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For an estimate, use the usual equal-device CMOS series-stack model: a $k$-input NAND gate or NOR gate has a worst-case propagation delay approximately $k$ times that of an inverter, because a conducting path contains $k$ series transistors. An AND gate or OR gate adds an output inverter. With a 1 ns inverter this gives, for $k\ge 2$,For $k=2$ it reproduces the stated 3 ns. This is a technology estimate under equal sizing and comparable loading, not a unique physical law inferable from a single measured delay. Actual propagation delay also depends on capacitance, fan-out and transistor sizing. If instead only the supplied two-input gates were allowed, a balanced tree would have worst-case depth $\lceil {\log }_{2}k\rceil$ and delay $3\lceil {\log }_{2}k\rceil$ ns; that is a different construction from a single $k$-input gate.

For the explicit circuit in part (a), the longest combinational logic path starts at $A$, passes through the shared inverter, either direction-match exclusive or, the AND gate, and the final exclusive or feeding $D_2$:The path $A\to D_1$ takes 5 ns, the paths from $B0$ or $B1$ to $D_2$ take at most 7 ns, and $B0\to D_0$ takes 1 ns. Both direction-match exclusive or gates act in parallel; their delays are not added together. The 8 ns excludes clock-to-Q delay, setup time, hold time and clock-distribution effects, because the question asks only for the combinational logic delay.

Patch-clamp series resistance, also called patch-clamp access resistance, is the effective electrical resistance between the pipette's sensed voltage and the cell interior. It includes tip access and any remaining obstruction after break-in. For injected electric current $I$ into the biological cell, an ideal resistive access path givesThus $1\mathrm{nA}$ through $20\mathrm{M\Omega }$ produces a $20\mathrm{mV}$ difference: a substantial possible voltage clamp error. The electric current in this equation is the access-path current, excluding any bypass seal leak current. In current clamp, the same access drop contaminates the pipette voltage during injection. Bridge compensation subtracts an estimated $I{R}_s$; an estimate that changes during recording leaves an error.

An independent passive voltage clamp calculation makes both speed and error visible. Write $v=V-E_L$ and $v_p=V_p-E_L$, with fixed leak reversal potential $E_L$, constant membrane electrical resistance $R_m$ and membrane capacitance $C_m$. Assume an ideal command at the pipette node and omit seal leak current and pipette capacitance. Electric charge conservation givesFor a step $v_p=\Delta V$ from $v(0)=0$, the ordinary differential equation givesWith the figure's parameters and $\Delta V=10\mathrm{mV}$, the time constant is $0.909\mathrm{ms}$, the final membrane rise is $9.091\mathrm{mV}$, and access-path electric current falls from $500$ to $45.45\mathrm{pA}$. Only when $R_m\gg R_s$ does the time constant approach $R_s{C}_m$ and the final voltage approach the command. Actual voltage clamp also depends on amplifier feedback control and the space clamp problem.

Photovoltaic mode uses no external reverse bias: a transimpedance amplifier can hold zero terminal electric potential difference, while an open electrical circuit develops a photovoltage. Photoconductive mode biases the diode cathode positive relative to the diode anode, generally reducing junction capacitance and increasing speed and dark current.^[ref]

A simple calculation separates the readout from the conversion. If a linear transimpedance amplifier has feedback electrical resistance $R_f=1.00\mathrm{M\Omega }$, the $650\mathrm{nm}$ model above produces an output change of magnitude $R_f{I}_{\mathrm{ph}}=0.393195\mathrm{V}$. The output sign depends on the connection. Increasing $R_f$ increases that electric potential difference, but does not create more collected charge carriers.

