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Induction and circuit approximations — Examples

1 · Prerequisites

2 · Summary

The companion collects the three leaf items of the pair in their dependency order. The sliding-bar thought experiment fixes an ideal sliding-contact model with separate segment velocities: the segmentwise electromotive force emf=−B0lx′(t) and the rectangular flux derivative are computed directly, a finite-volume bar force integral is evaluated from the adopted local matter coupling, and the signed instantaneous mechanical/electrical power balance is exhibited; self-fields, contacts, friction and bar dynamics are excluded. The two-circuit example proves the reciprocal inductive energy and power identity, the positive-semidefinite bound ∣M∣≤L1L2 for a smooth compact conserved-profile realization together with the null direction at perfect linkage, the explicitly adopted common-flux ideal-transformer relations and their reversed-winding sign, and it proves by a derivative-vector argument that the perfect common-flux voltage ratio is not a universal identity under strict linkage. The RL example verifies the constant-voltage exponential transient, its time constant L/R, the unique stationary solution and the qualified limit statement, with uniqueness inherited from the RL theorem. Each computation is exact and conditional on its declared idealisation, and no measurement is involved.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

Thought experimentStatement: AI-adaptedProof: AI-adaptedOpen item page →

Sliding-bar sliding-contact model: segmentwise emf, force integral and power balance

Physical scope

A hypothetical conditional sliding-contact model in a uniform static external test field: four separate segment velocities at sliding contacts, the direct segmentwise emf and flux computation, an adopted finite-volume bar force integral and the signed instantaneous mechanical/electrical power balance. Self-fields, radiation, contact resistance and dynamics are neglected; no measurement, no bar equation of motion and no total Maxwell or matter-dynamical solution is asserted.

Statement

Assume ACω for the explicitly used Lebesgue bar-volume interfaces (The Axiom of Countable Choice (ACω)). Work in the laboratory frame and units of Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields and Framework-local SI units, the vacuum constants and the exact/measured status of their values, choosing Ω=R3 and an open laboratory time interval containing J. Fix l>0, B0≥0, a nonempty interval J with a specified C1 neighbourhood extension for motion and its velocity jets when J is not open (in particular singleton data are read through that supplied extension), and a prescribed C1 motion x:J→(a0,a1) with a0<a1 and ∣x′(t)∣<c. The fixed rails have centerlines (s,±l/2,0), a0≤s≤a1, and the left closing side is (a0,w,0), ∣w∣≤l/2. At time t the ideal oriented contour C(t) comprises only the rail portions a0≤s≤x(t), that left side, and the sliding bar centerline (x(t),w,0). Give it normal e3, hence traverse the bar in the +e2 direction. The rails and closing side have material velocity zero; the bar velocity is vb(t)=x′(t)e1. These are separate segment velocities with sliding contacts; endpoint values at the contacts are taken independently on each segment. Define the contact-model emf as the sum of its four segment integrals ∑γ∫γ(E+vγ×B)⋅dr in Scalar line integrals with respect to arc length and vector-field line integrals. This is not a single-valued smooth moving material sheet of C2 moving material circuit: swept patch, single-valued material velocity, flux and electromotive force. Prescribe an arbitrary finite-valued signed current I:J→R, in amperes; no time regularity of I is needed for these instantaneous identities. The external test fields are E(y,t)=0, B(y,t)=B0e3 on the laboratory region; all self-fields are neglected.

For a measure space (X,A,μ), a measurable function f:X→R and K∈A, write ∫Kf dμ:=∫(fχK) dμ using the signed Lebesgue integral of Integrable real and complex functions, and their integrals: (fχK)+=max⁡{fχK,0} and (fχK)−=max⁡{−fχK,0}, and the integral is their integral difference when both are finite. For a measurable vector map G:X→R3 with all three components integrable on K, define ∫KG dμ:=(∫KG0 dμ,∫KG1 dμ,∫KG2 dμ) in the ordered spatial coordinates of the laboratory frame; the stored coordinate indices are 0,1,2 and the corresponding physical basis labels are e1,e2,e3.

