How to calculate Cell's Total Capacitance (not cm)?

Moderator: wwlytton

Post Reply
Krishna Chaitanya
Posts: 70
Joined: Wed Jan 18, 2012 12:25 am
Location: University of Pavia

How to calculate Cell's Total Capacitance (not cm)?

Post by Krishna Chaitanya » Wed Jan 18, 2012 12:59 am

Hi,

I am interested in knowing about the calculation of cell's total capacitance (not specific membrane capacitance (cm)). Under the biophysical properties of a section, cm is specified but I wanted to know the calculation of total membrane capacitance.

Could you kindly let me know how to do this? As I am a learner, detailed explanation could be very helpful.

ted
Site Admin
Posts: 5632
Joined: Wed May 18, 2005 4:50 pm
Location: Yale University School of Medicine
Contact:

Re: How to calculate Cell's Total Capacitance (not cm)?

Post by ted » Wed Jan 18, 2012 1:14 am

Good question. In pseudocode the algorithm would be

Code: Select all

cmtotal = 0
for each section
  for each segment
    cmtotal += segment_area * segment_cm
In hoc I'd implement this as a function

Code: Select all

func totalcm() { local tmp
  tmp = 0
  forall for (x,0) tmp += area(x)*cm(x)
  return tmp
}
The NEURON Programmer's Reference contains informative entries about the following:
func
local variables in functions
forall
for (x,0) and for (x)
area()
cm
syntax
expression

"Why did your pseudocode use a variable called cmtotal, but your function used a different name?"
No reason, actually.

Krishna Chaitanya
Posts: 70
Joined: Wed Jan 18, 2012 12:25 am
Location: University of Pavia

Re: How to calculate Cell's Total Capacitance (not cm)?

Post by Krishna Chaitanya » Wed Jan 18, 2012 3:15 am

Thank you Ted for the quick reply.

ted
Site Admin
Posts: 5632
Joined: Wed May 18, 2005 4:50 pm
Location: Yale University School of Medicine
Contact:

Re: How to calculate Cell's Total Capacitance (not cm)?

Post by ted » Wed Jan 18, 2012 9:25 am

Thanks for using NEURON in your research, and also for asking a question whose answer will, in one way or another, probably be of interest to many other users.

Post Reply