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A half-wave rectifier with one diode and a resistor load passes one half of each input cycle.^[ref]

For an original ideal example, let an ideal voltage source supply the sinusoid $v_S(t)=V_p\sin (\omega t)$, with $V_p=5\mathrm{V}$ and angular frequency $\omega >0$. Connect its positive reference terminal to the diode anode, the diode cathode to the upper terminal of a load resistor with electrical resistance $R_L=1\mathrm{k\Omega }$, and the lower terminal to the source return. Let $v_L$ be the upper-minus-lower load electric potential difference, and $i_L$ the downward load electric current. Assume the ideal diode model, zero source electrical resistance, and no filter capacitor, inductor or switching delay.

The Kirchhoff cycle law gives $v_S=V_D+v_L$, and Ohm's law gives $v_L=R_L{i}_L$. If $v_S>0$, the blocking assumption would give $V_D=v_S>0$, contradicting the ideal diode model, so the diode conducts with $V_D=0$. If $v_S<0$, conduction would give $i_L=v_S/R_L<0$, also forbidden, so the diode blocks. At the crossings both values vanish. ThusThe peak load electric current is $5\mathrm{mA}$. The output is unidirectional but still varies in time, reaching zero throughout every negative half-cycle. Over a period of a function $2\pi /\omega$, write the wave phase as $\theta =\omega t$. Its average value of a function and root mean square follow from two integrals:Consequently the average load power is $v_{L,\mathrm{rms}}^2/R_L=6.25\mathrm{mW}$, not ${\overline{v}}_L^2/R_L$. The ideal diode has zero loss; the resistor converts supplied energy into Joule heating. The same switching argument predicts $v_L=\min (0,v_S)$ when the diode is reversed. Directional selection changes the waveform and its average value of a function; it does not by itself produce a constant output.

The Kirchhoff cycle law says the directed sum of voltage drops around every graph cycle is zero. For unit edge resistance this is the sum of signed currents around the graph cycle. With unit edge resistance, the current is a voltage gradient: define the voltage difference by summing drops along a graph path. The difference of the sums along two graph paths is a sum around closed walks; decomposing these into graph cycles makes it zero. Together with the Kirchhoff node law, the resulting graph Laplacian equation determines current uniquely on a finite connected network.

A monolithic integrated circuit has its circuit elements fabricated in or on one semiconductor die. Their connections are essential: an unconnected collection of transistors does not implement the intended electrical circuit.

An original ideal example separates function from fabrication. Connect two resistors, $R_1=10\mathrm{k\Omega }$ above $R_2=20\mathrm{k\Omega }$, in a series circuit across an ideal $3\mathrm{V}$ voltage source. With no load electric current drawn from their meeting point, Ohm's law and the Kirchhoff cycle law giveThese equations apply to either a discrete circuit or an idealized monolithic integrated circuit realizing those same electrical resistances. Integration changes construction; the electrical circuit equations still determine the output. A real load, unwanted capacitance or finite resistor tolerance changes the ideal calculation. Manufacturing many functions together also couples their fate: a fatal defect can discard an entire semiconductor die, motivating the Poisson defect yield model.

Orient the unique path from $s$ to $t$ in a uniform spanning tree $T$, and put $j_{xy}^T=1$ when it uses the edge from $x$ to $y$, $-1$ when it uses the reverse direction, and zero otherwise. Then the proposed current is $i_{xy}=\mathbb{E}{j}_{xy}^T$; set it to zero on nonedges. It is antisymmetric.

For each tree path, every intermediate graph vertex has one incoming and one outgoing path edge, while the source has one outgoing edge and the sink one incoming edge. HenceTaking expectations givesthe Kirchhoff node law for a unit flow.

To establish the Kirchhoff cycle law, it is not enough to check it separately for each tree path: those path currents need not satisfy the voltage law on the original graph. Instead, let $\mathcal{T}$ be all spanning trees and let ${\mathcal{F}}_{st}$ be all spanning forests with two components, one containing $s$ and the other $t$. For $F\in {\mathcal{F}}_{st}$ denote the source component by $C_s(F)$.

For an existing edge $xy$, deleting it from a tree whose $s$-$t$ path uses $x\to y$ produces precisely a forest $F\in {\mathcal{F}}_{st}$ with $x\in C_s(F)$ and $y\notin C_s(F)$. Conversely, adding $xy$ to any such forest produces exactly that tree and path orientation. This is a bijection. Applying it to both orientations givesDefine $h(x)=|\mathcal{T}{|}^{-1}{\sum }_F{1}_{x\in C_s(F)}$. Then $i_{xy}=h(x)-h(y)$ on every edge. Since the resistances are one, this is Ohm's law, and the voltage drops telescope around every oriented cycle:Together with the node law, this proves that the mean tree-path current is the electrical unit current. This argument is the mean spanning-tree path current identity.