For a reverse-biased photodiode feeding an electrical load $R_L$, Ohm's law gives a readout magnitude $V_L\simeq I_{\mathrm{ph}}{R}_L$ while the response remains linear. With negligible dark current and other series drops, the remaining reverse bias is $V_R\simeq V_b-I_{\mathrm{ph}}{R}_L$, where $V_b$ is the supply magnitude. Thus the chosen electrical load can consume the available bias: a larger readout does not automatically reduce measurement uncertainty. For the same constructed example, $V_b=5\mathrm{V}$ and $R_L=1\mathrm{M\Omega }$ leave about $4.60680\mathrm{V}$; this algebra alone does not establish a real device's linear range.

A polymer exhibiting piezoelectricity. A compliant, low-density film can accept larger bending or strain than a brittle ceramic, while low permittivity can yield a large voltage coefficient despite a modest charge coefficient. Extractable energy still depends on mechanical input, capacitance and electrical loading.

A transducer using stress-induced electrical charge. A uniformly loaded slab of thickness $t$ has ideal open-circuit voltage sensitivity $dt/({ϵ}_0\kappa )$; leakage and input capacitance change its measured response.

For circular plates of radius $a$, separation $d\ll a$, and a thin axial resistor $R$, the quasistatic capacitance is $C={\epsilon }_0\pi a^2/d$. During discharge, $I=V/R=-C\dot{V}$. A circular Ampere-Maxwell law contour between the plates encloses wire current $I$ and opposing displacement current $-I{r}^2/a^2$, giving $H=I(1-r^2/a^2)/(2\pi r)$. The electric field has magnitude $V/d$, so the Poynting vector points radially inward with magnitude $V^2(1-r^2/a^2)/(2\pi rdR)$. Its flux through the cylindrical side is the displayed inward power. The field inside that cylinder releases $V^2{r}^2/(R{a}^2)$ per unit time; adding the incoming flux gives the resistor's full Joule heating rate $V^2/R$. End fringing, wire radius and magnetic-energy corrections are neglected.

The electronic contribution to differential capacitance arising from a change of chemical potential as particles are added. For noninteracting electrons at zero temperature, it is $e^2$ times the density of states of the whole sample at the Fermi energy. The quantum capacitance of a graphene cone gives a worked example.

The quantum capacitance per unit area is $c_q=e^{2}D(E_F)$. Differentiating the gate equation givesThus the available electronic states act as a second contribution in series with geometric capacitance. At exact neutrality the ideal zero-temperature model has $c_q=c_{\mathrm{diff}}=0$; actual finite-temperature or spatially nonuniform samples need a broader description. A capacitance measurement can therefore probe the density of states, rather than geometry alone.

An RC low-pass filter places a resistor $R>0$ between input and output, with a capacitor $C>0$ from output to their common return. The ideal model prescribes the input electric potential difference and leaves the output unloaded. It is a first-order low-pass filter with unit eventual gain.^[ref]

For electrical resistance $R$ and capacitance $C$, Ohm's law gives the electric current through the resistor as $(v_{\mathrm{in}}-v_{\mathrm{out}})/R$. The capacitor relation gives $Cd{v}_{\mathrm{out}}/dt$. Equating them by Kirchhoff's current law yields the ordinary differential equationThe time constant $\tau$ has units of time. A rapid change in electric potential difference requires a large electric current into the capacitor; the finite electrical resistance limits that rate. A slowly maintained voltage instead lets the capacitor approach the input voltage. The same equation therefore predicts both smoothing of sustained oscillations and a rounded response to a sudden change.

For the original constructed example throughout this article, takeThese stipulated ideal values specify an analytical model, rather than measurements of particular components.

The University of Colorado Boulder's Lab 3: Filters, §§7.1–7.4, specifies measuring the resistor and capacitor, then driving the assembled RC low-pass filter with a function generator. A two-channel oscilloscope reads actual input and output using high electrical impedance. Sweep the frequency, record the voltage amplitude ratio, and locate $1/\sqrt{2}$. A control compares a cable with a $10\times$ oscilloscope probe on a resistive voltage divider, isolating measurement capacitance before using the probe for the filter. A slow square wave then lets discharge timing estimate the time constant independently. Agreement supports the RC low-pass frequency response; source output impedance, probe loading, and instrument signal bandwidth limit it. These are prescribed measurements, with no measured dataset supplied here.^[ref]

The two-channel setup can also test the predicted phase difference directly: for positive frequency $f$, measure the output's time lag $\Delta t$ between corresponding features of the two sinusoids. For this single-stage model the phase lies between $-\pi /2$ and $0$, so choose the lag within a quarter cycle and calculate $\varphi =-2\pi f\Delta t$. At the cutoff frequency the predicted lag is one eighth of a cycle. This tests the timing part of the frequency response as well as the voltage ratio.

