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For inner radius $a$ and outer conductor's inner radius $b$, Gauss's law gives $E(s)=\lambda /(2\pi {ϵ}_0{ϵ}_{r}s)$. Integrating across the dielectric gives voltage $V=\lambda \ln (b/a)/(2\pi {ϵ}_0{ϵ}_r)$. Thus the capacitance per unit length isThe stated outer radius is interpreted as the radius of the dielectric-facing surface. Magnetic permeability affects inductance, but does not enter this electrostatic capacitance.

A transmission line stores electric and magnetic energy along its length, so voltage and current depend on both position and time. A distributed circuit uses inductance $L^{\prime }$ and capacitance $C^{\prime }$ per length, together with resistance $R^{\prime }$ and leakage conductance $G^{\prime }$. Applying voltage and charge balance to a short segment gives the telegrapher's equations,For a lossless line, both variables obey wave equations with speed $v=1/\sqrt{L^{\prime }{C}^{\prime }}$ and characteristic impedance $Z_0=\sqrt{L^{\prime }/C^{\prime }}$. Forward and backward waves obey $I^{+}=V^{+}/Z_0$ and $I^{-}=-V^{-}/Z_0$. Matching a termination $Z_L$ gives the voltage reflection coefficientA matched load absorbs without reflection; an open circuit has ${\Gamma }_L=1$, and a short has ${\Gamma }_L=-1$. Reflections can produce standing waves. A coaxial line confines the fields between conductors and, for a uniform dielectric TEM wave, has $v=1/\sqrt{\mu ϵ}$. Finite loss attenuates waves; frequency-dependent losses or material properties can also distort pulses. A long line is a wave-carrying distributed system, so it cannot generally be replaced by one lumped capacitor.

For distributed inductance $L$ and capacitance $C$, a small segment obeys $V_z=-L{I}_t$ and $I_z=-C{V}_t$. Differentiation gives $V_{zz}=LC{V}_{tt}$ and $I_{zz}=LC{I}_{tt}$. For a forward wave,A backward voltage wave has current $I^{-}=-V^{-}/Z$. This sign matters at an impedance discontinuity.

The source figure is the stacked-strip parallel-plate transmission line geometry, despite its use of the word coplanar. Gauss's law gives $C={ϵ}_0{ϵ}_{r}a/d$. Opposite surface currents give magnetic field $B={\mu }_{0}I/a$ between the strips by Ampère's law; the magnetic flux per unit length is $Bd$, hence $L=Bd/I={\mu }_{0}d/a$. ThereforeNo numerical $a/d$ is given; the impedance is appropriately left in terms of that ratio. The line is taken to be nonmagnetic and lossless.

Increasing the gap by $\alpha$ changes $L$ to $\alpha L$ and $C$ to $C/\alpha$, so $Z_2=\alpha Z_1$ while $v$ is unchanged. voltage and current continuity at the step giveConsequently the voltage and energy-flux transmission coefficients areHere amplitudes are peak voltage amplitudes, with mean power $|V{|}^2/(2Z)$. The reciprocal gaps have $r_V(1/\alpha )=-r_V(\alpha )$: equal reflected energy but opposite voltage phase. A voltage transmission coefficient above one does not imply energy gain.

For a finite impedance section of a transmission line, put $\ell =l_2-l_1$ and $\delta =\omega \ell /v$. Multiple reflections inside the altered section interfere. Summing their geometric series, relative to the first boundary,Transmission is unity at $\delta =n\pi$ and minimal at $\delta =(n+1/2)\pi$, with minima $4{\alpha }^2/(1+{\alpha }^2{)}^2$. Resonances are separated by $\Delta f=v/(2\ell )$; the long-wavelength limit also transmits fully. Matching the far termination removes end echoes, not the two internal reflections.

The characteristic impedance of a lossless transmission line is the ratio of the voltage to the current amplitude of a single forward-travelling wave. If $\mathcal{L}$ and $\mathcal{C}$ denote inductance and capacitance per unit length, the lossless telegrapher equations giveA backward wave has $I_{-}=-V_{-}/Z$ when current is positive along $+z$. Thus characteristic impedance is not a resistor's electrical resistance: it describes travelling-wave amplitudes and determines impedance matching, power transfer and reflections. A semi-infinite line, or a line terminated by a matching load, presents that impedance at its input.

