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Assume that the transverse field is large enough to saturate through the thickness despite demagnetizing effects, and retain the thin-layer plane-stress and open-circuit assumptions from part(a). The two imposed in-plane strains now becomeTheir sum is $-\Lambda$, whereas it was $\Lambda /2$ for longitudinal saturation. The in-plane stress sum, and hence the induced open-circuit voltage, is multiplied by $-2$:Thus the magnitude is about $24\mathrm{mV}$ with the opposite polarity, under this model. The larger demagnetizing field for an out-of-plane direction means the same external field magnitude might not produce the same degree of saturation. Changes in thickness constraint or strain transfer would also modify the estimate.

Reversing a saturating longitudinal field gives approximately the same strain and the same voltage polarity relative to the original unmagnetized reference. In the isotropic orientational model,It is unchanged under $\mathbf{m}\to -\mathbf{m}$. Both positive and negative longitudinal saturation therefore contract the negative-magnetostriction nickel in the same axial direction. The resulting stress on the piezoelectric layer has the same sign.

During field reversal there can be a hysteretic transient as domains reorganize, and finite circuit leakage can alter the measured waveform. Those effects do not turn the saturated strain law into a response odd in the field.

If removal restores the original random-domain magnetic state, the magnetostrictive strain and induced voltage return toward their original values. Elastic strain transfer then relaxes the stress on the central layer.

A ferromagnetic nickel layer can retain remanent magnetization and an altered domain population, so removing the external field does not guarantee removal of all magnetostriction. The electrical time response also depends on leakage and the input impedance of the voltmeter: an open-circuit strain-induced voltage can decay through a finite resistance even while some strain remains. The final observed voltage is therefore history- and circuit-dependent.

Write $w_{xy}=w_{yx}>0$ for an edge conductance and $w_x={\sum }_{y\sim x}{w}_{xy}$. Take $s\ne t$, as implicit in a unit source-to-sink electrical network. An oriented current flow is antisymmetric, $i_{yx}=-i_{xy}$. Kirchhoff's first law is conservation of current: ${\sum }_y{i}_{xy}=0$ at every vertex other than the source and sink, with the divergences there equal to the prescribed total flow and its negative. Kirchhoff's second law says that the sum of oriented voltage drops around any closed cycle is zero. Ohm's law isThus the resistance of an edge is $1/w_{xy}$, and the cycle law reads ${\sum }_{j=0}^{m-1}{i}_{x_j{x}_{j+1}}/w_{x_j{x}_{j+1}}=0$ for $x_m=x_0$.

Let $\tau =\inf \{n\ge 0:X_n=t\}$. Its expectation is finite. Indeed, from each state there is a path to $t$ of length at most $|V|-1$ with positive probability; finiteness gives a common lower bound $a>0$ on reaching $t$ within that many steps. The Markov property then gives $\mathbb{P}(\tau >k(|V|-1))\le (1-a{)}^k$, proving the assertion. Define the expected occupation count before absorptionConditional expectation at every pre-stopping visit to $x$ gives ${\mathbb{E}}_s{N}_{xy}=G(s,x)p_{xy}$ for the number $N_{xy}$ of directed traversals. Hence the expected signed count isThis proves Ohm's law with voltage $v_x=G(s,x)/w_x$, grounded at $v_t=0$. Summing its voltage drops along a cycle telescopes, proving Kirchhoff's second law.

For Kirchhoff's first law, count entrances and exits on each stopped path. Their net difference at $x$ is the initial indicator minus the final indicator. Taking expectations,Thus the source emits exactly one unit, the sink receives one unit, and all other vertices conserve current.