A real electrical load can draw electric current from the output. For a load resistor $R_L>0$ across the capacitor, Kirchhoff's current law instead givesTherefore the loaded continuous-time frequency response isLoading reduces the eventual gain and shortens the time constant. With the constructed values and $R_L=10\mathrm{k}\Omega$, $K_L\approx 0.81967$, ${\tau }_L\approx 84.754\mu \mathrm{s}$, and the corner relative to the loaded low-frequency gain moves to $1.8778\mathrm{kHz}$. A $3\mathrm{V}$ step now approaches $2.4590\mathrm{V}$. A quicker rise alone would conceal the error in the final voltage.

With the ideal input voltage source set to zero, the output impedance seen by the electrical load isThis follows from Thévenin's theorem: the resistor and capacitor are then in parallel. A sufficient small-loading condition is $|Z_L(\omega )|\gg |Z_{\mathrm{out}}(\omega )|$ over the frequencies of interest. An oscilloscope is itself an electrical load: its input electrical resistance and cable or probe capacitance can both change the response. Pure extra parallel capacitance increases $C$ and lowers the corner, whereas a finite parallel electrical resistance gives the different gain-and-corner changes above.

The definition of input also matters. If voltage is prescribed inside a source with series electrical resistance $R_s$, the unloaded response to that internal source uses $R+R_s$. If the actual voltage at the filter input terminal is measured, the ratio of output to that measured input still uses $R$. Measuring both terminals avoids mistaking a source setting for the voltage delivered to the filter.

Connect one physical resistor $R$ across an ideal lossless capacitor $C$, with no applied source and an ideal readout that draws no electric current. Represent the resistor by its noisy Thévenin equivalent. The resulting RC low-pass frequency response gives the stationary capacitor voltage power spectral densityIts mean square follows by the effective noise bandwidth calculation:This is also the equipartition theorem: the mean capacitor energy is $\frac{1}{2}C⟨v_C^2⟩=\frac{1}{2}{k}_{\mathrm{B}}T$. Raising $R$ makes the low-frequency spectrum taller and its width narrower in exactly compensating proportions. It changes the equilibration time $RC$, not the final variance.

The integral to infinity belongs to the ideal classical model. For a physical electrical circuit, assume thermal equilibrium, linear components with negligible extra dissipation, and a corner far below the frequencies where parasitic effects, frequency-dependent electrical resistance, or $hf\sim k_{\mathrm{B}}T$ invalidate the model; the omitted high-frequency tail must be negligible. A finite-bandwidth instrument measures only part of this variance. Its input capacitance contributes to the total $C$, and its finite input electrical resistance introduces both loading and its own thermal electrical noise.

For $T=300\mathrm{K}$ and $C=1\mathrm{nF}$, the predicted equilibrium variance is $4.141947\times 10^{-12}{\mathrm{V}}^2$ and the root mean square voltage is $2.0352\mu \mathrm{V}$. At $R=10\mathrm{k}\Omega$, $RC=10\mu \mathrm{s}$ and $B_{\mathrm{n}}=25\mathrm{kHz}$. At $R=100\mathrm{k}\Omega$, $RC=100\mu \mathrm{s}$ and $B_{\mathrm{n}}=2.5\mathrm{kHz}$; the equilibrium root mean square voltage is identical.