At a junction with no lumped branch, voltage and electric current are continuous:Eliminating $V_t$ gives the voltage reflection coefficientFor real positive $Z_1$, incident and reflected waves occupy the same line, so their power ratio is $|r{|}^2$.

For the slightly mismatched coaxial cables, the change in characteristic impedance isConsequentlyThe requested power limit $0.01\%=10^{-4}$ implies $|\Delta b|/b<0.02\ln (b/a)$. Since $\ln (b/a)=2\pi nZ/Z_{\mathrm{vac}}=1.918$ for the stated line, the adjacent cables' outer radii must match within approximately $3.84\%$. If a tolerance is quoted separately as a symmetric manufacturing range $\pm \delta b$ and arbitrary worst-case pairs can differ by $2\delta b$, the corresponding individual tolerance is approximately $\pm 1.92\%$. The leading calculation is for small mismatches; the exact logarithmic expression gives slightly asymmetric positive and negative limits. The vacuum wave impedance used here is approximately $376.73\Omega$.^[ref]

A needle is a shunt electrical resistance $R_n$, so the load seen at its location is the continuation of the line in parallel with $R_n$: $Z_L=Z{R}_n/(Z+R_n)$. The local voltage reflection coefficient and transmission amplitude areFor periodic resistive loading of a transmission line, the reflection from the next needle travels an additional round trip $2L$ and crosses each previous needle twice. Taking the first needle as phase reference, the single-return approximation for periodic line reflections therefore sums toHere $v_p$ is the propagation phase velocity in the cable, $c/n$ for the stated nondispersive nonmagnetic filling. Squaring the amplitude gives the requested expressionFor a finite number $N$ of needles the amplitude has the additional numerator $1-[t^2{e}^{i\varphi }{]}^N$. The displayed semi-infinite result assumes sufficiently many needles, no terminal reflection, and neglects intrinsic propagation loss.

For real $r<0$, write $q=t^2$. The denominator of the power coefficient is $1+q^2-2q\cos \varphi$. It has constructive peaks at $\varphi =2\pi j$ and minima at $\varphi =(2j+1)\pi$, repeating with frequency spacing $\Delta \nu =v_p/(2L)$. Within this approximation,Weak individual shunts thus give narrow interference peaks, with phase width of order $|r|$.

The requested sum excludes paths that reflect repeatedly between different needles. It should not be read as an exact multiple-scattering result, particularly at the coherent peaks. For example the exact semi-infinite amplitude satisfiesAt $\varphi =2\pi j$ this equation reduces to $(R+1{)}^2=0$: the exact ideal infinite array has unit power reflection there, rather than the first-return approximation's value near $1/4$. This does not change the requested geometric-series derivation, but identifies its physical limitation.

An inductor stores magnetic energy and opposes changes of current through electromagnetic induction. For a fixed linear device, the flux linkage is $\Psi =N\Phi =LI$, the self-induced electromotive force is $-LdI/dt$, and the passive terminal voltage is $V=LdI/dt$. HenceThe voltage leads a sinusoidal current by a quarter cycle. A long solenoid has $B\simeq \mu NI/l$ and $L\simeq \mu N^{2}A/l$; its energy agrees with the volume integral of $B^2/(2\mu )$.

For two coupled coils, ${\Psi }_1=L_1{I}_1+\mathcal{M}{I}_2$, ${\Psi }_2=\mathcal{M}{I}_1+L_2{I}_2$, with signs set by coil orientations. Reciprocity gives the same mutual inductance in both equations. For the energy stored in a nonlinear inductor, $U=\int Id\Psi$ replaces the simple quadratic formula, and hysteresis can dissipate energy. Real coils also have resistance, core loss and parasitic capacitance, so an ideal inductance model has a finite useful frequency range.^[ref]