We also establish the uniqueness required for the conclusion. The cycle law implies that edge voltage drops are differences of vertex potentials: define a potential by summing drops along a path to $t$, and cycle sums make this independent of the path. If two unit flows obey both laws, their difference has the form $j_{xy}=w_{xy}(h_x-h_y)$ with zero divergence at every vertex. ThereforeStrict positivity of the conductances and connectedness force $h$ to be constant, so $j=0$. We conclude the electrical current from stopped random-walk traversals identity

Fix an edge $e=\{u,v\}$ and orient it from $u$ to $v$. For each spanning tree $T$, let ${\gamma }_T$ be its unique graph path from $u$ to $v$. Assign a signed unit graph path flow $j_T$ to oriented edges: value $+1$ in the direction used by ${\gamma }_T$, value $-1$ on the reversed orientation, and zero otherwise. Define its meanEach $j_T$ has net divergence one at $u$, minus one at $v$ and zero elsewhere, so $J$ is a unit flow. The mean spanning-tree path current also satisfies the Kirchhoff cycle law. Although its verification is permitted to be omitted, a short counting argument makes it explicit. Let ${\mathcal{F}}_{uv}$ be the two-component spanning forests separating $u$ and $v$, write $A_F$ for the component containing $u$, and let $N_T$ be the number of spanning trees. Deleting an edge used by the oriented tree graph path gives one such forest. Conversely adjoining an edge across its two components gives a unique tree with that edge on its $u$-to-$v$ graph path. ConsequentlyThe gradient form proves the Kirchhoff cycle law directly and is Ohm's law for unit edge resistance. The Kirchhoff node law then gives $Lh={\delta }_u-{\delta }_v$, and the electrical unit flow is unique: the difference between two such voltages is a harmonic function on a graph and hence constant on the connected finite graph.

If $e\in T$, its single-edge route is the unique tree graph path from $u$ to $v$, so $j_T(u,v)=1$. If $e\notin T$, that tree graph path does not use $e$, so $j_T(u,v)=0$. ThereforeBy Ohm's law, $J(u,v)=h(u)-h(v)$; for a unit terminal current, this voltage difference is exactly the effective resistance between the terminals. We have proved the requested edge-inclusion formula for a uniform spanning tree:Now use linearity of expectation and the fact that every spanning tree has exactly $n-1$ edges. This gives Foster's theorem:The sums here use unoriented edges exactly once. For unit conductance this is the displayed identity; with conductances $c_e$ the inclusion probability and summand become $c_e{R}_{\mathrm{eff}}(e)$.

Use simple random walk on the loopless undirected unweighted graph, with transition probability $1/\mathrm{deg}(x)$ to each graph neighbour, and take $a\ne z$. Set $T_z=\inf \{t\ge 0:X_t=z\}$. Returns to $a$ before visiting $z$ do not end the commute. After its first visit to $z$, stop at the subsequent first visit to $a$, so $\tau =T_z+T_a\circ {\vartheta }_{T_z}$. Count transitions with departure times $0\le t<\tau$, including the final step arriving at $a$. The expected hitting times are finite: in a finite connected graph there is a uniformly positive probability to reach any fixed target within a fixed number of steps, giving a geometric tail bound.

We prove the needed killed-walk occupation voltage identity from its visit balance equation. DefineFor $x\ne z$, counting arrivals before absorption givesHence, for the graph Laplacian $Lv(x)=\mathrm{deg}(x)v(x)-{\sum }_{y\sim x}v(y)$, we have $Lv(x)=1_{\{x=a\}}$ away from $z$. Since the sum of all Laplacian coordinates is zero, $Lv(z)=-1$. Thus $v$ is the electrical voltage of a unit flow from $a$ to $z$, grounded at $z$, and $v(a)=R_{\mathrm{eff}}(a,z)$.

Construct $w(x)=g_{za}(x)/\mathrm{deg}(x)$ for the opposite killed leg. It satisfies $Lw={\delta }_z-{\delta }_a$ and $w(a)=0$. Therefore $L(v+w)=0$. The harmonic maximum principle on a finite graph makes $v+w$ constant: at a maximum its value equals the average over its graph neighbours, forcing all of them to share that value, and connectedness propagates it. At $a$ the constant equals $R_{\mathrm{eff}}(a,z)$.

