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A connection that couples a measured electric potential difference into an oscilloscope. Its electrical resistance and capacitance contribute to instrument loading.

A connection in which branches share both terminal nodes and therefore the same electric potential difference. Branch electric currents add. For sinusoidal signals, the total electrical admittance is the sum of branch electrical admittances; ideal capacitances add.

Two broad perfect conductors of width $w$, separated by gap $d\ll w$, carry opposite steady electric currents $I$ and connect through a terminating resistor $R$. Away from edges, Ampère's circuital law and the superposition principle give gap magnetic field magnitude ${\mu }_{0}I/w$. Each conductor is an equipotential, and Ohm's law sets their electric potential difference to $IR$. Planar symmetry and charge-free Gauss's law then give uniform gap electric field magnitude $IR/d$. Injection parallel to the upper current makes the electric and magnetic Lorentz forces opposite. Their cancellation selects speed $Rw/({\mu }_{0}d)$ independent of nonzero electric charge and of operating current; nonrelativistic kinetic energy is $m[Rw/({\mu }_{0}d){]}^2/2$.

Consider an original electrical circuit model between two equipotential cross-sections. A normal path has electrical resistance $R_n>0$. Let $b\in \{0,1\}$ indicate whether an additional path connects those same cross-sections; when connected it has electrical resistance $r\ge 0$. Ideal voltage contacts measure those cross-sections. This collapses a spatial network to two parallel branches; it is not a fit to the Bednorz-Müller experiment.

For $r>0$, both branches have the same electric potential difference $V$. Ohm's law and the Kirchhoff node law give $I=V/R_n+bV/r$, henceThe last case follows directly from the ideal short's constraint $V=0$, provided it can carry the imposed electric current below its superconducting critical current. It does not follow by substituting $r=0$ into $V/r$. A normal bridge in series with otherwise superconducting pieces belongs in $r$; superconducting grains alone need not short the measured interval.

Choose $R_n=1\Omega$. A connected bridge with $r=0.01\Omega$ gives $R_{\mathrm{eff}}=1/101\Omega \simeq 9.901\mathrm{m\Omega }$, almost a hundredfold drop despite finite electrical resistance. An ideal connected superconducting filament instead gives zero in this model even when its volume is a small part of the specimen. Isolated superconducting islands can have the same superconducting phase fraction $f$ while leaving $b=0$. For the illustrative disconnected geometry we stipulate that the normal conducting network retains $R_n$.

To plot a chosen connectivity sequence, set $b(f)=0$ for $f<f_c$ and $b(f)=1$ for $f\ge f_c$, with $r=0$ on the connected branch. The two displayed values of $f_c$ describe different chosen architectures. A real percolation calculation would need an explicit network and spatial distribution; the scalar phase fraction $f$ does not determine $b$ by itself.

An instrument with unresolved electric potential difference bounded by $\delta V$ reports only $R_{\mathrm{eff}}\lesssim \delta V/|I|$. For chosen $I=1\mathrm{mA}$ and $\delta V=100\mathrm{nV}$ this floor is $10^{-4}\Omega$, assuming offsets and noise are included in the bound. The figure clips smaller model values to that floor. Increasing $|I|$ improves this formal bound only while Joule heating and the path's superconducting critical current remain controlled. None of these transport statements determines the superconducting phase fraction or establishes complete Meissner effect shielding.

A parallel-plate capacitor consists of two parallel electrical conductors with overlapping area $A$ and separation $d$. If the gap is small compared with both plate widths and is filled by a uniform material of positive permittivity $ϵ$, its capacitance is approximatelyThis neglects the fringing field of a capacitor near the plate edges. The interior electric field points from the plate of positive electric charge to the plate of negative electric charge, and is approximately uniform. ^[ref].

For a worked vacuum example, stipulate $A=0.01{\mathrm{m}}^2$, $d=0.001\mathrm{m}$, and vacuum permittivity ${ϵ}_0\approx 8.854\times 10^{-12}F/m$. This rounds the value in Fundamental Physical Constants — Complete Listing by National Institute of Standards and Technology and others. Within the approximation,At terminal electric potential difference $V=10\mathrm{V}$, the electric field magnitude is $E\simeq V/d=10^{4}V/m$, the plate electric charge magnitude is $Q=CV=0.8854\mathrm{nC}$, and the capacitor energy is $U=C{V}^2/2=4.427\mathrm{nJ}$. These numbers use the stipulated rounded vacuum permittivity; finite plates also have corrections from the fringing field of a capacitor.

The formula makes design changes easy to predict: doubling the overlapping area doubles the capacitance, and halving the gap doubles the capacitance, with permittivity fixed and the fringing field of a capacitor neglected. Doubling permittivity also doubles capacitance if the material uniformly fills the gap and remains linear. This lets a compact device store more separated electric charge at the same terminal electric potential difference.