A normal mode is a motion in which all coordinates have a fixed relative pattern and oscillate at one frequency. Near a stable equilibrium, let the quadratic kinetic and potential energies be $T={\dot{\mathbf{q}}}^{\mathsf{T}}\mathbf{M}\dot{\mathbf{q}}/2$ and $V={\mathbf{q}}^{\mathsf{T}}\mathbf{Kq}/2$. With positive-definite mass matrix $\mathbf{M}$, the equations and harmonic ansatz giveThe stiffness and mass matrices are symmetric. For distinct frequencies, their symmetry gives $({\omega }_s^2-{\omega }_r^2){\mathbf{a}}_r^{\mathsf{T}}\mathbf{M}{\mathbf{a}}_s=0$, so the mode vectors are orthogonal in the mass inner product. A basis can also be chosen in a degenerate eigenspace. Expanding $\mathbf{q}={\sum }_s{Q}_s{\mathbf{a}}_s$ decouples the equations into ${\ddot{Q}}_s+{\omega }_s^2{Q}_s=0$; initial displacements and velocities determine the coefficients. Positive stiffness gives real frequencies, a zero eigenvalue can describe a free symmetry motion, and a negative one signals instability rather than an oscillatory mode.^[ref]

For two equal masses between fixed walls with three equal springs, the stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and the mass matrix is $m1$. The patterns $(1,1)$ and $(1,-1)$ have frequencies $\sqrt{k/m}$ and $\sqrt{3k/m}$. An initial displacement of one mass excites both modes and can produce beats.

For a uniform taut string fixed at both ends, the wave equation ${\partial }_t^{2}y=c_s^2{\partial }_x^{2}y$, with $c_s=\sqrt{\mathcal{T}/{\mu }_{\ell }}$, has patterns and angular frequenciesThe boundary conditions select the spectrum. Fourier expansion of arbitrary admissible initial data provides the mode amplitudes, illustrating the extension from a finite set of coordinates to a field.

For the normal modes of inductively coupled LC circuits, take two identical circuits with self-inductance $L$, capacitance $C$ and mutual inductance $\mathcal{M}$, where $|\mathcal{M}|<L$. The capacitor charges obey $L{\ddot{q}}_1+\mathcal{M}{\ddot{q}}_2+q_1/C=0$ and the exchanged equation. In-phase and opposite-phase charge patterns haveThe positive inductance matrix for two identical coils is the physical counterpart of positive mass. Here inductance plays the role of the mass matrix. Mechanical, wave and electrical examples therefore share the same eigenvalue structure. Weak forcing excites modes according to its spatial overlap with each pattern; damping broadens the resonances, while strong nonlinearity can invalidate a fixed linear-mode decomposition.

Distributed circuit model. A transmission line must be treated as a distributed system when its propagation time is appreciable compared with the signal timescale. Divide a uniform two-conductor line into elements of length $dx$. Each has series resistance $R^{\prime }dx$ and inductance $L^{\prime }dx$, and shunt electrical conductance $G^{\prime }dx$ and capacitance $C^{\prime }dx$. The primes denote quantities per unit length, with units $\mathrm{\Omega }{\mathrm{m}}^{-1}$, $\mathrm{H}{\mathrm{m}}^{-1}$, $\mathrm{S}{\mathrm{m}}^{-1}$ and $\mathrm{F}{\mathrm{m}}^{-1}$, respectively.

Kirchhoff's voltage and current laws give the Telegrapher's equations:For a lossless line, $R^{\prime }=G^{\prime }=0$, differentiation gives $V_{xx}=L^{\prime }{C}^{\prime }{V}_{tt}$ and the same wave equation for $I$. The propagation speed and characteristic impedance areA general lossless signal has $V=V^{+}(t-x/u)+V^{-}(t+x/u)$ and $I=[V^{+}(t-x/u)-V^{-}(t+x/u)]/Z_0$. Thus characteristic impedance is the voltage/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no energy, because energy is carried away by propagation.

For sinusoidal quantities with convention $e^{i\omega t}$, the general lossy-line formulas areChoose the square-root branches for forward attenuation and positive forward power. The real part of ${\gamma }_{\mathrm{prop}}$ is attenuation per unit length; its imaginary part is the phase wavenumber. For small losses, attenuation is approximately $R^{\prime }/(2Z_0)+G^{\prime }{Z}_0/2$, using the lossless real $Z_0$ in this estimate.

Fields and geometry. Two separated conductors support a transverse electromagnetic mode, with electric and magnetic fields transverse to the propagation axis. For a homogeneous dielectric and negligible conductor penetration, both examples below have $L^{\prime }{C}^{\prime }=\mu ϵ$ and hence $u=1/\sqrt{\mu ϵ}$. The distributed circuit model follows from the field energy and charges in each slice.

