Where did you read that quote? It is my understanding that I(ionic) = -
I(cap) for a native neuron, or for a neuron which has an electrode
stuck in it I(ion) + I(elec) = -I(cap).
You just need to think that I(ionic) isn't responsible for changing
the membrane potential of the cell, it just supplies charge that loads
the membrane capacitor, which then changes the transmembrane voltage.
And the degree to which the charge supplied by I(ionic) changes the
transmembrane voltage depends on C, the capacitance of the cell
membrane, i.e. C=Q/V.
Another way to think of it, is instead of the classical membrane
circuit being shown as a resistance and capacitor in parallel, think
of the path that current will take as it moves through the neuron.
Current passes through an ion channel into the cell, and then it
passes from inside the cell over the capacitor and out of the cell,
completing the circuit. Hence, you can see that sum of all current
that enters the cell via ion channels, passes out through the membrane
capacitor. Hence, I(ion) = -I(cap). Thus, C dV/dt = gna*(Erev - V).
I hope this helps.
On Jul 31, 3:38 am, reza moghadam <reza_... from yahoo.com> wrote:
> Hello everyone,
>> dose anybody can give me a help to understand what people mean by "
> ...capacitive membrane current is larger than ionic membrane current at the interesting neurobiology frequency..."
>> what i understand from HH type of equation (cdv/dt=ionic_current+leak_current+Isyn+Iinj),
>> capacitive current defined as cdv/dt and ionic current is G*(Erev-V). am i right? then independent of frequency capacitive current should be larger than total ionic current! or
> dose capacitive current have another interpretation? or we have to use another formula?
>> thanks in advance