The short derivation uses surface charge density $\sigma =Q/A$. In the interior, Gauss's law gives electric field magnitude $E\simeq \sigma /ϵ=Q/(ϵA)$. The electric potential difference is $V\simeq Ed=Qd/(ϵA)$, so dividing $Q$ by $V$ gives $C\simeq ϵA/d$. A narrow curved capacitor gap has the same leading local approximation when its width is much less than the curvature radius.

A long coaxial cylindrical capacitor of length $L$ and radii $a,b$ has a dielectric-filled overlap of length $x$ and relative permittivity ${ϵ}_r$. Neglecting end and interface fringing, its filled and vacuum portions are in parallel because both share the same two conductors. The electric potential difference $V$ produces $E_r=V/[r\ln (b/a)]$ in either portion; the electric displacement field and electrode charge differ by the local permittivity. Adding their capacitances gives the title formula. The dielectric boundary is axial while the interior electric field is radial and tangential to that interface, consistent with the electrostatic interface boundary conditions.

A particle accelerator increases the kinetic energy of charged particles through the work of an electric field. A single electric potential difference $V$ gives an energy gain of magnitude $|q|V$ to a particle of electric charge $q$ crossing in the accelerating direction. Reusing a smaller electric potential difference many times can give a much larger final kinetic energy without requiring the whole apparatus to withstand the equivalent single voltage. A magnetic field can return the particles to the accelerating region: magnetic forces do no work, but they make repeated energy gains possible. The cyclotron turns this separation between bending and acceleration into a particularly elegant machine.

The passive sign convention defines terminal electric potential difference $V=\varphi (A)-\varphi (B)$ and positive electric current entering at $A$. Then $VI$ is absorbed electrical power: positive means the component receives energy, negative means it supplies energy. Reversing both reference directions leaves the product unchanged. For a positive electrical resistance obeying Ohm's law, both $V$ and $I$ reverse together and Joule heating remains $I^{2}R\ge 0$.

Since ${\mathbf{e}}_{\rho }\times {\mathbf{e}}_{\phi }={\mathbf{e}}_z$, the Poynting vector is axial. The magnetic permeability cancels from $\mathbf{E}\times \mathbf{B}/\mu$ in each material layer:The signed Poynting flux through the plane with unit normal $+{\mathbf{e}}_z$ is its surface integral with area element $\rho d\rho d\phi$:ThusThe transmitted electromagnetic energy per unit time is the circuit's terminal power. For $VI>0$, the Poynting flux points in the direction of the inner electric current. The electromagnetic energy travels through the insulating gap and reaches a load at the end of the coaxial cable. The radii and material constants redistribute the electromagnetic energy density and the local Poynting vector, while their integrated Poynting flux depends only on the terminal electric potential difference and electric current.

For this radially layered coaxial cable, defineUse the solution regular at the axis with bounded exterior electric potential, so that the empty interior and exterior have the constant electric potentials fixed by their adjacent conducting cylinders.

There is no free charge in either electrical insulator or on their interface. Gauss's law and continuity of normal electric displacement field give $D_{\rho }=C/\rho$ in both layers. The electric potential difference fixes the constant becauseThe electric potential, including the empty regions bounded by the conducting cylinders, is thereforeBoth inner and outer Dirichlet boundary conditions hold, and the electric potential is continuous at the material interface, whereTaking the negative gradient of the electric potential yieldsThe radial electric field generally jumps at $\rho =c$, but $ϵE_{\rho }(c^{-})={ϵ}^{\prime }{E}_{\rho }(c^{+})$, as required by the absence of free surface charge density there. The conducting cylinders carry opposite linear charge densities $\lambda =2\pi C$ and $-\lambda$.

For an Amperian loop of radius $\rho$, the enclosed electric current is zero, $I$, or zero according as $\rho <a$, $a<\rho <b$, or $\rho >b$. The magnetostatic Ampère law therefore gives $H_{\phi }=I/(2\pi \rho )$ in the gap, and zero outside. Using the two magnetic permeabilities,The tangential magnetic field strength is continuous at the material interface, while the tangential magnetic field need not be. At each ideal conducting surface and at the material interface, the displayed electric field and magnetic field formulas specify one-sided limits; a single interface value is unnecessary when the limiting values differ.

Orient each edge both ways. A unit flow $j$ from $x$ to $y$ is antisymmetric, $j(u,v)=-j(v,u)$, with divergence of a flow equal to $+1$ at $x$, $-1$ at $y$ and zero elsewhere. Its energy iswhere each unoriented edge is counted once. Put $c_e=1/r_e$. The electrical unit current is $i(u,v)=c_{uv}(V(u)-V(v))$, where the potential $V$ solves the Kirchhoff node law with those source and sink divergences. Because $G$ is a connected graph, this potential is unique up to an additive constant. The effective resistance is the electric potential difference needed for unit current,Discrete summation by parts gives ${\mathcal{E}}_r(i)={\sum }_{u}V(u)\mathrm{div}i(u)=R(x,y)$.