For a coaxial cable with inner radius $a$ and inner surface of the outer conductor at $b$, Gauss's law and Ampère's law giveThe electric field is radial and the magnetic field circles the axis. Integration of voltage and field energy givesThe equal and opposite return current on the outer conductor confines the ideal external field to the annulus.

For wide parallel plates of width $w$ separated by $d\ll w$, neglect fringing. The electric field crosses the gap, $E\simeq V/d$, while the magnetic field lies in the gap parallel to the plates and perpendicular to propagation, $H\simeq I/w$. ThereforeFor a forward wave, $\mathbf{E}\times \mathbf{H}$ points along the line. Integrating this Poynting vector over either cross-section gives instantaneous power $VI$. For complex peak amplitudes, average power is $\mathrm{Re}(V{I}^{\ast })/2$, or $|V^{+}{|}^2/(2Z_0)$ for a single forward wave on a lossless line. The conductors guide an energy flux predominantly in the dielectric between them.

Terminations and reflection. At a load $Z_L$, require $V_L=Z_L{I}_L$. If $V^{+}$ and $V^{-}$ are the incident and reflected voltage amplitudes there, the total current is $(V^{+}-V^{-})/Z_0$. ThusA matched load, $Z_L=Z_0$, absorbs the incident power with no reflection. An open circuit has ${\Gamma }_V=+1$, a current node and a doubled voltage at the end. A short circuit has ${\Gamma }_V=-1$, a voltage node and a doubled current there. A purely reactive load on a lossless line has $|{\Gamma }_V|=1$ and accepts no average power. For real $Z_0$, the reflected power fraction is $|{\Gamma }_V{|}^2$ and the accepted fraction is $1-|{\Gamma }_V{|}^2$.

Incident and reflected waves produce a standing-wave pattern with voltage standing wave ratio $(1+|{\Gamma }_V|)/(1-|{\Gamma }_V|)$. The input impedance of a lossless length $\ell$ isThis gives $i{Z}_0\tan (k\ell )$ for a short and $-i{Z}_0\cot (k\ell )$ for an open. A quarter-wave section gives $Z_{\mathrm{in}}=Z_0^2/Z_L$, enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return time is $2\ell /u$. Real conductor resistance, dielectric loss and departures from a TEM field profile limit the ideal model at high frequency.

Magnetic field and force. A steady electric current produces a magnetic field. In vacuum the Biot-Savart law givesThe Lorentz force on the charge carriers transfers force to the material carrying them. The resulting force on a current element of another circuit is $d{\mathbf{F}}_2=I_{2}d{\mathbf{\ell }}_2\times {\mathbf{B}}_1$, or, for a continuous current distribution, the magnetic force per unit volume is $\mathbf{f}=\mathbf{J}\times \mathbf{B}$.

For two long parallel straight wires a distance $r$ apart, Ampère's law gives $B_1={\mu }_0{I}_1/(2\pi r)$. ThereforeThe force is attractive for parallel currents and repulsive for opposite currents, by the cross product rule. Its units are $\mathrm{N}{\mathrm{m}}^{-1}$. This approximation neglects wire thickness and end effects; a complete circuit includes its return current. Element-by-element forces need not form an isolated Newton's-third-law pair, but complete stationary circuits have balanced mutual forces when the entire system is included.

Energy description. For two linear coupled circuits the magnetic energy isHere $q$ is a mechanical coordinate and $M$ the signed mutual inductance. At prescribed currents, the generalized force is $I_1{I}_2\partial M/\partial q$ if the self-inductances do not vary. This positive derivative is obtained from magnetic coenergy, or by including the electrical sources' work; differentiating $W_B$ with a minus sign at fixed currents would omit that source work. At prescribed flux linkages the force instead equals the negative derivative of stored magnetic energy with those linkages fixed. The geometry thus determines both inductance and mechanical force.

For an ideal electrical transformer, let $n=N_p/N_s$. The reflected secondary load is $R_{\mathrm{in}}=n^2{R}_L$. Matching a $4\Omega$ load to an $8\Omega$ source therefore requiresWith the generator replaced by its internal resistance, its resistance referred to the secondary isThis is the source output resistance relevant to matching and maximum power transfer.