For any oriented edge $(x,y)$, each visit to $x$ before the target hit produces a transition to $y$ with conditional probability $1/\mathrm{deg}(x)$. The expected traversals during the first killed leg are therefore $v(x)$, and those during the second leg are $w(x)$ by the Strong Markov property at $T_z$. This proves directed edge occupation in a random-walk commute:In particular, both directions of every unoriented edge have this same expected value, although their counts on an individual commute need not coincide. Summing over the $2|E|$ directed edges also recovers the commute time identity ${\mathbb{E}}_a\tau =2|E|R_{\mathrm{eff}}(a,z)$.

The distinct-terminal convention is necessary for the printed “returns” interpretation. If $a=z$ and $\tau$ is the first positive return, a graph with two graph vertices and one edge gives $S(a,y)=1$ while $R_{\mathrm{eff}}(a,a)=0$. Thus either take $a\ne z$, as intended, or explicitly define the degenerate commute as $\tau =0$. A general nonuniform random walk is not covered by the unweighted simple-walk formula.

A voltage with boundary values $u(a)$ and $u(b)$ is a function $u:V\to \mathbb{R}$ that is harmonic at every other vertex:A current flow is an antisymmetric function $i(x,y)=-i(y,x)$ satisfying Kirchhoff's node law ${\sum }_{y}i(x,y)=0$ away from its source and sink. Voltage and current are related by Ohm's law,The effective resistance is the voltage drop divided by the total current from $a$ to $b$. Equivalently, it is the voltage drop generated by a unit current flow:

The SE annotation identifies a secondary electron image. The specimen has an open network of ridges, thin walls and irregular cavities, with lateral features on the scale of tens to hundreds of nanometres when compared with the 500 nm bar. Exposed edges can emit secondary electrons from more than one nearby surface; inclined faces also have a larger near-surface escape fraction than a face perpendicular to the beam. These mechanisms make protruding rims conspicuous. Recesses can be dark because emitted secondary electrons are intercepted or collected less efficiently. A through-lens detector changes the collection geometry, so a simple one-sided illumination analogy is incomplete.^[ref]

An irregular porous relief with bright rims and darker recesses is consistent with the image. The cross-section below is an original illustrative geometry, rather than a reconstruction of a particular image line. A single secondary electron intensity map does not supply calibrated heights, prove that every dark patch is a through-hole, or distinguish charging from all material contrast. Repeat with specimen tilt, altered detector collection or reduced accelerating voltage, and compare backscattered electron or elemental signals before assigning a compositional cause.

The image annotation gives probe electric current $I=43\mathrm{pA}=4.3\times 10^{-11}\mathrm{A}$ and working distance $L=2.4\mathrm{mm}$. Interpret $d_o$ as the probe diameter and $\alpha$ as its convergence semiangle, in radians. To estimate that angle from the available dimensions, approximate the effective aperture plane as a distance $L$ from the specimen. With aperture radius $r_a=25\mu \mathrm{m}$,The electron beam brightness $\beta$ measures electric current per probe area per solid angle; the supplied convention corresponds to a circular probe and a paraxial cone of solid angle approximately $\pi {\alpha }^2$. The aperture diameter must be halved, and the annotated current is in picoamperes. Treating the full aperture angle as $\alpha$ would introduce a factor-of-two error in $d_o$.

This is a conditional optical estimate: the working distance is not necessarily the actual electron-optical distance from the effective aperture to the focused probe. The answer assumes that approximation, the quoted electron beam brightness at the operating voltage, and negligible optical aberrations. The 30 kV annotation gives a much shorter de Broglie wavelength than 4 nm, so a nanometre-scale probe is plausible. Actual electron microscope resolution also depends on diffraction, scattering in the specimen, signal generation and collection, signal-to-noise ratio, drift and specimen charging. A 4 nm probe does not establish that every measured edge width is 4 nm.

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.

