threshold (~-50mV), an action potential is produced – strong depolarisation up to 40mV
Cell membrane can be seen as a diffusion barrier or a resistance
The high concentration gradient between the intra- and extracellular space for $K^+$ establishes a chemical driving force that causes the efflux of $K^+$
The continued efflux of $K^+$ builds up an excess of positive charge on the outside of the cell and leaves behind an excess of negative charge inside the cell. This buildup of charge leads to a potential difference across the membrane that impedes the further efflux of $K^+$ - electrical driving force
The membrane potential at which the efflux of $K^+$ due to concentration gradient is exactly balanced by an opposing membrane potential
The pump exchanges 3 $Na^+$ with 2 $K^+$
To re-establish a concentration gradient, Sodium-Potassium Pump is required
The third of the 3 phosphate residuals in ATP stores a high amount of energy
Hydrolysis of the third phosphate releases energy to induce a confirmational change in the pump
The more active a neuron, the higher demand for ATP as more energy is required to re-establish the resting concentration gradient
At a certain equilibrium potential, $E_x$, the chemical driving force, $W_A$, is equal to the electrical driving force, $W_B$:
$$ W_A = RT\ln \left(\frac{X_o}{X_i}\right) \\ W_B = E_xzF \\ \therefore RT\ln \left(\frac{X_o}{X_i}\right) = E_xzF $$
Rearrange to obtain Nernst Equation:
$$ E_x=\frac{RT}{zF}\ln\left( \frac{X_o}{X_i}\right) $$
where:
$X_o$ - outer ion concentration
$X_i$ - inner ion concentration