To find the windings, use the inductance of a toroidal winding. The mean magnetic path is $\ell =2\pi (0.10)\mathrm{m}$, the core area is $A=30\times 10^{-4}=0.003{\mathrm{m}}^2$ and $\mu =500{\mu }_0$. ThusThe ideal secondary turn count is $N_s=500/\sqrt{2}=353.55$, giving the practical nominal designWith integer turns the irrational ratio cannot be exact: $500:354$ reflects the $4\Omega$ load as $7.98\Omega$, an approximately $0.25\%$ resistance mismatch. These turn counts use the stated $0.75\mathrm{H}$ self-inductance; no operating frequency was specified from which to set a separate minimum inductance.

The $200\Omega$ condition must refer to the unloaded magnetizing reactance, not to the matched loaded input resistance. With $L_p=0.75\mathrm{H}$,The secondary self-inductance is $L_s=L_p/n^2=0.375\mathrm{H}$, so its corresponding unloaded coil impedance satisfiesIf output impedance means this transformed coil reactance, $100\Omega$ is its minimum magnitude. It is distinct from the $4\Omega$ transformed source resistance. This is the magnetizing reactance and reflected resistance in a transformer distinction. A finite magnetizing inductance adds a parallel branch, $Z_{\mathrm{in}}=(8\Omega )\parallel (i\omega L_p)$, so exact purely resistive matching is an ideal large-reactance approximation; interpreting the loaded input itself as both $8\Omega$ and at least $200\Omega$ would be impossible.

A transmission line must be treated as a distributed circuit when its propagation time is significant on the signal timescale. Let $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$ denote resistance, inductance, leakage conductance and capacitance per unit length. Applying Kirchhoff's voltage law and Kirchhoff's current law to a short length $dx$ gives the telegrapher's equationsFor a lossless line, $R^{\prime }=G^{\prime }=0$. Differentiating again gives wave equations withThe travelling solutions can be written $V=V_{+}(t-x/v)+V_{-}(t+x/v)$ and $I=[V_{+}(t-x/v)-V_{-}(t+x/v)]/Z_0$. The minus sign for the backward wave is essential to power transport.

At a load $Z_L$, continuity of voltage and current gives the voltage reflection coefficientA matched termination $Z_L=Z_0$ absorbs the incident signal without reflection. An open circuit has ${\Gamma }_L=+1$ and a short circuit has ${\Gamma }_L=-1$. Superposed incident and reflected waves produce standing waves. A sinusoidal lossless line of length $\ell$ transforms the termination intofor the $e^{i\omega t}$ convention. At a quarter wavelength, $Z_{\mathrm{in}}=Z_0^2/Z_L$, allowing impedance transformation. A coaxial line supports a transverse electromagnetic mode whose ideal $L^{\prime },C^{\prime }$ are calculated in A4. Its energy travels in the fields between conductors; the characteristic impedance is not the wire's DC resistance. Losses introduce attenuation and make the characteristic impedance and propagation constant complex.

Faraday's law relates induced electromotive force to the rate of change of magnetic flux linkage. For a coil with $N$ turns and a consistent right-hand orientation,Flux can change through field variation or circuit motion. Lenz's law is the minus sign: the induced current's field opposes the flux change, consistent with energy conservation. It opposes the change rather than necessarily opposing the original field.

For a linear fixed-geometry inductor, $N{\Phi }_B=LI$ and the self-induced electromotive force is $-LdI/dt$. The passive terminal voltage drop is instead $V=LdI/dt$. The stored magnetic energy isInductance is measured in henries. Transformers use mutual inductance: changing one winding's current induces an electromotive force in another.

Electromagnetic induction is the production of an electromotive force by changing magnetic flux or by moving a conductor through a magnetic field. With the loop and surface orientations related by the right-hand rule, Faraday's law of induction isHere $\mathbf{v}$ is the local wire velocity. For a stationary loop, the induced electric field can have a nonzero closed-path integral; its differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a straight rod moving perpendicularly to itself and to a uniform magnetic field, the motional electromotive force has magnitude $B\ell v$.

The minus sign expresses Lenz's law: the induced electric current produces a magnetic field that opposes the change in linked magnetic flux. Pulling a conducting loop through a magnetic field therefore requires work against an electromagnetic resisting force; that work supplies the resulting heat or stored energy.