For clarity, these laws determine the current uniquely. The difference of two solutions is a potential gradient $k(x)-k(y)$ with zero divergence. Multiplying its divergence by $k(x)$ and summing over graph vertices gives ${\sum }_{\{x,y\}}(k(x)-k(y){)}^2=0$. Thus every difference current is zero.

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 $c_e{R}_{\mathrm{eff}}(e)$.

For a p-n junction at fixed temperature, the Shockley diode equation models its electric current asHere $I_S>0$ is the diode saturation current, $n>0$ the diode ideality factor, $V_T$ the thermal voltage, $k_B$ the Boltzmann constant, and $q>0$ the elementary charge.^[ref]

For an original model example, choose $I_S=1\mathrm{nA}$, $n=2$, and $T=300\mathrm{K}$. Using $k_B=1.380649\times 10^{-23}J/K$ and $q=1.602176634\times 10^{-19}\mathrm{C}$ gives $V_T=25.8520\mathrm{mV}$. Then $V_D=0.600\mathrm{V}$ gives $I_D=0.109591\mathrm{mA}$, whereas $V_D=0.800\mathrm{V}$ gives $I_D=5.24450\mathrm{mA}$. These are predictions of the declared model. There is no universal turn-on electric potential difference: solving the Shockley diode equation for the value needed at a chosen electric current givesAt $I_D\gg I_S$, increasing $V_D$ by $0.100\mathrm{V}$ multiplies the electric current by approximately $e^{0.100/(n{V}_T)}=6.91772$. At $V_D=-0.300\mathrm{V}$ the same model gives $I_D=-0.996979\mathrm{nA}$. Its reverse branch approaches $-I_S$; it cannot describe diode reverse breakdown.

A constructed extension adds internal series electrical resistance $R_s$ to a junction whose electric potential difference is $V_j$. Applying the Kirchhoff cycle law givesFor $R_s=20\Omega$ and the same $I_S,n,T$, sustaining $I_D=5.24450\mathrm{mA}$ needs $V_D=0.904890\mathrm{V}$: $0.800\mathrm{V}$ across the junction plus $0.104890\mathrm{V}$ across its internal electrical resistance. The derivativeshows why the current-voltage characteristic becomes nearly a straight line when the internal series electrical resistance dominates. This is still a fixed-temperature model; self-heating can change $I_S$, $n$ and $V_T$. Real diodes also have leakage and charge-storage effects omitted here.^[ref]

A spark gap can rapidly discharge a charged capacitor through an inductor. After excitation, the stored energy runs down through Joule heating, radiation and other losses. A key held down can permit repeated discharges: each discharge starts a damped train of rapid oscillations. The spark repetition frequency, the electrical oscillation frequency within each train, and the rate of keyed symbols are different quantities. Repetition does not make each train a maintained sinusoid.

An original lumped model replaces the conducting spark gap and other losses by a constant total electrical resistance $R>0$ in an isolated series RLC circuit. For electric charge $q$ on the capacitor, the Kirchhoff cycle law givesFor $Q>1/2$, its characteristic roots are $-\alpha \pm \mathrm{i}{\omega }_d$, with ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$, proving the stated damped form. The electric current $\dot{q}$ has the same exponential decay scale. Its wave envelope falls by $\exp (-2\pi \alpha /{\omega }_d)$ per cycle; the energy envelope falls twice as fast in the exponent. A maintained sinusoid instead requires continuing input power balancing losses.

With $L=5\mathrm{mH}$, $C=2\mathrm{nF}$ and $R=100\Omega$, the model gives $f_0={\omega }_0/(2\pi )=50.329\mathrm{kHz}$, Q factor $Q=15.811$, amplitude time constant $1/\alpha =100\mu \mathrm{s}$, and damped frequency ${\omega }_d/(2\pi )=50.304\mathrm{kHz}$. These are chosen component values. A real spark gap switches nonlinearly, an antenna radiates, and coupling exchanges energy between circuits; one constant-$R$ free decay does not model the complete transmitter. Rapid damping also broadens the frequency spectrum, which helps explain why short trains make sharp receiver selection difficult.