Question

After the switch has been closed for a long time, so that the capacitor is completely charged, the switch then opens so that the capacitor discharges through the membrane resistor. This models "repolarization" of the axon. Here, after the potential of the inside the axon as compared to the outside becomes quite large and positive, the membrane becomes more permeable to potassium ions, which then rush from the inside of the axon (top capacitor plate) to the outside of the axon (bottom capacitor plate) because of the positive potential of the inside with respect to the outside as well as there being, inherently, a larger concentration of potassium ions inside the axon as compared to the outside. S./ Ra VLC.RM For the membrane capacitor discharging through the membrane resistor, the charge of the capacitor and, hence, the voltage across the capacitor as well as the current through the membrane capacitor-resistor loop all decay exponentially. For example, the voltage on the capacitor changes in time as V = Voe-t/T where r is the time constant, i.e., the time it takes for the voltage to decay to 1/e of its initial value, of this circuit. What is the form of this time constant in terms of the membrane resistance and capacitance? We will find that this time constant is relevant for determining the speed of the pulse.

a) 1/RmCm

b) Cm/Rm

c) R_mC_m

d)R_m/C_m

Answer 1

The capacitor dischaged thngh he membrane Yuis hor R there fre the time umstant s T Rm m