For force bookkeeping choose an actual finite-volume homogeneous bar, separate from the ideal centerline: fix d,h>0 with cross-section area A=dh and set Kt=[x(t)−d/2,x(t)+d/2]×[−l/2,l/2]×[−h/2,h/2], so λ3(Kt)=Al. For each frozen time prescribe the uniform current density jt=(I(t)/A)e2 on its interior, with the ideal local conductor-matter coupling of The Lorentz coupling: external force on a test charge and the continuum force and power densities(ii) applied there. Its extension to the boundary is used only for the volume integral; contacts/end faces have zero volume and no additional contact force is included. The same signed lumped current runs around the contour. The local force-law adoption and prescribed current do not assert a total Maxwell or matter-dynamical solution.

(a) The rectangular flux is Φ(t)=B0l(x(t)−a0) and the segmentwise emf is emf(t)=−B0lx′(t)=−Φ′(t). Its magnitude is B0l∣x′∣, including either motion direction and B0=0.

(b) The finite-volume force is F(t)=∫Ktjt×B dλ3=I(t)B0le1.

(c) The signed instantaneous balance is F(t)⋅vb(t)+emf(t)I(t)=I(t)B0lx′(t)−B0lx′(t)I(t)=0. It is ideal mechanical/electrical transfer bookkeeping with prescribed motion and current; no dissipative, radiation or field-storage term is included. A support enforcing the motion may supply the opposite mechanical power.

(d) This is a hypothetical conditional sliding-contact model with no measurement. Self-fields, radiation, friction, contact resistance, deformation and constitutive details are neglected; no bar equation of motion, physical realization, or total Maxwell solution is asserted. The strict C2 material-sheet theorem remains a separate result under its own hypotheses.

Facts & Assumptions

Given: Assume ACω, the laboratory frame and units, the constants l>0, B0≥0, d,h>0, A=dh, the C1 motion x:J→(a0,a1) with ∣x′∣<c, the prescribed current I:J→R and the contour and bar data of the Statement.

[A1]

Countable Choice, written ACω, says every sequence of nonempty sets has a choice function. (The Axiom of Countable Choice (ACω)).

[F1]

The line integral of a continuous vector field along a piecewise C1 path is the sum of the integrals ∫titi+1⟨F(γ(t)),vi(t)⟩ dt over an admissible partition. (Scalar line integrals with respect to arc length and vector-field line integrals).

[F2]

The cross product in coordinates is (u×v)x=uyvz−uzvy and cyclically; the standard basis satisfies e1×e3=−e2. (The cross product in R3).

[F3]

The Euclidean inner product is ⟨x,y⟩=∑kxkyk, in particular ⟨e2,e2⟩=1. (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn).

[F4]

For a nonnegative measurable f:X→[0,+∞] and measurable E, ∫Ef dμ=∫fχE dμ. (Integral over a measurable subset).

[F5]

For a nonnegative simple measurable function s=∑jcjχEj, the simple integral is ∑jcjμ(Ej), and the nonnegative integral of a nonnegative simple function equals its simple integral. (The integral of a nonnegative simple function, The nonnegative integral agrees with the simple integral on simple functions).

[F6]

The Lebesgue integral is linear on L1. (The Lebesgue integral is linear on L1(μ)).

[F7]

Every set between the open and closed box forms is Lebesgue measurable with λn(R)=∏i(bi−ai) ; in particular closed boxes have the product volume and their faces are null when some side degenerates. (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included).

[F9]

Fields are maps Ω×I→R3; E carries V m−1, B carries T. (The electric and magnetic fields as fixed-time spatial vector fields).

[F10]

The framework adopts the SI constants including c>0 and the units V, T; the emf unit is the volt. (Framework-local SI units, the vacuum constants and the exact/measured status of their values).