The Thomson principle says that this current minimizes energy among all unit flows. Indeed write any other unit flow as $j=i+k$, where $\mathrm{div}k=0$. ThenConsequentlyThis also proves uniqueness of the minimizing current.

If $r_e^{\prime }\ge r_e$ for every edge, then ${\mathcal{E}}_{r^{\prime }}(j)\ge {\mathcal{E}}_r(j)$ for every admissible flow. The admissible set is unchanged, so taking its minimum proves the Rayleigh monotonicity principle:Deletion is the limiting operation $r_e\to \infty$: finite-energy flows must then carry zero current on that edge. If the terminals become disconnected, the effective resistance is infinite.

Use $T_y=\inf \{n\ge 0:X_n=y\}$ and the positive return time $T_x^{+}=\inf \{n\ge 1:X_n=x\}$. The walk has transitions $p_{uv}=c_{uv}/d(u)$, $d(u)={\sum }_v{c}_{uv}$; its transition matrix defines a reversible Markov chain, although the transition probabilities need not be symmetric.

The voltage $h(u)=P_u(T_x<T_y)$ equals one at $x$, zero at $y$, and is a harmonic function on a graph away from the terminals: $h(u)={\sum }_v{p}_{uv}h(v)$. For this unit electric potential difference the electric current isThe first-step analysis therefore proves the return probability from effective resistance identity:Finite irreducibility ensures eventual hitting of one of the terminals, so no missing escape event appears in this finite-network calculation.

For an infinite locally finite connected graph, exhaust it by finite sets and wire their exteriors to one sink. The event of exiting before the first return decreases to the event of never returning. Thuswith $1/\infty =0$. In particular recurrence is equivalent to infinite effective resistance to infinity.

After deleting an edge, consider any infinite component and choose a root $o$ away from that edge's endpoints. Its vertex degree $d(o)$ is unchanged. On every finite wired exhaustion the Rayleigh monotonicity principle makes the resistance no smaller; the limit of a sequence does too. A recurrent original network consequently leaves a recurrent component. Finite components give recurrent walks as well, with an isolated graph vertex viewed as an absorbing state. This proves that edge deletion preserves recurrence. The phrase “no less recurrent” refers to this recurrence comparison: at a deleted-edge endpoint the vertex degree factor also changes, so monotonicity of its first-return probability alone should not be inferred.

With no electrical resistance or applied electric potential difference, divide the equation by the inductance $L$ to get a simple harmonic oscillator with angular frequency ${\omega }_0=(LC{)}^{-1/2}$. Its general real solution isEach nonzero solution has period $2\pi \sqrt{LC}$. Here ${\omega }_0$ is an angular frequency; the ordinary frequency in cycles per unit time is ${\omega }_0/(2\pi )$.

For $\theta =30^{\circ }$, the result in b gives $1/3=2E_0/(E+E_0)$, hence $E=5E_0$. Acceleration from rest supplies only the change in kinetic energy:The accelerating electric potential difference is a magnitude; its sign depends on which electrode is taken as the reference, since the electron has negative electric charge.

Energy–momentum transformations and invariant. Let the primed inertial frame move at signed velocity $u$ along the unprimed $x$ axis. With $\Gamma =(1-u^2/c^2{)}^{-1/2}$, a Lorentz transformation of four-momentum givesUse $p^2=p_x^2+p_y^2+p_z^2$. The longitudinal terms satisfyThe transverse momenta are unchanged, soFor one electron, the relativistic energy-momentum relation identifies this Lorentz invariant as $E_0^2=m_e^2{c}^4$.

Symmetric electron scattering. Assume an isolated elastic collision of the two electrons, with no emitted radiation. The incident electron has total energy $E$ and momentum magnitude $p$, so $p^2{c}^2=E^2-E_0^2$; the target electron initially contributes rest energy $E_0$ and zero momentum. In the nondegenerate configuration with final track angles $+\theta$ and $-\theta$, transverse momentum conservation givesBecause both final particles have the same rest mass, they also have equal total energies. Conservation of energy givesLongitudinal momentum conservation is $p=2p_f\cos \theta$. Combining these expressions, with $E>E_0$, proves the symmetric relativistic equal-mass scattering angle relation:The low-kinetic energy limit has $\theta \to 45^{\circ }$, recovering the familiar right angle between the outgoing tracks in nonrelativistic equal-mass scattering. Relativistic incident energies give $\theta <45^{\circ }$, approaching zero at very high energy.

