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10 0223 BME402HW3

# 10 0223 BME402HW3 - BME402Spring2010 [email protected]=0,gNA=5nS...

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BME 402 Spring 2010 Homework 3 Solutions Question 1 @t=0, gNA=5nS C=12 pF [email protected]=5 ms? Remember for a differential equation of the following form: dV/dt= (‐V + V )/ Tau the solution is: V(t)= De ‐t/Tau +A We will solve the KCL equation for dV/dt (V m ‐E l )g l + (Vm‐E Na )g Na + C(dV m /dt) = 0 dV m = ‐V m + (E l *g l /(g l +g Na )) + (E Na *g Na /(g l +g Na )) dt (C/(g l +g Na )) V m (t=0 )= D+A= ‐75 mV V m (t=∞)= A = E l *g l + E Na *g Na = 33.3333 g l + g Na ‐75 mV= D + 33.33333 D=‐108.33 V m (t)= ‐108.33* exp(‐t/.002 sec) +33.33 V m (t=5ms)= 24.441 mV (+/‐ 0.005 was accepted)

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BME 402 Spring 2010 Homework 3 Solutions Question 2 i=1; v(i)= ‐100; vmax=25; %%whole itteration while (v(i) <= vmax) %% m_inf if (abs(v(i) + 45) <=0.01) a_m(i)=1; else a_m(i) = (‐0.1*(v(i) + 45))/(exp(‐(v(i) + 45)/10) ‐ 1); end; %% n_inf if (abs(v(i) + 60) <=0.01) a_n(i) = .1;
BME 402 Spring 2010 Homework 3 Solutions else a_n(i) = (‐0.01*(v(i) + 60))/(exp(‐(v(i) +60)/10) ‐ 1); end; %% a_h(i) = (0.07*(exp(‐(v(i)+70)/20))); b_n(i) = (0.125*exp(‐(v(i)+70)/80)); b_m(i) = (4*exp(‐(v(i)+70)/18)); b_h(i) = 1/(exp(‐(v(i)+40)/10)+1); n_inf(i)= a_n(i)/(a_n(i)+b_n(i)); Tau_n(i)= 1/(a_n(i)+b_n(i)); m_inf(i)= a_m(i)/(a_m(i)+b_m(i)); Tau_m(i)= 1/(a_m(i)+b_m(i)); h_inf(i)= a_h(i)/(a_h(i)+b_h(i));

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10 0223 BME402HW3 - BME402Spring2010 [email protected]=0,gNA=5nS...

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