[F11]

The adopted continuum matter coupling gives the force density ρE+J×B (N m−3) for matter interacting with the field through an explicitly stated matter model. (The Lorentz coupling: external force on a test charge and the continuum force and power densities).

[F12]

A moving material circuit of the strict branch is a C2 embedded sheet with a single-valued C1 material velocity; the strict theorem on this page applies only to such sheets. (C2 moving material circuit: swept patch, single-valued material velocity, flux and electromotive force).

[F13]

The laboratory region, its chart and the spatial frame are those of the laboratory geometry; laboratory fields are maps on Ω×I with the spatial coordinates of the frame. (Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields).

[F14]

For a measurable real function f, put f+=max⁡{f,0} and f−=max⁡{−f,0}. It is integrable when ∫∣f∣<∞, equivalently both part integrals are finite; then ∫f=∫f+−∫f−. (Integrable real and complex functions, and their integrals).

Proof

technique · direct
1.1F1F2F3algebra

Fix t∈J. By [F1] the contact-model emf is the sum of the four segment integrals of E+vγ×B with the segmentwise velocities. On the two rail portions and the left closing side the material velocity is zero and E=0, so the integrand vanishes identically and each of those three integrals is 0. On the bar, parametrized by s↦(x(t),s,0) with s from −l/2 to l/2 in the +e2 direction, the velocity is vb=x′(t)e1, so [F2] gives vb×B=x′(t)e1×B0e3=−B0x′(t)e2 and [F3] gives ⟨vb×B,e2⟩=−B0x′(t); the bar integral is therefore ∫−l/2l/2(−B0x′(t)) ds=−B0lx′(t). Hence emf(t)=−B0lx′(t) for every t∈J.

1.2F2F4F5F6F7F11F14algebra

For the force choose the constant extension of jt=(I(t)/A)e2 to Kt and zero outside Kt. This boundary convention adds no contact force; the faces are null by [F7]. Put qt:=I(t)B0/A∈R. On the interior the adopted coupling [F11] gives jt×B0e3=qte1 by [F2], so its volume-integrand extension has coordinate tuple (qtχKt,0,0) in the Statement's convention. By [F7], Kt is measurable and λ3(Kt)=dlh=Al<∞. By [F4] and [F5], ∫χKt=Al and ∫∣qtχKt∣=∣qt∣Al<∞, so [F14] gives scalar integrability for every sign of I(t). Applying [F6] to this scalar multiple of the integrable indicator and then the componentwise vector-integral definition yields F(t)=qtAle1=I(t)B0le1. Thus the vector integral is finite component by component. This is the prescribed local matter coupling applied to the declared bar model; it asserts no force on an arbitrary source without such a matter model.

2.1F3F7F8step 1.1algebra

Put Dt=[a0,x(t)]×[−l/2,l/2]⊆R2 and ψ(s,w)=(s,w,0). Then ψs×ψw=e3, so the parametrized rectangular flux is Φ(t)=∫Dt⟨B0e3,e3⟩ dλ2=B0λ2(Dt)=B0l(x(t)−a0) by [F7]. Thus λ2 is used on the two-dimensional parameter rectangle; its pushforward by ψ is the plane-area convention. With the prescribed C1 neighbourhood motion, [F8] gives Φ′(t)=B0lx′(t), with one-sided endpoint values and the specified extension jet at a singleton. Step 1.1 then gives emf=−Φ′ and ∣emf∣=B0l∣x′∣.

2.2F3step 1.1step 1.2algebra

The signed power balance is immediate from steps 1.1 and 1.2 and the Euclidean pairing: with vb=x′(t)e1 and ⟨e1,e1⟩=1 by [F3], F(t)⋅vb(t)=I(t)B0lx′(t) and emf(t)I(t)=−B0lx′(t)I(t), so the sum is 0 for every sign of I(t) and of x′(t), including the zero values.

