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Model a spatially uniform cell membrane as a capacitor of membrane capacitance $C_m$ in parallel with membrane electrical resistance $R_m$ and a fixed leak reversal potential $E_L$. Exclude active ion channels, seal leak current and pipette capacitance. For electric current $I$ injected into the biological cell, electric charge conservation yieldsStarting at $V(0)=E_L$ and applying constant $I$ gives the ordinary differential equation solutionThe steady state rise is $I{R}_m$; the time constant is $R_m{C}_m$. With $I=50\mathrm{pA}$, $R_m=200\mathrm{M\Omega }$, $C_m=50\mathrm{pF}$ and $E_L=-70\mathrm{mV}$, the rise is $10\mathrm{mV}$ and ${\tau }_m=10\mathrm{ms}$. At one time constant the rise is $10(1-e^{-1})\mathrm{mV}\approx 6.321\mathrm{mV}$. The figure's uncompensated pipette voltage adds $I{R}_s=1\mathrm{mV}$ for $R_s=20\mathrm{M\Omega }$. The membrane voltage is continuous at step onset in this model, while the pipette voltage immediately acquires that access drop.

Checking the units is instructive: electrical resistance times capacitance has units $\Omega \mathrm{F}=\mathrm{s}$, and electric current times electrical resistance has units of voltage. Unlike the passive voltage clamp transient under patch-clamp series resistance, this ideal current clamp membrane time constant is set by $R_m{C}_m$, not $R_s{C}_m$. Strong current steps can recruit voltage-gated ion channels; the passive model then fails and cannot generate an action potential. Small pulses of both signs, stable access and repeated baseline measurements help test whether the passive approximation is appropriate.

The spatial panel uses two compartments with membrane potentials $V_s,V_d$, one representing the soma and one a dendrite. Set $v_s=V_s-E_L$, $v_d=V_d-E_L$, with leak reversal potential $E_L=-65\mathrm{mV}$, each membrane capacitance $C=100\mathrm{pF}$, each leak electrical conductance $g_L=5\mathrm{nS}$ and axial electrical conductance $g_c=10\mathrm{nS}$. Electric charge conservation givesThis chosen passive model has two decay time constants, $C/g_L=20\mathrm{ms}$ for the equal-voltage mode and $C/(g_L+2g_c)=4\mathrm{ms}$ for the voltage-difference mode. Starting from rest, a $100\mathrm{pA}$ step into the soma gives, with $u$ the time since onset,while the same step into the dendrite gives$u$ is expressed in $\mathrm{ms}$. Their long-time rises are $12$ and $8\mathrm{mV}$ respectively. Applying both inputs gives $20(1-e^{-u/20})\mathrm{mV}$ exactly: spatial attenuation coexists with linear summation of prescribed electric currents. After the steps end, both modes decay with their respective time constants. This deliberately coarse model is not a reconstructed dendrite or a fit to the experiment above.

The conductance panels use the existing one-compartment synaptic potential equation, with $C=200\mathrm{pF}$, $g_L=10\mathrm{nS}$, $E_L=-65\mathrm{mV}$ and excitatory reversal potential $E_e=0\mathrm{mV}$. For a constant excitatory synaptic conductance $g_e$, the steady state voltage rise and membrane time constant areEach isolated $2\mathrm{nS}$ input has a limiting rise $130/12\approx 10.833\mathrm{mV}$. Adding those isolated rises predicts $21.667\mathrm{mV}$, but opening both together gives $260/14\approx 18.571\mathrm{mV}$. The two conductances share a changing driving force. This is sublinear input summation even though the voltage ordinary differential equation is linear for each prescribed conductance waveform.