For an inductor with constant self-inductance, $\mathcal{E}=-LdI/dt$ and its stored magnetic energy is $L{I}^2/2$. A changing electric current in one coil can induce an electromotive force in another through mutual inductance, the operating principle of a electrical transformer. These examples connect the flux law, the direction of the response and conservation of energy.

A Josephson junction weakly couples two superconducting condensates. Its coupling energy is $-E_J\cos \phi$, with a gauge-invariant phase difference across the junction. Choose current and voltage orientations consistently so thatThe DC Josephson effect permits a stationary zero-voltage supercurrent up to $I_c$. A constant nonzero voltage produces the AC Josephson effect, with frequency $f=2e|V|/h$. Resistive and capacitive currents are additional circuit contributions, not part of the ideal sinusoidal supercurrent alone.

In a symmetric DC SQUID with negligible loop self-inductance, the two junction phases differ by $2\pi \Phi /{\Phi }_0$. Adding their currents and maximizing over the common phase givesFlux changes therefore modulate the critical current or a suitably biased output voltage. This converts phase interference into sensitive flux measurement. Finite loop inductance, unequal junctions, noise and readout conditions alter the simple modulation; the loop flux then includes screening currents. A SQUID is useful in magnetometry, but achieving sensitivity requires both a suitable circuit and controlled environmental noise.

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].

An RL circuit with the output measured across its resistor is a first-order low-pass filter. For an ideal inductor, an ideal voltage source and a high-resistance output instrument, the electrical impedances $i\omega L$ and $R$ giveThe gain approaches one at low angular frequency and $R/(\omega L)$ at high angular frequency; at ${\omega }_c$ it is $1/\sqrt{2}$. The time-domain equation is $L\dot{I}+RI=V_{\mathrm{in}}$ and $V_{\mathrm{out}}=RI$, so the time constant is $L/R$. Instrument loading, source resistance and nonideal inductance can change the response.

Self-inductance is inductance associated with a coil's own electric current, rather than magnetic flux supplied by another coil. In the linear model, fixed geometry and magnetic response give constant $L$ in $\Lambda =LI$. With the turn orientations defined in magnetic flux linkage, Faraday's law of induction gives the self-induced electromotive forceThe final equality requires constant $L$; if the geometry changes, use $-d\Lambda /dt$. The minus sign expresses Lenz's law: the induced electromotive force opposes the change in electric current. It does not always oppose the electric current itself. When positive electric current decreases, the induced electromotive force acts in its direction. ^[ref].

For terminal sign bookkeeping, label the ideal inductor's terminals $a,b$, choose positive electric current entering $a$ and leaving $b$, and define $v_L$ as the terminal electric potential difference drop from $a$ to $b$ in the lumped electrical circuit model. This passive convention givesThus a coil with self-inductance $0.5\mathrm{H}$ and increasing electric current $dI/dt=12A/s$ has terminal drop $v_L=6\mathrm{V}$ and induced electromotive force ${\mathcal{E}}_{\mathrm{ind}}=-6\mathrm{V}$. These signs describe the same process with different reference conventions. The RL circuit uses the terminal-drop convention throughout.

A finite terminal electric potential difference cannot make the electric current of an ideal inductor with constant $L>0$ jump. Integrating its differential equation givesIf $v_L$ stays bounded, the integral tends to zero as the interval shrinks, ensuring continuity of electric current. A step in terminal electric potential difference changes $dI/dt$ immediately, while electric current remains continuous. At constant electric current, the self-induced electromotive force is zero, although the inductor's magnetic energy can remain nonzero.

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.

In a series RLC circuit, the same electric current $i=q^{\prime }$ flows through the resistor with electrical resistance $R$, the inductor, and the capacitor. Choose $q$ as the signed electric charge on the capacitor plate into which positive $i$ flows. With terminal drops measured in that direction, Kirchhoff's voltage law givesHere $V(t)$ is the applied electric potential difference; assume constant $L>0$, $C>0$, and $R>0$. With the applied source set to zero and the loop kept closed, the free response satisfiesThe angular frequency ${\omega }_0$ is the value for lossless free oscillation; $\alpha$ is the decay parameter. The three possibilities are underdamped RLC response for $\alpha <{\omega }_0$, critically damped RLC response for $\alpha ={\omega }_0$, and overdamped RLC response for $\alpha >{\omega }_0$. ^[ref].

