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Seal leak current bypasses the membrane patch through an imperfect seal. For an ideal ohmic seal to a zero-voltage bath, $I_{\mathrm{seal}}=V_p/R_{\mathrm{seal}}$, positive from pipette to bath. At $|V_p|=50\mathrm{mV}$ and $R_{\mathrm{seal}}=1\mathrm{G\Omega }$ its magnitude is $50\mathrm{pA}$, far larger than the synthetic $0.8\mathrm{pA}$ single-channel step. A stable baseline can be subtracted, but changing leakage can mimic a biological response. Intrinsic membrane leak through ion channels is distinct from this bypass: interpreting both as the same electrical conductance obscures whether the biological cell or the seal changed.

For N independent equivalent ion channels with open probability $P_O$ and approximately Ohmic open-state electrical conductance $\gamma$, the mean electric current is $I=N{P}_O\gamma (V-E)$. A numerical Electrical conductance cannot be inferred without the relevant reversal potential or driving voltage.

For an ideal open ion channel with constant electrical conductance $g$, membrane potential $V$ and reversal potential $E_{\mathrm{rev}}$, define outward electric current as positive:This is an Ohm's law model of the open state, not a universal channel law or a rule for opening probability. In the synthetic figure, $g=10\mathrm{pS}$ and $E_{\mathrm{rev}}=0\mathrm{mV}$. Because $1\mathrm{pS}\cdot 1\mathrm{mV}=10^{-3}\mathrm{pA}$, the open-state electric currents at $-80$ and $-40\mathrm{mV}$ are $-0.8$ and $-0.4\mathrm{pA}$. Zero electric current at the reversal potential does not imply a closed ion channel. Real ion channels can have nonlinear responses, so the straight line is an explicit idealization.

The electrical conductance of skin, often measured between two recording electrodes. Sweating changes the conductive properties of the skin. Baseline, movement and recording conditions must be considered before interpreting an event-related response as evidence about emotion.

On a finite undirected graph, stationary mass is proportional to vertex degree.

Synaptic conductance describes how strongly activated synaptic ion channels conduct electric current for a given driving force. In the passive approximation, $I_{\mathrm{syn}}=g_{\mathrm{syn}}(E_{\mathrm{syn}}-V)$, with inward-positive electric current. It is the electrical conductance contribution of a synapse; its postsynaptic voltage response also depends on the driving force, the other membrane electrical conductances, and the capacitance. The synaptic potential model and shunting inhibition example show why a nonzero synaptic conductance can have zero electric current at rest.

To determine the Thévenin resistance, remove the external electrical load and set each independent electrical source to zero in the circuit model. A zero ideal voltage source enforces zero electric potential difference, so replace it by a short connection. A zero ideal current source enforces zero electric current, so replace it by an open connection. Retain every resistor, including any explicitly modeled supply electrical resistance. Turning an actual power supply's front-panel control to zero is not evidence that it physically becomes an ideal wire.

Retain every dependent electrical source and its controlling variable: its response to an electrical test source is part of the network's electrical resistance. Impose terminal voltage $v_t$ with an electrical test source and calculate $j_t$ entering the suppressed network at $a$. For a nonzero test yielding a finite ratio,Alternatively inject a known entering electric current and solve for $v_t$. The test electric current is opposite to the delivered-current convention of the original Thévenin terminal response. ^[ref]

For an original illustration, put a positive resistor $R$ across $a,b$ and a dependent electrical source in parallel that injects electric current $g{v}_t$ from $b$ toward $a$, where $g$ has units of electrical conductance. Its control is the same terminal voltage $v_t$. Kirchhoff's current law givesWith $R=1\mathrm{k\Omega }$ and $g=0.5\mathrm{mS}$, a $1\mathrm{V}$ test draws $0.5\mathrm{mA}$, giving $R_{\mathrm{th}}=2\mathrm{k\Omega }$. Removing the dependent electrical source would falsely give $1\mathrm{k\Omega }$. With $g=2\mathrm{mS}$, the model instead has $R_{\mathrm{th}}=-1\mathrm{k\Omega }$; with $g=1\mathrm{mS}$ it draws zero test electric current for any imposed voltage and has no finite ratio. An active physical realization requires a supply and has a limited operating range.

For a finite nonzero $R_{\mathrm{th}}$ and a well-defined short-circuit response, $I_{\mathrm{sc}}=V_{\mathrm{oc}}/R_{\mathrm{th}}$. Thus $R_{\mathrm{th}}=V_{\mathrm{oc}}/I_{\mathrm{sc}}$ is usable when $I_{\mathrm{sc}}\ne 0$. A source-free positive resistor has both $V_{\mathrm{oc}}=0$ and $I_{\mathrm{sc}}=0$, so this quotient cannot reveal its electrical resistance; an electrical test source can.

Thomson principle says that effective resistance is the minimum energy among unit flows from $a$ to $z$:

The change of output electric current per change of input electric potential difference, evaluated at an operating point with the other terminal conditions specified. In a small-signal model, $i_{\mathrm{out}}=g_m{v}_{\mathrm{in}}$ describes this contribution; transconductance has the dimensions of electrical conductance.

Two-state ion-channel gating models an individual gate as a random switch between closed and open states. Write $p(t)$ for its open probability; this is the $X$ in the header equation. An individual gate's state is always closed or open, rather than partly open. A population of identical gates can have a fraction open near $p(t)$, and averaging repeated identically prepared trials estimates the same probability. A finite population's observed fraction fluctuates. See Purves and colleagues, Neuroscience, 2nd edition (2001): Ion Channels Underlying Action Potentials for single-channel steps and their population averages.