Add $g_i=20\mathrm{nS}$ with reversal potential $E_i=E_L$ to the joint $4\mathrm{nS}$ excitation. The limiting rise becomes $260/34\approx 7.647\mathrm{mV}$ and the membrane time constant becomes $200/34\approx 5.882\mathrm{ms}$. The shunt alone leaves rest unchanged. At the joint-excitation-only steady state $V\approx -46.429\mathrm{mV}$, however, its inward electric current would be $20(-65+46.429)\approx -371.429\mathrm{pA}$: it opposes that excursion. The finite $50\mathrm{ms}$ traces in the diagram approach, but need not reach, these limiting voltages. See shunting inhibition for the constant-electric current comparison.

Conservation of total probability gives $A(t)=1-B(t)$. The two-state master equation is thereforeSolving this linear ordinary differential equation givesThe middle panel in d sketches this solution for the explicitly illustrative ratio $k_f/(k_f+k_b)=0.6$. The curve starts at zero with slope $k_f$, is increasing and concave down, and approaches the horizontal limit $B_{\infty }=k_f/(k_f+k_b)$ on time constant $(k_f+k_b{)}^{-1}$. Its initial tangent meets the asymptote at that time constant. The complementary limit is $A_{\infty }=k_b/(k_f+k_b)$, so $B_{\infty }/A_{\infty }=k_f/k_b=e^{-\Delta G_{AB}/(k_{B}T)}$, the Boltzmann distribution ratio. These are occupancy probabilities; $B(t)$ is not the cumulative first-passage probability when reverse transitions remain possible.

Ion-channel deactivation and ion-channel inactivation are different. Ion-channel deactivation closes an activation gate when the voltage returns toward rest. Ion-channel inactivation makes a channel unavailable even while depolarization remains favorable for activation. A minimal kinetic picture is a closed state, an open state and an unavailable state, with recovery required before reopening. This helps terminate inward electric current and produces a neuronal refractory period.

In rapid voltage-gated potassium channel N-type inactivation, a cytoplasmic terminal segment, sometimes supplied by an auxiliary subunit, blocks the open inner pore. The classic ball-and-chain inactivation picture is supported by loss of rapid ion-channel inactivation after removing the relevant terminal segment and restoration by an appropriate soluble peptide. This links a mobile structural element to open-pore blockade; it does not imply that all potassium channels use that mechanism.^[ref]

Slower C-type inactivation involves rearrangement near the selectivity filter and external pore. Dependence on pore structure, ion occupancy and extracellular ions distinguishes it from a simple cytoplasmic plugging particle. Different channels can combine several unavailable states, so a single recovery time constant need not describe every trace.

Fast ion-channel inactivation in canonical animal voltage-gated sodium channels involves the intracellular linker between domains III and IV, including its hydrophobic IFM motif, and is coupled to domain-IV voltage-sensor activation. The modern structural interpretation is an allosteric latch acting on the gate, rather than a universal literal particle sitting in the ion pathway. The older hinged-lid analogy is useful only with that qualification.^[ref]

Functionally, a voltage clamp trace shows electric current declining during a maintained step; a second test pulse measures recovery of availability. Holding-potential dependence separates an altered fraction of available channels from a mere change in maximal open electrical conductance. Persistent electric current after defective ion-channel inactivation, or delayed recovery that reduces subsequent responses, illustrates why both entry into and exit from unavailable states matter for excitability. A decaying macroscopic electric current alone does not uniquely identify the molecular mechanism.

For the strong Hill autorepression response, in the requested approximation, $\dot{Y}=\beta K^n{Y}^{-n}-\alpha Y$. Put $z=Y^{n+1}$. Thenso the formal zero-initial-concentration solution isSetting $Y(T_{1/2})=Y_{\mathrm{st}}/2$ givesThe formal ratio decreases with positive $n$; for $n=1$ it is about $0.208$, and for $n=2$ about $0.0642$. Its large-$n$ asymptote is $2^{-(n+1)}/[(n+1)\log 2]$. This is a nonlinear half-response time, not simply the near-equilibrium relaxation time $1/[(n+1)\alpha ]$.