To obtain the growth curve, prepare $v_C(0)=0$ exactly and connect the resistor at $t=0$ while maintaining its temperature. The RC low-pass filter's causal impulse response is $h(t)=e^{-t/(RC)}/(RC)$ for $t\ge 0$. Classical ideal white noise with two-sided power spectral density $2k_{\mathrm{B}}TR$ givesThe factor $2k_{\mathrm{B}}TR$ here is the two-sided time-domain noise intensity; the equilibrium spectral integral above used the one-sided $4k_{\mathrm{B}}TR$. They give the same long-time limit. The variance approaches equilibrium with time constant $RC/2$, although a deterministic initial voltage decays with time constant $RC$. If the capacitor already starts in thermal equilibrium, its initial variance supplies the missing exponentially decaying term and the total variance stays constant.

A reversal potential $E_{\mathrm{rev}}$ is a membrane potential at which a specified open ion channel, or specified combination of ion channels, carries zero net electric current, with the surrounding conditions fixed; the electric current changes sign across that voltage. Specifying which current matters: an individual ion contribution and a sum of several contributions can reverse at different voltages. Zero net electric current does not mean that all ions have stopped moving. See Purves et al., Neuroscience, 2nd edition (2001): reversal of a mixed-ion channel current.

Throughout these examples, membrane potential is $V=V_{\mathrm{inside}}-V_{\mathrm{outside}}$, and conventional current is inward-positive. Positive electric current corresponds to net positive electric charge entering or net negative electric charge leaving. Negative electric current corresponds to the reverse charge transfer. A convenient Ohmic approximation iswhere $g$ is the available electrical conductance. For fixed $g>0$, $V<E_{\mathrm{rev}}$ gives positive electric current, $V=E_{\mathrm{rev}}$ gives zero, and $V>E_{\mathrm{rev}}$ gives negative electric current; the line has slope $-g$. If $g=0$, the modeled pathway carries zero electric current at every voltage: closing it does not locate its reversal potential. 3.1 Synapses (§3.1: conductance-based synaptic current) uses outward-positive $g(V-E)$; the formula here reverses that sign to match this wiki's inward-positive convention.

For the stipulated single-pathway example from ion channel, $E_{\mathrm{rev}}=-80\mathrm{mV}$ and $g=10\mathrm{nS}$ giveThe open ion channel carries zero net electric current at $-80\mathrm{mV}$ even though $g$ remains positive. Near that point a change of membrane potential changes the electric current, which is why zero current at one voltage does not make an open pathway electrically irrelevant; see shunting inhibition.

For an ion channel passing only one ion species by passive transport, its reversal potential can equal that species' Nernst equilibrium potential, where the chemical and electrical contributions to electrochemical potential balance. At that thermodynamic equilibrium, passage in both directions can continue with no net transport. For an ion channel admitting several ion species, opposing contributions can cancel while each species continues to have net transport. Such a reversal potential is not generally any one species' Nernst equilibrium potential. See the Purves reference above.

To work out that cancellation explicitly, put two Ohmic ion channel populations across the same cell membrane. Stipulate constant electrical conductances and fixed reversal potentials:The potassium and sodium contributions add because both pathways experience the same membrane potential:Thus $I_{\mathrm{total}}=12\mathrm{nS}(E_{\mathrm{eff}}-V)$. This weighted mean follows simply by expanding and regrouping the two current terms; it assumes additive Ohmic contributions and a positive total electrical conductance. At $V=E_{\mathrm{eff}}$, the inward-positive individual electric currents areThe potassium contribution is outward and the sodium contribution inward. Equal and opposite electric currents give $I_{\mathrm{total}}=0$; neither pathway has stopped transporting its ions. The calculation treats two distinct pathways. Applying it to one mixed-species ion channel would additionally require a justified decomposition into independent Ohmic contributions; a general mixed-channel reversal potential cannot be obtained by assuming those hypotheses silently.