Accelerating potential. For $\theta =30^{\circ }$, ${\tan }^2\theta =1/3=2E_0/(E+E_0)$, so $E=5E_0$. The acceleration from rest supplies kinetic energy, not total energy. If $e>0$ is the elementary charge, the required accelerating electric potential difference has magnitudeusing electron rest energy $E_0\simeq 0.511\mathrm{MeV}$. The electric potential difference is oriented to accelerate the negative electron; the result above specifies its magnitude.

Place charges $+Q$ and $-Q$ on the inner and outer conductors. In the vacuum gap, Gauss's law gives $E(r)=Q/(4\pi {ϵ}_0{r}^2)$. Their electric potential difference isConsequently the spherical capacitor hasFor a narrow gap $b-a\ll a$, this approaches ${ϵ}_0(4\pi a^2)/(b-a)$, the local parallel-plate approximation. For $b\to \infty$, it becomes the capacitance $4\pi {ϵ}_{0}a$ of an isolated sphere.

For a fixed number $N$ of identical bulbs, first approximate each as a resistor of electrical resistance $R$ connected to a fixed ideal battery electric potential difference $V$. A series circuit has electrical resistance $NR$, while a parallel circuit has electrical resistance $R/N$. ThusParallel connection therefore maximizes dissipation in this model: every bulb receives the full source electric potential difference instead of sharing it. For $N=1$ the two arrangements coincide.

An actual incandescent light bulb changes electrical resistance as its filament heats, so a single constant $R$ need not describe both operating points. The same qualitative conclusion holds for identical stable bulbs with increasing current-voltage function $i(U)$: parallel operation gives $P_p=NVi(V)$, whereas symmetric series operation gives $P_s=Vi(V/N)$. The former is larger for $N>1$. This comparison assumes a fixed collection and an ideal source as specified; an unlimited number of parallel bulbs on a zero-internal-resistance ideal battery would have no finite total-power maximum.

Let $R_1<R_2$ and put charge per unit length $+{\lambda }_q$ on the inner shell and $-{\lambda }_q$ on the facing outer conductor. For a long coaxial Gaussian surface of radius $r$ in the gap, Gauss's law givesThe electric potential difference between the shells isWith total inner charge $Q={\lambda }_{q}L$, the capacitance isThis assumes a vacuum or approximately air-filled gap and negligible fringing at the ends. A uniform linear dielectric replaces ${ϵ}_0$ by its permittivity. The expression depends on the shell-to-shell potential difference, not on the arbitrary common potential reference.

For an ideal inductor of constant self-inductance $L$, the passive-sign electric potential difference is $V_L=LdI/dt$. The supplied power is $V_{L}I$, so increasing the electric current quasistatically from zero storesAny electrical resistance would additionally dissipate energy as heat; it is not included in this stored magnetic energy.

For a long vacuum solenoid with $N$ turns, axial length $\ell$ and cross-sectional area $A$, Ampère's law gives $B={\mu }_0(N/\ell )I$. Its flux linkage is $N\Phi =NBA$, soThe nearly uniform field occupies volume $A\ell$, with negligible external-field energy in the long-solenoid approximation. Dividing its stored energy by this volume yieldsThe vacuum and linear-inductance assumptions matter; a nonlinear magnetic material requires the appropriate field-energy integral rather than simply substituting a varying inductance into $L{I}^2/2$.

Label the capacitor's left terminal $a$ and right terminal $b$, with the battery's negative node at zero. The source diagram has $2R$ from $a$ to ground, $R$ from $b$ to ground, and a source-side resistor $R$ feeding $a$ from electric potential difference $V$. The capacitor electric current $I_2$ flows from $a$ to $b$; $I_1$ flows through $2R$.

With the capacitor removed, $V_a=2V/3$ and $V_b=0$, so the Thévenin equivalent seen by the capacitor has $V_{\mathrm{th}}=2V/3$. Suppress the ideal electric potential difference source to find its electrical resistance: the left terminal sees $R\parallel 2R=2R/3$ to ground and the right sees $R$, giving $R_{\mathrm{th}}=5R/3$ between the terminals.

The initially uncharged capacitor electric potential difference $V_C=V_a-V_b$ therefore obeysSolving gives $V_C=(2V/3)(1-e^{-t/\tau })$, with $\tau =R_{\mathrm{th}}C$. The capacitor electric current is $I_2=C{\dot{V}}_C$, henceAt $t=0^{+}$, the short-like uncharged capacitor makes $V_a=V_b=2V/5$, so $I_2=2V/(5R)$ directly from the right resistor. At long times the capacitor carries no direct electric current. These endpoint checks agree with the exponential solution.