Approximation qualification. The omitted initial layer is precisely the one the source permits us to ignore: the approximate derivative diverges at $Y=0$, whereas the full Hill equation has initial slope $\beta$. The approximation is not uniform as $n\to \infty$ at fixed $R=\beta /(\alpha K)$. Strong repression already at half of the final concentration needs $R^{1/(n+1)}/2\gg 1$. The actual Hill equation obeys $\dot{Y}\le \beta$, so $T_{1/2}\ge Y_{\mathrm{st}}/(2\beta )$. Its steep-switch limit approaches the logic-response scale in a; the formal vanishing large-$n$ result above cannot be extrapolated past its regime of validity.

The equivalent circuit has membrane capacitance in parallel with electrical conductance–battery branches: voltage-dependent sodium and potassium conductances in series with their reversal potentials, a chloride electrical conductance in series with $E_{Cl}$, and other leak conductances. For externally supplied electric current,The model figure in solution includes this circuit and an illustrative passive response. Although chloride electric current is small when $V\simeq E_{Cl}$, a perturbation produces the restoring electric current $\delta I_{Cl}=g_{Cl}\delta V$. A large chloride electrical conductance therefore shunts perturbations, lowers input resistance and shortens the passive time constant, $R_{\mathrm{in}}\simeq 1/\sum g_j$ and ${\tau }_m\simeq C_m/\sum g_j$.

Reducing that electrical conductance increases the voltage response to a given electric current or extracellular-potassium disturbance and weakens stabilization near rest, making repeated excitation easier. There need not be an instantaneous shift of resting voltage if $E_{Cl}$ equals the old rest exactly. In skeletal muscle, loss of ClC-1-mediated stabilization is associated with myotonic hyperexcitability.^[ref]

Later cessation of contraction can be explained by sustained depolarization causing voltage-gated sodium channel ion-channel inactivation and failure to recover: fewer available channels support progressively smaller or less frequent spikes until depolarization block occurs. Accumulated extracellular potassium in restricted spaces and incomplete recovery between spikes can promote that depolarization. This requires the nonlinear voltage-gating dynamics; a passive chloride resistor and capacitor alone cannot generate or terminate an action-potential train. The stated membrane permeability change does not uniquely determine the time to block or prove one particular ionic accumulation history.

Detailed balance means that at thermodynamic equilibrium each elementary transition and its reverse carry equal probability flux. For normalized state probabilities $A_{\mathrm{eq}}$ and $B_{\mathrm{eq}}$,With the printed convention $\Delta G_{AB}=G_A-G_B$, the sign is thereforeThe $G$ values are free energies per system, including internal degeneracies; for molar Gibbs free energies, replace $k_{B}T$ by $RT$. If B has lower free energy, its forward-to-reverse rate ratio exceeds one, as it should.

For constant rates, conservation gives $A=1-B$, so the master equation is $\dot{B}=k_1(1-B)-k_{2}B$. Its steady state is $B_0=k_1/(k_1+k_2)$, and separation of the decaying deviation yieldsfor the specified initial state A. The relaxation time constant is $(k_1+k_2{)}^{-1}$.

In the Hodgkin-Huxley model, $V$ is the intracellular-minus-extracellular membrane potential, $C_m$ is the membrane capacitance per unit area, and $I_{\mathrm{app}}$ is the applied inward electric current per unit area. A space clamp makes $V$ spatially uniform, leaving temporal ordinary differential equations. With outward ionic electric current positive, the capacitance balance is $C_m\dot{V}=I_{\mathrm{app}}-I_i$.

The variables $m,n,h$ lie between zero and one. They describe sodium activation, potassium activation and availability against sodium inactivation, respectively. Thus larger $h$ means less inactivation. Their voltage-dependent limiting values are $m_{\infty },n_{\infty },h_{\infty }$, and ${\tau }_m,{\tau }_n,{\tau }_h$ are their positive relaxation time constants.

