# Homework Help: Kirchov’s Current Law

1. Dec 3, 2016

### Alex Katko

1. The problem statement, all variables and given/known data
Using Kirchov’s Current Law (KCL) at the top node of the circuit, write the differential equation for the membrane voltage (Vm). Solve for Vm(t) symbolically,assuming that all other voltages, all resistances, and the capacitance are constant

2. Relevant equations
Current law (sum of currents at a node equals 0)

3. The attempt at a solution
I think after i set it up i would be able to solve the differential equation. What I have is:
Cm dVm/dt + (Vm-Eth)/Rth + (vm-Vc)/Rc = 0.

Let me know if im on the right track

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2. Dec 4, 2016

### Staff: Mentor

Hi Alex. Yes, your work looks fine; You're on the right track.

3. Dec 4, 2016

### Alex Katko

Simplifying my setup gives me
dVm/dt + Vm(1/Rth + 1/Rc) = (Eth/Rth) + (Vc/Rc)

From here i use integrating factor to solve.
My integrating factor would be: u(t) = exp(((1/CmRth) + (1/CmRc))t)

Finishing some algebra, i get: Vm = (Rth + Rc)((Eth/Rth) + (Vc/Rc)) + Cexp(-((1/CmRth)+(1/cmRc))t)

Part B of the problem says: Solve for Vm at time, t = ∞. From this equation and the data for the long time point (t=1000s), solve for Rth. Your parameters from the experiment are Vc = -25mV, Rc =1.75kΩ, and ETh = -65mV. Vm(1000) = -43.9 mV

At t = ∞, the exponential term should drop off and i would be left with: Vm = (Rth + Rc)((Eth/Rth) + (Vc/Rc))
However, plugging in the values given to me gives me imaginary values for Rth.

I'm really struggling with this problem and have spent a long time on it. Any guidance is much appreciated. It's either a problem with the setup in the original post, or my math went wrong somewhere (through i went through the entire think 3 times)

4. Dec 4, 2016

### Staff: Mentor

I'm finding that the circuit definitions are a bit odd. For example, the two sources are shown with their negative terminals upwards and their specified potential differences are given as negative values. So really they are sources that are positive on top? Is this just meant to add confusion or is there some physical reason why the circuit is modeled in this way? Also Vm is defined on the diagram to be the potential at the bottom node with respect to the top node, yet you're asked to use the top node for writing the node equation. I'm wondering if there's a point to all this? What is the significance of the dashed lines for the Vc branch? Is this branch only connected at t = 0? This would have implications for the initial conditions of the circuit.

5. Dec 4, 2016

### Alex Katko

I too was confused on why all of the given voltages were negative. If it helps, there was a bit of information at the top of the page:

You are conducting a voltage clamp experiment on a cell in which all of the gated channels have been inactivated. The circuit model of the cell membrane and voltage clamp is given below, as well as the data of the membrane voltage measured over time

I think that it is okay that Vm is the difference between the top and bottom node. You can still use the top node to find Vm (because Vm shows up in the differential equation). Also, the dashed lines can be read as a solid line. The professor mentioned that he made them dashed lines just to show it's the voltage clamp and not the cell, but functionally it is a solid line. The branch is connected for all t.

6. Dec 4, 2016

### Staff: Mentor

Okay.

Going back to your differential equation solution, I think you need to revisit it. The units don't look right:
Note that if C is a capacitance then you are equating volts to farads.

7. Dec 4, 2016

### Alex Katko

C was the constant that came out of my integration. I actually solved for C with the initial value V(0) = -64.3
C = -64.3 - (Rth+Rc)((Eth/Rth)+(Vc/Rc))

8. Dec 4, 2016

### Alex Katko

But the question asked me for what Vm is as t approaches infinity. So solving for C wasn't really necessary here because I believe that term should just go to 0

9. Dec 4, 2016

### Staff: Mentor

I see. Where did this V(0) come from ? It wasn't mentioned previously.

10. Dec 4, 2016

### Alex Katko

Yeah I should have mentioned it. There is a table of values, but i don't really see how they are useful, besides V(1000) which i gave to you and possibly V(0). I attached that table.

#### Attached Files:

• ###### Capture.PNG
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11. Dec 4, 2016

### Staff: Mentor

Alright. Looks like output voltage readings taken from the moment that the "clamp" is applied.

Getting back to determining Rth then, my first instinct would be to use the steady state conditions to find it, bypassing the differential equation entirely. Remove the capacitor (since at steady state its current will be zero) and find an expression for the output voltage. Equate it to the given Vm for t = 1000s. You should not run into any square roots doing it this way.

12. Dec 4, 2016

### Alex Katko

I thought of doing something like this, but i felt the instructions sortof implied i was supposed to use the solved differential equation to find this. And if my solution to the equation is correct, then it should give me the right value for Rth. Assuming the equation is correct, am i right to say that the exponential term would drop off, leaving only Vm = (Rth + Rc)((Eth/Rth) + (Vc/Rc))? If so, solving for Rth gives imaginary values, witch means my solution is incorrect. Part A (solving for Vm at any time t) was worth almost all of the credit, so i am trying to focus my efforts in seeing what i did wrong.