This is an instantaneous balance of the specified passive electric currents. It would be the stationary voltage of a passive capacitance model containing only these constant pathways and no applied electric current, but it is not a complete prediction of a living cell's resting membrane potential. Other electric currents, voltage-dependent gating, and pumps involved in active transport can matter. Pumps that transfer net electric charge are electrogenic and contribute their own electric current. See Purves et al.: functional properties of the sodium/potassium pump.

The toy model holds ion concentrations and therefore the stipulated reversal potentials fixed. Continuing inward sodium and outward potassium transport would change finite reservoirs unless active transport or external control maintained them. Transport can be a small fraction of the reservoir's ions over a short interval, making fixed concentrations a useful approximation; it is not a claim that the reservoirs never change. See Gerstner et al., Neuronal Dynamics (2014), §2.1: equilibrium potentials, channel currents and maintained gradients.

A reversal potential makes the effect of opening an ion channel predictable from the present membrane potential. The same open pathway can carry inward or outward electric current as $V$ crosses its reversal potential. The two-pathway example reveals a particularly useful distinction: zero net charge transfer can coexist with continuing exchange of different substances. That distinction connects membrane transport to electric current measurements without mistaking electrical balance for thermodynamic equilibrium.

An RLC circuit combines electrical resistance $R$, inductance $L$, and capacitance $C$. The capacitor stores energy in its electric field; the inductor stores energy in its magnetic field. Their exchange can produce an oscillation, while Joule heating removes energy. A series RLC circuit gives a particularly clear example: changing $R$ turns ringing into a direct return to equilibrium, and changing $L$ or $C$ tunes a resonance.

These circuits connect two useful viewpoints. In time, they describe how stored energy is released after switching. In frequency, they select a range of signals: the resonant series RLC response explains radio tuning and simple filters. The same mathematics describes a damped harmonic oscillator, so insight about mechanical ringing transfers directly to an electrical circuit. ^[ref]; ^[ref].

Scanning-probe mapping of local capacitance response. Conventional semiconductor dopant profiling measures a bias-modulated $dC/dV$ signal involving an insulating layer and a changing semiconductor depletion region; an electrically isolated conducting flake on glass is not automatically that device geometry.

Consider a stipulated series circuit containing an ideal resistor of electrical resistance $R=100\Omega$ and an ideal capacitor of capacitance $C=1\mu \mathrm{F}$. The same electric current passes through both. Their terminal electric potential differences add by Kirchhoff's voltage law, soA quick derivation uses $i=Cd{v}_C/dt$: taking the derivative of the complex exponential multiplies the capacitor voltage phasor by $\mathrm{i}\omega$, hence ${\hat{V}}_C=\hat{I}/(\mathrm{i}\omega C)$. Dimensionally, $1/(\omega C)$ has units $s/F=V/A=\Omega$, matching $R$.

At frequency $f=1000\mathrm{Hz}$, the angular frequency is $\omega =2\pi f$. The signed capacitive electrical reactance, modulus and complex argument of the total electrical impedance areApply $V_{\mathrm{rms}}=1.00\mathrm{V}$ across the complete series circuit. Its root mean square electric current isWith voltage as the zero-phase reference, the current phasor has complex argument $+57.858092^{\circ }$: electric current leads. The phasor drops have moduli $V_{R,\mathrm{rms}}=0.532018\mathrm{V}$ and $V_{C,\mathrm{rms}}=0.846733\mathrm{V}$; these add as complex numbers, not as positive magnitudes. Their squared magnitudes sum to $(1.00\mathrm{V}{)}^2$.

The mean absorbed power uses the root mean square values:The ideal capacitor has zero cycle-averaged absorbed power; all the mean absorbed power becomes Joule heating in the resistor.^[ref] To see why the phase factor appears, average $p(t)=v(t)i(t)$ over a full cycle with $v(t)=\sqrt{2}{V}_{\mathrm{rms}}\cos \omega t$ and $i(t)=\sqrt{2}{I}_{\mathrm{rms}}\cos (\omega t-\varphi )$. The cosine addition formula reduces the average to $V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi$; the term at $2\omega$ averages to zero. Independently, $p_C=d(C{v}_C^2/2)/dt$, and the capacitor energy returns to its starting value after each cycle.

