Unformatted text preview: Some useful equations
Current I = (Coulombs/second) = Amperes (A) Ohm’s law V= IR Capacitance C = Q/V (Coulombs/Volts) = (F) Voltage across capacitor V = Q/C Changing the voltage
in a capacitor ΔV = ΔQ/C We change the charge
by passing current Ic = ΔQ/Δt The change in V depends
on the duration of Ic ΔV = Ic∗Δt/C Friday, September 10, 2010 Capacitor
+ + _ A= Area
Q
++ _ _
++
_
+
d = distance (separation of
the plates) Q
+ Capacitance C = Q/V Coulomb/Volt or Farads (F)
C =ε ∗A/d
ο Friday, September 10, 2010 εο electrostatic
permittivity A= C d= C Capacitor
rubber membrane Water pressure Small Capacitance Friday, September 10, 2010 Release of pressure Capacitors in parallel a dd
larger Capacitance Ion channels as resistors
(conductors) Internal conducting
solution
(ions) External conducting
solution
(ions) Ion channels
Voltagegated, NTgated
etc.
Friday, September 10, 2010 A = Area
l = length
Ohm’s law
R = V Ohms (Ω)
/I
l= R R = ρ∗l/A ρ resistivity
A= R For the same current, a larger R
produces larger V
Friday, September 10, 2010 Resistors
R1 As the # of Rs in parallel increases RT For ion channels is better to
think in terms of conductance R1 = 1/g1 decreases! R1 1/RT = 1/R1 +1/R2
More (open) channels in the
membrane more conductance R2 g T = g 1 +g 2
R1 R2 RT = R 1 + R 2
Friday, September 10, 2010 Long, thin parts of a
neuron have
large resistance! Also remember... Current likes to ﬂow through the path
w ith less resistance
R = 100 Ω IT And I2
I1
R=1Ω IT = I 1 + I 2
Friday, September 10, 2010 Speciﬁc membrane resistance
cross section of a cell To compare cell with different sizes
The speciﬁc membrane resistance
(resistance per area)
2 RM = Ω∗cm depends on the # of channels per cm
More channels make RM smaller For a spherical cell
2 Rin = RM /4πa a = radius Rin determines how much the cell depolarizes in response
to a steady current
Friday, September 10, 2010 2 Example same RM = 2000 2
Ω∗cm a a Cell diameter is 5 μm
Cell diameter is 50 μm
4 a = 25 μm = 25∗10 cm
Rin = 2
RM/4πa 2
2/4π(25∗10 4
Ω∗cm
cm) Rin = 2000
Rin = 25 MΩ
Friday, September 10, 2010 your numbers here...
Rin = 637 MΩ Rin is larger in a
smaller cell Speciﬁc membrane capacitance of
biological membranes
2 CM = 1 μF/cm CM = Q/V
6
Q = 10 C/V∗ 0.08 V
8 2 Q = 8∗10 C/cm
5 Faraday constant ≈ 10 Coulombs/mole
23
1
Avogadro’s number = 6.02∗10 mole
11 2 Then this is 4.8∗10 ions/cm
Friday, September 10, 2010 Speciﬁc membrane capacitance of
biological membranes
2 CM = 1 μF/cm For a cell at 80 mV how many ions
is this? CM = Q/V
6
Q = 10 C/V∗ 0.08 V
8 2 Q = 8∗10 C/cm
5 Faraday constant ≈ 10 Coulombs/mole
23
1
Avogadro’s number = 6.02∗10 mole
11 2 Then this is 4.8∗10 ions/cm
Friday, September 10, 2010 Speciﬁc membrane capacitance of
biological membranes
2 CM = 1 μF/cm For a cell at 80 mV how many ions
is this? CM = Q/V
6
Q = 10 C/V∗ 0.08 V
8 2 Q = 8∗10 C/cm
5 Faraday constant ≈ 10 Coulombs/mole
23
1
Avogadro’s number = 6.02∗10 mole
11 2 Then this is 4.8∗10 ions/cm
Friday, September 10, 2010 Is this a lot??? Let’s assume the cell is 50 μm in diameter
4 a = 25 μm = 25∗10 cm 50 μm 2 surface of sphere A = 4πa
5 cm2
A = 7.85∗10
11 2 4.8∗10 ions in 1 cm 7 So total is 4∗10 ions
8 3 The volume of this cell is 6.55∗10 cm
Then this number of ions is 6 10 M If KCl inside is 120 mM this means that only
1/120,000 ions is in excess!
Friday, September 10, 2010 Small cell
small capacitance Large cell
large capacitance “CM” is the same
(same membrane)
For a spherical cell, the input capacitance
2 Cin = CM∗4πa a = radius More charge (current) is required to change
the voltage across a larger cell
Friday, September 10, 2010 Solving for Vm a) INa + IK = 0
b) VinV0ut = Ek+ IK/gk
c) VinV0ut = ENa+ INa/gNa http://heart.med.upatras.gr/Prezentare_adi/electric_memb.jpg Vout +++
I
 INa K Vin VinV0ut =Vm  http://www.unm.edu/~toolson/equivcirc.gif + d) Ik = gk ( Vm EK)
 e) INa = gNa ( Vm ENa) Vm = (ENagNa + EKgK)
gNa + gK
Friday, September 10, 2010 In summary Rin = RM /4πa 2 Cin = CM∗4πa 2 τ = Rin∗Cin The product of input Capacitance and Resistance (τ)
determines the time it takes to change the potential
Notice that (τ) is not affected by “a”
Friday, September 10, 2010 Current in axons and dendrites
Current pulse rm
Section of axon or dendrite of
determined length (x). In this
case 1 cm. ra
ra axial resistance (Ω/cm)
rm membrane resistance (Ω∗cm)
Friday, September 10, 2010 Axial resistance increases with distance (x)
Total axial resistance ra1 ra2 ra3 ra4 rX = ra∗x (x = 4) (remember RT = R1 + R2) Near the site of injection, the current ﬂows through
rm (less resistance)
VO
Then VO = Im ∗ rm Friday, September 10, 2010 For 1 cm of cytoplasm (dendrites or
axon)
ρ (Ω∗cm) 3
resistive property of 1 cm of cytoplasm (dendrites or axon) 2 ra = ρ/(πa ) rm = Rm/(2πa)
a = radius 1 cm λ =√ (rm/ra) λ =√ R ∗a
m 2ρ If R and ρ are
constant
m λ =√ K a The length constant is proportional to
the
square root of the radius of the process
For neurons is usually 0.1 to 1 mm
Friday, September 10, 2010 Changes in length constant
Given a dendrite or an
axon with the same diameter
(a, b, c) If the length constant
increases, the potential
decreases less with
distance. Friday, September 10, 2010 ...
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This note was uploaded on 09/21/2010 for the course NACS 641 taught by Professor Areneda during the Fall '10 term at University of Maryland Baltimore.
 Fall '10
 ARENEDA

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