Each ionic term has the Ohm's law form conductance times driving potential. The sodium term has instantaneous electrical conductance $g_{\mathrm{Na}}{m}^{3}h$, the potassium term has instantaneous electrical conductance $g_{\mathrm{K}}{n}^4$, and the leak term has constant electrical conductance $g_{\mathrm{L}}$. Here the $g$ constants are maximal electrical conductances per area, and $V_{\mathrm{Na}},V_{\mathrm{K}},V_{\mathrm{L}}$ are the corresponding reversal potentials. The factors $m^{3}h$ and $n^4$ arise from independent permissive gates in an ion channel. A conducting channel carries zero net ionic electric current at its reversal potential.

Use the three standard reduction assumptions: $m$ relaxes much faster than $V,n,h$ and can be replaced by $m_{\infty }(V)$; the two slower gates have the same time constant, ${\tau }_n(V)={\tau }_h(V)=\tau (V)$; and their limiting open fractions have a voltage-independent sum, $n_{\infty }(V)+h_{\infty }(V)=K$. The constant-sum assumption is essential: equality of two time constants by itself does not preserve a moving voltage-dependent sum.

For $q=n+h$, addition of the two gate ordinary differential equations givesThe specified initial condition is $q(0)=K$. Uniqueness for this first-order linear differential equation gives $q(t)=K$, hence $h=K-n$ for all time. The remaining two ordinary differential equations areThese assumptions give a planar reduction of the Hodgkin-Huxley model. The FitzHugh-Nagumo model keeps its fast activation and slow recovery variable geometry, approximating the voltage response by a cubic polynomial and the recovery response by a linear function after nondimensionalization. The polynomial equations in part (c) are a further model simplification; they are not an exact algebraic consequence of arbitrary original gate functions.

Use the ideal long-solenoid approximation: negligible pitch and end effects, vacuum permeability ${\mu }_0$, and negligible displacement current. By cylindrical symmetry the magnetic field is axial. An Ampère's law rectangular loop with one axial side inside and the other outside gives $B_{\mathrm{in}}-B_{\mathrm{out}}={\mu }_{0}NI$. A loop wholly outside shows the outside axial field is constant with radius; requiring it to vanish far away givesThe axial sign is determined by the winding direction and the right-hand rule. The small axial component of current from the spiral pitch is neglected in this idealization.

There are $Nl$ turns, and the magnetic flux through each is $\pi a^2{\mu }_{0}NI$. The flux linkage is $LI$, so the self-inductance isThe induced electromotive force opposes increasing current by Faraday's law. Together with Ohm's law, the circuit equation is $L\dot{I}+RI={\mathcal{E}}_0$. Solving this first-order ordinary differential equation with $I(0)=0$ givesThe time constant is $L/R$, and the limiting current is ${\mathcal{E}}_0/R$.

A thermoelectric generator maintains hot and cold junctions of two materials with different Seebeck coefficients. Mobile charge carriers redistribute in the temperature gradient, producing an open-circuit voltage of magnitude $|V|=|S_p-S_n|\Delta T$ for constant coefficients. A load permits electric current; pairs of p-type and n-type legs are commonly connected electrically in series and thermally in parallel. Maintaining the output requires continuous heat transfer from the hot reservoir and rejection of heat to the cold reservoir. The thermoelectric figure of merit $ZT=S^2\sigma T/\kappa$ expresses the favorable combination of large Seebeck coefficient, large electrical conductivity and small thermal conductivity. Contact losses, Joule heating and parasitic heat leaks reduce practical efficiency.