3.1A1F9F10F12F13step 1.1step 1.2step 2.1step 2.2∎

The three displayed conclusions are established by steps 1.1, 1.2, 2.1 and 2.2. The prescribed constant fields are C∞, and their types and units are those of [F9] and [F10]; the model is the hypothetical sliding-contact model of the Statement, not a single-valued smooth material sheet of [F12], and its strict theorem is not used; the laboratory reading is that of [F13]. No measurement, self-field, radiation, dissipative or dynamical conclusion is asserted, and ACω is used only through the explicitly cited box-measure and integral interfaces.

Source notes

Griffiths §7.1.3, printed pp. 294-296, presents the primitive generator: a loop pulled at speed v through a uniform field, emf =vBh, the flux rule E=−dΦ/dt for motional emf and the work/power discussion; Tong §4.1.1 and §4.1.3, printed pp. 69-77, gives the sliding bar E=vBd, the Lorentz force F=IBd on the bar and the coupled motion with the resistor. The present thought experiment extracts only the instantaneous idealised statements: the segmentwise emf, the flux function and its derivative, the adopted finite-volume force integral and the signed power balance. The coupled bar dynamics and the Joule-heating energetics of the sources are not reproduced, because they require a mechanical mass model and a dynamical circuit solution that this item does not adopt.

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Two-circuit mutual inductance, the perfect-linkage bound and ideal-transformer limitations

Physical scope

Conditional two-circuit bookkeeping and smooth-profile realization: the reciprocal energy and instantaneous power, the exact positive-semidefinite bound ∣M∣≤L1L2 and the sign-correct perfect-linkage null direction for either winding orientation, the positive common-flux model and its signed reversal, and failure of a universal voltage ratio under strict linkage with accidental histories retained. No physical leakage or loss mechanism and no measurement is asserted.

Example

Assume the lumped circuit model of Lumped circuit approximation: branch data, terminal laws and the declared quasistatic regime with the units of Framework-local SI units, the vacuum constants and the exact/measured status of their values, and let L=(L1MML2) be a constant symmetric block of its inductance matrix with L1,L2>0. Let I1,I2:J→R be C1 currents on a nonempty interval J (each extending to a C1 map on an open interval containing J, as in the model's conventions) with terminal voltages Vα=∑βLαβI˙β. Assume Countable Choice (ACω, The Axiom of Countable Choice (ACω)) for the realization and bounded-linking clauses (b), inherited from the linkage and energy suppliers.

(a) [Energy and instantaneous power] the stored inductive bookkeeping energy is W(t)=12L1I1(t)2+12L2I2(t)2+MI1(t)I2(t), and V1I1+V2I2=W˙ by part (a) of Restricted inductor energy LI^2/2 and its identification with the frozen magnetic field energy; the mutual term increases the energy for currents of the same sign and decreases it for opposite signs (as the sign of the product I1I2 indicates when M>0).

(b) [Perfect linkage and the bound] if the two currents are realised by fixed compact smooth conserved profiles of the class of The steady-current (magnetostatic) regime, the compact smooth current class and the declared ideal noncompact models(b) with linkage matrix L as in Reciprocal linkage coefficients and magnetic energy of smooth compact conserved currents, then W(t)≥0 for all currents, and evaluating the quadratic form on both (L2,−L1) and (L2,+L1) gives the bound ∣M∣≤L1L2; moreover if M=sL1L2 with s∈{+1,−1} (perfect linkage), the null direction I=(L2,−sL1) has JI=0 pointwise by the vanishing clause of Reciprocal linkage coefficients and magnetic energy of smooth compact conserved currents(c).