At fixed applied $V_{\mathrm{rms}}$, the limit $\omega \to 0^{+}$ gives $|Z|\to \infty$, $\varphi \to -\pi /2$ and $I_{\mathrm{rms}}\to 0$: the ideal capacitor blocks steady electric current. The limit $\omega \to \infty$ gives $Z\to R$, $\varphi \to 0$ and $I_{\mathrm{rms}}\to V_{\mathrm{rms}}/R$. These are limits of the ideal lumped electrical circuit; unwanted lead inductance and component losses can invalidate that model at high frequency. At $\omega RC=1$, the equal legs of the impedance right triangle give $|Z|=\sqrt{2}R$ and $\varphi =-\pi /4$ without numerical evaluation. Measuring both the magnitude and the sign of the phase difference distinguishes this capacitor response from an equally large positive electrical reactance.

MIT's guide uses an illuminated vacuum photodiode, with collecting anode positive relative to the emitting cathode. A transimpedance amplifier gives $V_1=R_{F}I$; a multimeter reads its mean. The fluctuation path rejects below $100\mathrm{Hz}$, adds roughly tenfold voltage gain, then nominal hundredfold preamplification and a $1$–$50\mathrm{kHz}$ band-pass filter; the readout measures root mean square voltage. With diode bias off, a sinusoid at the test port calibrates $g(f)$. Restore bias, retain gain settings, vary lamp brightness, and repeat both readings. The guide warns about overload; it supplies a procedure, not measured results.^[ref]

The inference can be expressed without relying on nominal passband edges. Let $g(f)$ denote the calibrated voltage transfer after the $R_F$ conversion, so $K(f)=R_{F}g(f)$ and define $\mathcal{K}={\int }_0^{\infty }|K(f){|}^{2}df$. Under independent arrivals and an unchanged additive background,The slope of voltage variance against mean electric current can therefore estimate $q$. A free intercept accounts for a constant background; forcing the line through zero generally biases the slope. The root mean square readings must have their mean removed and then be squared. Subtracting two root mean square amplitudes does not subtract independent variances. Estimate the variance from successive records; averaging the waveforms first would suppress the fluctuation being measured.

Additional controls sharpen this interpretation. Block the light with detector bias, temperature, and gain unchanged; this preserves dark current and electronics while removing photocurrent. Repeat this dark comparison between illumination settings to check drift. If independent dark arrivals have mean magnitude $I_d$ and illuminated arrivals add $I_{\mathrm{ph}}$, their spectra add: $2q(I_d+I_{\mathrm{ph}})$. Subtracting the unchanged dark variance isolates $2q{I}_{\mathrm{ph}}\mathcal{K}$. This subtraction model assumes unchanged transfers and independence; switching off bias for gain calibration is a different control and need not preserve the dark background.

Vary measured $I_{\mathrm{ph}}$ rather than assuming lamp knob position is proportional to photocurrent. Check collection versus bias to identify a range where incomplete collection is negligible; a space charge limited tube can correlate transfers. Verify that the transimpedance amplifier remains linear and that its frequency response is unchanged over the illumination range. Changing signal bandwidth should change the excess variance through $\mathcal{K}$; compare inferred input spectra in several bands. Correlated fluctuations of the light source can add a contribution scaling approximately as $I_{\mathrm{ph}}^2$, while an electronics floor can hide the linear term at small electric currents. A fitted line over a short range alone does not establish independent transfers.

The gain calibration must include the same detector-to-readout path used for noise, or explicitly supply any omitted transfer. Detector capacitance, finite collection time, amplifier voltage and current noise, thermal electrical noise, pickup, clipping, and finite recording precision limit the inference. Current-dependent electrical impedance or heating can make a background current-dependent too. Since $\mathcal{K}$ contains squared gain, a small fractional error in gain produces approximately twice that fractional error in the predicted variance scale. Knowing a nominal gain and two cutoff settings is insufficient.

