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I have read from several sources that in the leaky integrate-and-fire model of a single neuron resistor (leakage) and capacitor (membrane potential) are arranged in parallel. Look at the following RC circuit, taken from the book Neuronal Dynamics:

RC circuit

The author says:

The cell membrane acts like a capacitor in parallel with a resistor which is in line with a battery of potential $u_{rest}$.

But I think capacitor and resistor are arranged in series. Can anybody please explain why it is said that they are arranged in parallel?

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    $\begingroup$ Why do you think they are arranged in series? $\endgroup$ – Fizz Jul 15 '18 at 22:27
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    $\begingroup$ @Fizz resistor and capacitor are not connected from both sides to the battery! Should I consider the injected current (I(t)) as battery? $\endgroup$ – m.taheri Jul 16 '18 at 14:59
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I see your confusion is caused by $u_{rest}$. Indeed that diagram is somewhat confusing because $u_{rest}$ is not the main source relative to which to consider the topology of the circuit. The main source is actually a current source designated by $I(t)$ in your diagram. With respect to that source R and C are in parallel. If that's not convincing enough, consider another (information) source which presents a different (but still leaky) integrate and fire model, sans a $u_{rest}$:

enter image description here

In the latter diagram, $I_{dc}$ is the equivalent of $I(t)$ in yours. In the alternative circuit, the firing actually comes from a voltage-controlled switch, which is indeed in series with the RC circuit (to the left of it).

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  • $\begingroup$ Thank you @Fizz. Please consider the circuit that I posted. It is said that the battery is providing resting potential. So when there is not input current, the potential difference between two sides of the capacitor (u(t)) should be equal to u_rest. But the resistor that is in series with the capacitor causes u(t) to be less than u_rest. right? $\endgroup$ – m.taheri Jul 19 '18 at 7:31
  • $\begingroup$ @m.taheri: there's going to be a voltage drop across R, correct... but it's going to go down over time as the capacitor charges: en.wikipedia.org/wiki/RC_circuit#Time-domain_considerations I think at this point your (subsequent) questions are better addressed on electronics.stackexchange.com $\endgroup$ – Fizz Jul 19 '18 at 8:18
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The resistor is the conductance of the ion channels in the membrane. The capacitor is the membrane itself, which acts as a capacitor (a non-conductive center sandwiched between electrically active materials). There are two ways for "current" to flow between the outside and inside of the cell. One is through ion channels (the resistor), and one is for charges to collect on one side of the membrane and to push charges away from the other side of the membrane (the capacitor). These are in parallel, as they are independent ways for electrical activity on one side of the membrane to affect electrical activity on the other side of the membrane.

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    $\begingroup$ Please add some sources; some mention about Hodgkin & Huxley would be great imo. For now, +1, and a banner ;-) $\endgroup$ – AliceD Jul 16 '18 at 6:36
  • $\begingroup$ @honi I understand your explanation from viewpoint of neuroscience. My question is about the RC circuit that is a neuron model. If you look at the circuit, resistor and capacitor are arranged in series with respect to battery (u_rest). $\endgroup$ – m.taheri Jul 16 '18 at 12:05
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    $\begingroup$ i answered the question that you asked "why it is said that capacitor and resistor are in parallel". if you have another question, could you be explicit about what that question is? $\endgroup$ – honi Jul 16 '18 at 13:13
  • $\begingroup$ if you are asking for an explanation of the battery, the answer is that this is a non-standard simplified formulation. "If the driving current I(t) vanishes, the voltage across the capacitor is given by the battery voltage urest . For a biological explanation of the battery we refer the reader to the next chapter. Here we have simply inserted the battery ‘by hand’ into the circuit so as to account for the resting potential of the cell." $\endgroup$ – honi Jul 16 '18 at 13:15

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