A pyroelectric generator instead exploits a change of spontaneous electric polarization with temperature. If the relevant component of the pyroelectric coefficient is $p=d{P}_s/dT$, an electrode of area $A$ receives a short-circuit electric current approximately $I=pAdT/dt$. A heating stroke and a cooling stroke produce opposite charge changes. Rectification or a synchronized electrical cycle can transfer energy to storage. A spatial temperature difference drives the thermoelectric device; a temporal temperature change drives the pyroelectric device. Uniform, constant temperature gives no continuing current from pyroelectricity even if the material is hot. Reported nanowire generators deliberately use changing temperature, rather than infer continuous generation from a static hot specimen.^[ref]

Both approaches can be compact, silent and free of macroscopic moving parts. The thermoelectric generator gives continuous output under a sustained gradient, but needs two effective thermal contacts and sufficient heat flow; very small $\Delta T$ gives a small voltage and thermal contacts can consume much of the available gradient. Thin films exhibiting pyroelectricity can respond to intermittent heating without a permanently maintained hot/cold separation across their thickness, but require repeated thermal cycling, low electrical leakage and suitable charge extraction. Their thermal time constant limits useful cycling frequency, and sensible heating/cooling can dominate the heat budget. A steady hot pipe with an available cold sink favors a thermoelectric generator; periodic radiative heating can favor a pyroelectric generator. Neither converts equilibrium thermal motion at a single fixed temperature into sustained useful work.

Put $e_t=e^{-t/\tau }$. From $u=V/2-R{I}_1$ and $I_0=I_1+I_2$,At $t=0^{+}$ the three values are $I_1=0$ and $I_2=I_0=V/(4R)$. At late times they are $I_1=V/(4R)$, $I_2=V/(8R)$, and $I_0=3V/(8R)$. Thus $I_1$ and $I_0$ increase monotonically while $I_2$ decreases; every transient has the same time constant. By $10\tau$, the exponential correction is only $e^{-10}\simeq 4.54\times 10^{-5}$ of its initial amplitude.

Let $C$ denote the capacitance omitted after the word “capacitance” in the source sentence. For the ideal series RC charging circuit, Kirchhoff's voltage law gives $RI+Q/C=V$, with $I=dQ/dt$ and $Q(0)=0$. ThereforeThe electric current falls from $V/R$ to zero, while the electric charge rises from zero to $CV$. Their initial slopes are $-V/(R\tau )$ and $V/R$, respectively.

Reducing $RC$ shortens the approach to the relevant final values. Which heights change depends on the component changed: reducing $R$ at fixed $C$ increases $I(0)$ but leaves $Q(\infty )$ unchanged; reducing $C$ at fixed $R$ leaves $I(0)$ unchanged but reduces $Q(\infty )$. A shorter time constant alone does not specify both endpoint amplitudes.

The RC low-pass filter satisfies $RC{\dot{V}}_{\mathrm{out}}+V_{\mathrm{out}}=V$ for $t>0$. With zero initial capacitor voltage, the first-order linear differential equation givesThe output is continuous at switching and approaches $V$ with time constant $RC$.

When $\omega RC\ll 1$, each half-cycle lasts many time constants, so $M\simeq V_0$. The output nearly follows the two input levels, with exponentially rounded transitions lasting order $RC$. It remains continuous rather than making ideal vertical jumps.

Let $v_C$ denote the capacitor voltage with the same polarity as the applied voltage. By Kirchhoff's voltage law, Ohm's law and $I=Cd{v}_C/dt$,Solving this first-order differential equation gives the RC low-pass step responseThe capacitor voltage is continuous at switching because an instantaneous finite jump would require an impulsive electric current.

For the square-wave response of an RC low-pass filter, take equal positive and negative half-periods. Let $A=10\mathrm{V}$, $T=1/f$, $h=T/2$ and $\alpha =e^{-h/\tau }$. After the transient, symmetry makes the capacitor voltage alternate between $-v_p$ and $+v_p$. Over the positive half-period,Requiring $v_C(h)=v_p$ givesThe next half-period is the sign-reversed copy. These exact expressions explain the requested sketches:

Position A disconnects the battery and closes the resistor-inductor loop. Taking clockwise electric current as positive, Kirchhoff's voltage law for this RL circuit givesThe electric current through an ideal inductor remains continuous across a switch with no impulsive voltage. Thus $I(0^{+})=I_0$, and separation of variables givesIf the switch remains at A, the same result holds for every $t>0$. The time constant is $L/R$; the stored magnetic energy is dissipated by Joule heating.