For a fair comparison, start each circuit with the same electric charge $q(0)=q_0$ and zero electric current $q^{\prime }(0)=0$. Set $\tau ={\omega }_{0}t$ and $a=\alpha /{\omega }_0$; then $x=q/q_0$ obeys $d^{2}x/d{\tau }^2+2adx/d\tau +x=0$. The plot compares $a=0.5,1,2$. These dimensionless curves apply to any $L,C$ after rescaling time and electric charge.

The stored energy is the sum of magnetic energy and capacitor energy:Differentiating and using the circuit equation gives the complete power balance,In the free response, $E^{\prime }=-R{i}^2\le 0$: Joule heating accounts exactly for the lost energy. For the plotted initial conditions, $E(0)=q_0^2/(2C)$ and $E/E(0)=x^2+(dx/d\tau {)}^2$. Since every free solution tends to zero for $R>0$,In the lossless limit $R=0$, energy stays constant and shuttles between the capacitor and inductor. Thus an oscillation of electric charge and a decrease of total energy are compatible statements.

The analogy with a damped harmonic oscillator is exact within these ideal models: electric charge corresponds to displacement, electric current to velocity, inductance to mass, $1/C$ to spring constant, and $R$ to the coefficient of a resisting force proportional to velocity. The two stored energy terms play the roles of kinetic energy and elastic potential energy. ^[ref].

Differentiating the driven equation also gives $L{i}^{\prime \prime }+R{i}^{\prime }+i/C=V^{\prime }$. When using this equation to solve for electric current, retain the original constraint $L{i}^{\prime }(0)+Ri(0)+q(0)/C=V(0)$: the initial electric charge fixes the initial rate of change of electric current.

For a long vacuum-core or air-core solenoid with fixed geometry and linear magnetic response, neglecting end effects gives self-inductanceHere $N$ is the total number of turns, $n=N/\ell$ is the solenoid turn density, and $A$ is the interior cross-sectional area. The reason for the square of $N$ is physical: more turns produce more magnetic field, and more turns link the resulting magnetic flux. Approximately equal magnetic flux $\Phi \approx BA$ through each turn gives magnetic flux linkage $\Lambda \approx N\Phi \approx {\mu }_0{N}^{2}AI/\ell$; division by the electric current yields $L$. ^[ref].

For the same $N=1000$, $\ell =0.50\mathrm{m}$, and $a=0.025\mathrm{m}$, the area is $A=\pi (0.025{)}^2{\mathrm{m}}^2\approx 1.9635\times 10^{-3}{\mathrm{m}}^2$. The long-coil estimate isAt $I=0.50\mathrm{A}$, each turn links approximately $2.4674\times 10^{-6}\mathrm{Wb}$ and the total magnetic flux linkage is approximately $2.4674\times 10^{-3}\mathrm{Wb}$. Their ratio to $I$ is the stated self-inductance. A real finite coil need not give precisely these values: the axial centre correction above is not by itself a correction formula for inductance, which depends on magnetic flux across all the turns.

At fixed $A$ and $\ell$, doubling $N$ quadruples $L$. At fixed solenoid turn density $n$ and area $A$, doubling $\ell$ doubles both $N$ and $L$. These are different changes to the geometry, so there is no contradiction between $L\propto 1/\ell$ at fixed $N$ and $L\propto \ell$ at fixed $n$. A magnetic core requires its own magnetic permeability model; nonlinear magnetization can make a constant-$L$ description inadequate.

A distributed transmission line with per-unit-length series resistance $R^{\prime }$, series inductance $L^{\prime }$, shunt conductance $G^{\prime }$ and shunt capacitance $C^{\prime }$ obeys the displayed coupled first-order equations by applying Kirchhoff's voltage law and Kirchhoff's current law to a short element. With $R^{\prime }=G^{\prime }=0$, both voltage and current satisfy a wave equation with speed $1/\sqrt{L^{\prime }{C}^{\prime }}$ and characteristic impedance $\sqrt{L^{\prime }/C^{\prime }}$. The quantities $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$ are distributed parameters, not a single lumped resistor or capacitor.