At fixed membrane potential $V$, let constant rates $\alpha ,\beta >0$ govern closed-to-open and open-to-closed transitions respectively, in ${\mathrm{s}}^{-1}$. This is a two-state continuous-time Markov chain. Over a short interval $\Delta t$, an initially closed gate opens with probability $\alpha \Delta t+o(\Delta t)$, and an initially open gate closes with probability $\beta \Delta t+o(\Delta t)$. Gain minus loss gives the deterministic ordinary differential equation for the probability:Here $p_{\infty }$ is the fixed-voltage limiting open probability, and $\tau$ is a time constant. The ratio of the rates fixes the limit, while their sum fixes the speed. With initial probability $p(0)=p_0\in [0,1]$,For a short derivation, subtract $p_{\infty }$: the deviation obeys $d(p-p_{\infty })/dt=-(p-p_{\infty })/\tau$, yielding the exponential function. After one time constant, $1/e$ of the initial deviation remains. The solution stays between $p_0$ and $p_{\infty }$, so it remains a valid probability. The rate and time constant formulations are described in Gerstner and colleagues, Neuronal Dynamics, §2.2.2: transition rates and channel kinetics.

For a worked toy condition A, choose $\alpha =\beta =50{\mathrm{s}}^{-1}$ and $p_0=0.1$. Then $p_{\infty }=0.5$, $\tau =0.01\mathrm{s}=10\mathrm{ms}$, andIn a separately prepared condition B, choose $\alpha =200{\mathrm{s}}^{-1}$, $\beta =50{\mathrm{s}}^{-1}$ and the same $p_0=0.1$. Now $p_{\infty }=0.8$, $\tau =4\mathrm{ms}$, and $p(10\mathrm{ms})=0.8-0.7e^{-2.5}\approx 0.742540$. Condition B approaches a higher limit more quickly. These are two stipulated constant-rate conditions with $V$ held fixed in each experiment, not an empirical function relating rates to voltage or a sequential switch between conditions. The initial $p_0=0.1$ need not be the limiting probability at the newly imposed voltage.

In the Hodgkin-Huxley model, the rates depend on membrane potential: ${\alpha }_X(V)$ and ${\beta }_X(V)$. Its $X_{\infty }(V)={\alpha }_X(V)/({\alpha }_X(V)+{\beta }_X(V))$ is the limit for voltage held at that $V$. During an action potential, changing $V$ changes the rates and gates feed back onto $V$ through ionic electric currents; substituting the present voltage into the constant-rate solution does not solve that coupled system. Gating describes availability of a conducting pathway. Even an open ion channel carries zero net ionic electric current at its reversal potential.

The useful bridge is from abrupt random molecular events to a smooth population prediction. A channel requiring several independent permissive gates in an ion channel has an open probability equal to their product; this can make channel availability much smaller than each gate's availability. The Hodgkin-Huxley model uses ${\bar{g}}_{\mathrm{Na}}{m}^{3}h$ and ${\bar{g}}_{\mathrm{K}}{n}^4$ for sodium and potassium electrical conductances, respectively.

A voltage clamp uses electrical feedback to hold a preparation's membrane potential near a commanded value while measuring the electric current required to maintain it. Holding voltage separates changes in electrical conductance from changes in electrical driving force. Imperfect voltage control and finite recording bandwidth limit the inference. The muscle experiment under ionotropic receptor used separate intracellular electrodes for voltage control and a sealed patch pipette for recording local electric current.^[ref]

A channel supporting ballistic electron transport carries incident populations set by different source and drain chemical potentials, rather than an everywhere-equilibrated local distribution with an Ohmic voltage gradient. Its electrostatic barrier responds self-consistently to contacts, gates and electrostatic screening. A symmetric rigid-barrier model places reservoir energies at $E_F\pm eV/2$ relative to the barrier. Asymmetric partition or bias-dependent transmission changes the weights in the differential electrical conductance, and must be included when extracting an energy scale.

A weighted graph assigns a symmetric nonnegative conductance ${\mu }_{xy}$ to each edge. With killing weights ${\kappa }_x$, the associated discrete-time walk moves from $x$ to $y$ with probability ${\mu }_{xy}/{\mu }_x$ and is killed with probability ${\kappa }_x/{\mu }_x$, where ${\mu }_x={\kappa }_x+{\sum }_y{\mu }_{xy}$.

For a finite undirected graph without self-loops, give each edge a symmetric nonnegative weight $w_{ij}=w_{ji}$. In the weighted graph, set $(A_w{)}_{ij}=w_{ij}$ for an edge and zero otherwise, and $(D_w{)}_{ii}={\sum }_j{w}_{ij}$ with all other entries zero. The Weighted graph Laplacian isEach edge appears once in the quadratic form. For the path graph $A-B-C$ with weights $w_{AB}=2$ and $w_{BC}=3$,The Weighted graph Laplacian is still a positive semidefinite matrix. Its kernel consists of vectors constant on each connected component of a graph after zero-weight edges are removed: an edge of weight zero imposes no constraint. In an electrical network, taking the weights to be electrical conductances makes $x^T{L}_{w}x$ the total power dissipated by Joule heating when $x$ gives the voltages at the vertices. This physical interpretation explains why stronger connections penalize disagreement more heavily.^[ref]

Wilson's algorithm samples a uniform spanning tree by starting from a root and successively attaching loop-erased random walks from vertices not yet in the tree.