Resting membrane potential is maintained by

Propagation of action potential is very fast in nerve fibres which have

Given is the diagrammatic representation of impulse conduction through an axon (at points $A$ and $B$). View the diagram and arrange the steps of impulse conduction.
$I.$ The polarity of the membrane at site $A$ is reversed and depolarized,i.e.,the outer surface becomes negatively charged and the inner side becomes positively charged,generating a nerve impulse.
$II.$ $A$ stimulus causes a disturbance to the membrane at site $A$ of the nerve fibre,resulting in the leakage of $Na^+$ ions inside the nerve fibre.
$III.$ On the outer surface,current flows from site $B$ to site $A$ to complete the circuit of current flow. Hence,the polarity at the site is reversed,and an action potential is generated at site $B$. The impulse (action potential) generated at site $A$ arrives at site $B$. The sequence is repeated along the length of the axon and consequently,the impulse is conducted.
$IV.$ Immediately ahead,the axon (e.g.,site $B$) membrane has a positive charge on the outer surface and a negative charge on its inner surface. As a result,a current flows on the inner surface from site $A$ to site $B$.

When an impulse passes,the membrane is depolarized and the charge of the cells is

The state in which the axonal membrane is positively charged on the outside and negatively charged on the inside is known as:

When a neuron is in resting state,$i.e.$,not conducting any impulse,the axonal membrane is