2. When i a nerve impulse travels down an axon membrane, sections of the membran

Question

2. When i a nerve impulse travels down an axon membrane, sections of the membrane experience "depolarization" olarization" in which the potential difference across the membrane reverses. We have seen how to el this reversal using a circuit with two batteries and a switch. We could think of the axon membrane as a and "rep ese circuits next to each other a nerve impulse traveling down the axon would correspond to closing and then re-opening the switches in each circuit in sequence. This model is incomplete, however, because it doesn't include a mechanism for triggering the closing and opening of the switches (ie., how does the next circuit in the sequence know when to close and open its switch?). In other words, we haven't modeled the R1 = 100 k traveling" part of the nerve impulse. To accomplish this, we will use a different circuit that involves capacitors, shown here a. Identify any circuit elements that are in series, and any that are in parallel. b. Write an equation that relates the potential difference across capacitor 1, JAVcl,to the potential differences across resistor 2, aVyl, and capacitor 2, 11al. Write an equation that relates the potential difference between points P and Q. AVpol, to the potential differences across resistor 1, IaVyl, and capacitor 1, lavel, c. We'll begin with the capacitors initially uncharged (1Q1 0). Suppose we connect a 10 V battery betweer points P and Q, so IvP0|-10 V. The following questions will analyze what happens right when we connect the battery-current has just begun to flow but no charge has built up on the capacitor plates. d. What is the potential difference across C1? What about C2? e. What are the potential differences across R1 and R2?

Explanation / Answer

2. a. the circuit elements R2 and C2 are in series

C1 and Series combionation of R2, C2 are in parallel

R1 and parallel combination of C1 and series combination of R2,C2 are in series

b. potential diff across capacitor 1 = dVc1

potential diff across capacitor 2 = dVc2

potential diff across resistor 2 = dVr2

from kirchoff's loop law

dVc1 = dVc2 + dVr2

c. again from kirchoff's loop law

dVpq = dVr1 + dVc1

d. dVpq = 10 V

Q = 0 at t = 0

as soon as the battery is connected, the capacitors act like closed switches

so dVc1 = dVc2 = 0

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