Shunting inhibition is suppression of another input's effect through increased synaptic conductance: opening additional ion channels lowers effective electrical resistance, reducing the voltage response to a given added electric current. When the added reversal potential equals the resting membrane potential, there can be no voltage change at rest and still be inhibition of a simultaneous input. See 1.4 Limitations of the Leaky Integrate-and-Fire Model.

Use the passive synaptic potential model below with leak $g_L=10\mathrm{nS}$, $E_L=-65\mathrm{mV}$ and capacitance $C=200\mathrm{pF}$. A sustained inward test electric current of $I_{\mathrm{ext}}=100\mathrm{pA}$ alone gives a steady voltage increase of $I_{\mathrm{ext}}/g_L=10\mathrm{mV}$, so $V_{\infty }=-55\mathrm{mV}$. Effective electrical resistance is $100\mathrm{M\Omega }$ and the time constant is $20\mathrm{ms}$.

Now hold a shunting conductance $g_i=20\mathrm{nS}$ open with $E_i=E_L=-65\mathrm{mV}$. With no test electric current, rest stays at $-65\mathrm{mV}$. With the same $100\mathrm{pA}$ test electric current, total conductance becomes $30\mathrm{nS}$, so the steady voltage increase is only $100/30\simeq 3.333\mathrm{mV}$: $V_{\infty }\simeq -61.667\mathrm{mV}$. Effective electrical resistance is $33.333\mathrm{M\Omega }$ and the time constant is $200/30\simeq 6.667\mathrm{ms}$. A faster response can therefore be smaller. At the original rest the shunting electric current is zero; at $V=-55\mathrm{mV}$ it is $20(-65+55)=-200\mathrm{pA}$, opposing the voltage excursion.

The factor of three reduction applies to this constant-electric current, constant-conductance example. A finite pulse, conductance-driven input or spatially extended dendrite need not have the same reduction. The interesting feature is conditional action: the shunt can leave a quiet neuron's voltage unchanged yet reshape its response when another input arrives. A model that represents all inhibition as a fixed outward electric current misses this behavior.

MIT's documented slotted-line waveguide measurement shorts one end of a rectangular electromagnetic waveguide and excites it through a probe coupled by capacitance. A second such probe moves through a longitudinal slot, aligned with the wall electric current to reduce disturbance. Its output locates axial standing wave nulls. The book's standing-wave setup chooses a frequency above the first waveguide cutoff frequency and below higher-mode cutoffs. Lowering it below the first cutoff instead reveals spatial decay away from the source. The text supplies apparatus and an observable comparison, rather than a table of measured null positions.^[ref]

The inference can be checked without assuming that detector output is proportional to electric field amplitude. At fixed measured frequency, record several minima at coordinates $z_0,\dots ,z_N$; then estimate ${\lambda }_g=2(z_N-z_0)/N$ when the minima are consecutive. This uses positions alone. Compare the result with the guided wavelength prediction using independently measured $a$ and $f$. The $40\mathrm{mm}$ spacing above is a synthetic example, not a reading from the MIT apparatus.

Additional proposed controls are to keep probe insertion fixed, repeat the scan in both directions, recheck source frequency, and compare shallow probe settings for shifts in the minima. These test probe loading, mechanical backlash and drift. A deliberately reflecting termination supplies the standing wave; varying its position should translate the pattern without changing adjacent-null spacing in a uniform single-mode region. Source-adjacent evanescent waves, unintended reflections, additional propagating normal modes, detector noise and finite positional resolution can spoil that inference. A disappearing remote signal alone does not distinguish below-cutoff decay from loss or poor excitation, and a scan at one frequency does not measure a pulse's group velocity.

Capacitance normalized by a stated mass or volume. Mass-specific differential capacitance is $C_{\mathrm{sp}}=m^{-1}dQ/dV$; electrode and full-device mass conventions must be distinguished.