The desired net process transfers copper from an impure anode to a purer cathode. In the idealized common-electrolyte approximation, both interfaces use the same ${\mathrm{Cu}}^{2+}$/Cu equilibrium potential. Setting solid copper activities to one and interfacial ion activities equal gives $E_{\mathrm{rev}}=0$ V: subtracting two equal 0.34 V reduction potentials yields zero, rather than 0.34 V.

A strictly thermodynamic refining voltage need not be exactly zero. At equal dissolved-ion activity, the required reversible applied voltage isDilution of copper in the impure phase can therefore require positive reversible work. Weight percentage alone does not specify that phase's thermodynamic activity or impurity molar composition. The zero result is the intended pure-solid approximation, not a precise chemical-potential measurement from the stated weight purities.

Gas formation is decided by each interfacial electrode potential and reaction kinetics, rather than by comparing the terminal cell voltage directly with 1.23 V. On the conditional model above, the cathode potential is $E_c\simeq 0.34-0.12854=+0.21146$ V versus the standard hydrogen electrode. At unit proton activity and unit hydrogen pressure this is above the hydrogen equilibrium potential, so hydrogen evolution is not favored there. The anode is at $E_a\simeq 1.0188$ V, below the unit-activity oxygen equilibrium potential of 1.23 V, so oxygen evolution is likewise not favored at the intended point. Neither gas is expected in that model, despite a terminal voltage near 2 V; 1.2 V is lost in the electrolyte.

Without the missing cathodic kinetic information, the exact cathode potential is not fixed, so the hydrogen conclusion is conditional. Locally depleted copper ions, increased current, electrode blocking or strong passivation can move an interface into gas evolution. Actual proton/oxygen/hydrogen activities shift equilibrium through the Nernst equation, and gas nucleation and activation barriers further affect observable bubbles. A reference electrode at each interface and direct gas observation distinguish these effects from an increased solution-resistance drop.

Four moles of lithium react per mole of iron disulfide. Using molar masses $M_{\mathrm{Li}}=6.94\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, $M_{\mathrm{Fe}}=55.845\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and $M_{\mathrm{S}}=32.06\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ givesThus the stoichiometric mass ratio is Li:FeS₂ = 1:4.322, or approximately 18.79 wt% lithium and 81.21 wt% iron disulfide within the combined active reactants. This excludes electrolyte, current collectors, casing and any deliberate excess of one reactant. The underlying conversion reaction, rather than the nominal cell voltage, sets that ratio.^[ref]

Each lithium atom supplies one electron. If all of the stated lithium is electrochemically accessible and there is enough iron disulfide, the ideal electric charge isThe voltage is not needed to calculate charge. At an assumed constant 1.5 V the corresponding ideal energy would be $VQ\simeq 9.45\mathrm{kJ}=2.63\mathrm{Wh}$, but an actual discharge voltage varies with load and state of charge. Side reactions, incomplete utilization and the practical cutoff reduce accessible capacity. The preceding optimal ratio would require approximately 1.96 g of iron disulfide for this amount of lithium.

A material exhibiting piezoelectricity converts cyclic mechanical stress into electric charge; its mechanical and electrical boundary conditions jointly set the extractable energy. piezoelectric ceramics such as lead zirconate titanate are stiff and often have large charge coefficients and electromechanical coupling, but are brittle and can be difficult to match to large-strain flexible motion. piezoelectric polymers such as polyvinylidene fluoride are compliant, light and compatible with flexible large-area films. Their lower charge coefficient is partly offset in voltage-sensitive applications by lower permittivity: the voltage coefficient is $g=d/ϵ$.

Ceramics often favor strong charge generation and coupling; polymers favor flexibility and strain accommodation. The larger open-circuit voltage of a low-capacitance film does not itself imply more useful power. For a simple fixed-stress comparison the electrical energy scale involves $d^2/ϵ$, while coupling also depends on mechanical compliance. At fixed imposed strain, different stiffnesses change the available mechanical energy. resonance tuning, fatigue, mechanical mounting, rectifier losses and load matching can reverse a simplistic material ranking; compare complete devices under the same actual vibration source.