Use an unloaded RC low-pass filter: a series resistor between the source and the output node, with a capacitor from that node to ground, and measure the output across the capacitor. The components are ideal in this first-order model; “ideal” here does not mean a brick-wall frequency response.

The time constant is $\tau =RC$, henceThe capacitive electrical impedance gives the transfer function $H(i\omega )=1/(1+i\omega RC)$, soOn the magnitude Bode plot, $20{\log }_{10}|H|=-10{\log }_{10}[1+(f/f_c{)}^2]$. The gain approaches $0\mathrm{dB}$ for $f\ll f_c$, is $-3.01\mathrm{dB}$ at $f_c$, and falls with slope $-20\mathrm{dB}$ per decade for $f\gg f_c$. These results assume negligible source resistance beyond the specified $R$ and a sufficiently large load impedance.

Put ${\omega }_0^2=1/(LC)$. The initial conditions giveThe printed electric current arrow points toward the positively labelled capacitor plate, so its signed electric current is $I=\dot{q}$, not the positive discharge magnitude. HenceFor the stated electrical resistance,The voltage starts at $6\mathrm{V}$ with zero slope; the signed electric current starts at zero and initially becomes negative. Its initial slope is $\dot{I}(0)=-V_0/L=-6\mathrm{A}{\mathrm{s}}^{-1}$. Both oscillate with an $e^{-25t}$ envelope and a $40\mathrm{ms}$ amplitude time constant. The phase difference is close to a quarter period, with the small damping correction displayed explicitly in $V_C$.

Faraday's law of induction relates an induced electromotive force to the rate of change of magnetic flux. Choose a positive orientation around a loop and its associated surface normal by the right-hand rule. For a stationary conducting loop,Unlike an electrostatic field, the induced electric field generally has nonzero curl; the differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a moving circuit, magnetic flux transport for a moving circuit includes the magnetic Lorentz force on the charges:where $\mathbf{u}$ is the local wire velocity. Thus changing the field or moving a conductor can produce induction.

The minus sign is Lenz's law: an induced electric current, when a closed conducting path permits it, produces a magnetic field opposing the change of linked flux. It need not oppose the original field itself. For example, increasing flux through a loop produces a current tending to reduce that increase; decreasing the same flux reverses the current. The opposition is consistent with conservation of energy: external mechanical or electrical work is needed to maintain a change against its inductive reaction.

For a coil of $N$ turns, define the magnetic flux linkage $\Lambda ={\sum }_{j=1}^N{\Phi }_j$, reducing to $N{\Phi }_B$ when every turn links the same flux. For fixed geometry and a linear magnetic response, self-inductance $L$ is defined by $\Lambda =LI$, and the coil's induced electromotive force isThe corresponding terminal voltage drop, with the passive sign convention, is $V_L=LdI/dt$. It opposes changes in electric current; it does not require the current itself to vanish.

Building the current from zero requires power $P=I{V}_L=LIdI/dt$. Integrating gives the energy stored by an ideal inductor:This energy is stored in the magnetic field and can be returned when the current falls. In a series RL circuit driven by a constant voltage $V_0$, Kirchhoff's voltage law gives $L\dot{I}+RI=V_0$. Starting with zero current, $I=(V_0/R)(1-e^{-tR/L})$, with time constant $L/R$. This makes the resistance to rapid current changes directly observable. Two coils also have mutual inductance: changing primary current induces secondary electromotive force $-Md{I}_1/dt$ with a polarity determined by the chosen winding orientations, the principle behind a transformer.