For a clean, adiabatically connected GaAs channel at low temperature and small bias, opening successive spin-degenerate subbands givesConductance therefore shows a staircase of plateaux, rather than the continuously geometry-dependent classical Ohmic result. Each fully transmitted mode contributes approximately $77.5\mu \mathrm{S}$. More positive, or less negative, depletion-gate voltage opens a conventional GaAs constriction; more negative voltage closes it.

Require subband spacings larger than thermal and bias broadening and a mean free path exceeding the constriction length. The general Landauer formula is $G=(e^2/h){\sum }_{\alpha }{T}_{\alpha }$ with each spin/valley channel counted separately. A material with additional valley degeneracy need not have the same step sequence as GaAs, and imperfect transmission or contact resistance blurs the ideal plateaux.

One suitable platform is a monolayer molybdenum disulfide semiconductor. An electrostatic accumulation gate supplies conduction electrons whose motion is confined to the atomic sheet; source and drain contacts act as reservoirs. A clean insulating substrate and dielectric or encapsulation reduce disorder and prevent gate leakage. Two split gates deplete the adjacent sheet and leave a narrow conducting gap, thereby confining the in-plane transverse direction while retaining transport along the gap.

Apply a small source-drain voltage, measure the current and determine linear-response conductance while tuning the split-gate voltage and carrier density. A graphene sheet is another two-dimensional carrier system, but its lack of an ordinary gap complicates purely electrostatic confinement; etched boundaries or a suitable gapped architecture may be needed. A two-dimensional material does not by itself guarantee ballistic transport or resolved subbands.^[ref]

In a vibrating-sample magnetometer, a magnetized specimen vibrates sinusoidally in a nominally uniform applied field. Its motion changes the magnetic flux through stationary pickup coils, producing an induced voltage by Faraday's law of induction. The component at the vibration frequency is proportional to the sample magnetic moment times vibration amplitude and frequency, for a calibrated coil geometry.

A lock-in detector referenced to the vibration extracts this small signal. Calibrate with a known moment, stabilize the vibration amplitude and sweep the applied field to measure a magnetization loop. Dividing moment by film volume gives $M$ in $\mathrm{A}{\mathrm{m}}^{-1}$; the silicon substrate and holder background must be subtracted for such a thin film.^[ref]

Relative to conventional crystalline-silicon technology, many organic absorbers offer low-temperature thin-film processing and lightweight flexible devices, but have more strongly bound excitons and more disorder-limited charge transport. These are common material trends, not universal statements about every organic or inorganic device.

Relevant semiconductor properties include the absorption spectrum and optical gap, donor/acceptor energy levels and offsets, dielectric screening and exciton dissociation, electron/hole mobilities, recombination lifetimes, traps and morphological stability. Contacts must extract the appropriate carrier selectively. These properties determine the device quantities inA favourable gap alone does not fix efficiency: current generation, voltage loss and fill factor must all be controlled.

The intended double-barrier resonant-tunnelling diode has a bias-dependent well level. Initially increasing bias can bring a quasi-bound level into alignment with available occupied emitter states, sharply increasing current. At larger bias the resonant level moves out of the effective injection window, for example below the emitter conduction-band edge, so resonant transmission and current decrease despite increasing voltage. This produces a peak-to-valley interval with $dI/dV<0$. Further bias can increase nonresonant or higher-level current again.

The right panel of the preceding figure is a qualitative example. The device can supply small-signal negative differential conductance to an appropriate oscillator circuit while its current and voltage remain positive. A bias-independent transmission function alone does not guarantee NDR: the voltage-dependent potential profile, contacts and carrier supply must be included. The source's reference to part (ii) appears to point to the earlier barrier matching; the resonant device mechanism uses the added barrier of part (iii).