(c) [Ideal common-flux model] if M=L1L2, choose any Λ>0 in henries and set N1:=L1/Λ, N2:=L2/Λ; then L1=ΛN12, L2=ΛN22 and M=ΛN1N2. Writing Φc:=Λ(N1I1+N2I2) for the common flux linkage, the terminal voltages are V1=N1Φ˙c,V2=N2Φ˙c,i.e.V1N1=V2N2=Φ˙c, and the stored energy is W=Φc2/(2Λ); equivalently the induced emfs eα:=−Vα satisfy e1/N1=e2/N2=−Φ˙c. Here N1,N2>0 are dimensionless real effective winding factors, rather than asserted integer turn counts; Φc has unit Wb. This is the explicitly adopted common-linkage ('ideal transformer') model with effective ratio N1:N2.

(d) [Limitations] for an imperfect realized linkage, ∣M∣<L1L2, the perfect common-flux voltage ratio is not an identity for all current histories; a particular history can nevertheless satisfy that ratio accidentally (including both voltages zero). Reversing the second winding retains s=−1: with M=sL1L2 put Φc=Λ(N1I1+sN2I2) and obtain V1/N1=V2/(sN2)=Φ˙c, with the same energy Φc2/(2Λ). The positive-orientation model (c) remains as stated. Winding resistance, hysteresis and core loss, stray capacitance, magnetising current and finite permeability are outside the declared model and are not inferred from it. No measurement is involved; the statements are conditional computations in the ideal model. Regularity on a non-open interval uses the stated C1 neighbourhood extension; if the interval is a singleton its terminal-voltage jets are those of a specified extension, not derivatives inferred from its value restriction.

Verification

technique · direct expansion, the reciprocity theorem, and an explicit derivative-vector contradiction for the strict-linkage non-identity

Given: The lumped model data, the constant symmetric block L=(L1MML2) with L1,L2>0, the interval J, the C1 currents I1,I2 with open-neighbourhood extension, the terminal voltages Vα=∑βLαβI˙β and ACω for clauses (b).

[A1] Countable Choice, written ACω, says every sequence of nonempty sets has a choice function. (The Axiom of Countable Choice (ACω)).

[F1] The adopted lumped model carries the Faraday branch relation V=LI˙: each terminal potential is the rate of change of the branch flux linkage LI; the matrix and terminal data are declared model data. (Lumped circuit approximation: branch data, terminal laws and the declared quasistatic regime).

[F2] For constant symmetric L and a C1 current I with open-neighbourhood extension, ∫t0t1⟨V,I⟩ dt=12⟨I(t1),LI(t1)⟩−12⟨I(t0),LI(t0)⟩ with V=LI˙, and for a smooth compact conserved-profile realization the quadratic form equals the frozen magnetic field energy 12μ0∫∣BJI(t)∣2 at each time. (Restricted inductor energy LI^2/2 and its identification with the frozen magnetic field energy).

[F3] For fixed compact smooth conserved profiles j1,…,jn with linkage matrix L=(Lαβ) one has U(I)=12∫⟨JI,AJI⟩=12∑α,βLαβIαIβ=12μ0∫∣BJI∣2≥0, and U(I)=0 exactly when JI=∑αIαjα=0 pointwise; the matrix is positive definite exactly when the profiles are pointwise linearly independent. (Reciprocal linkage coefficients and magnetic energy of smooth compact conserved currents).

[F4] The compact smooth current class consists of the currents J∈Cc∞(Ω;R3) with div⁡J=0, extended by zero to R3. (The steady-current (magnetostatic) regime, the compact smooth current class and the declared ideal noncompact models).

[F5] The Euclidean inner product is ⟨x,y⟩=∑kxkyk; in particular for I=(I1,I2) and the block L one has ⟨I,LI⟩=L1I12+2MI1I2+L2I22. (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn).

[F7] The SI units are those of the framework: V in volts, I in amperes, t in seconds, L in henries and M in henries. (Framework-local SI units, the vacuum constants and the exact/measured status of their values).

1.1F1F2F5F6algebra

Clause (a): by [F5], 12⟨I(t),LI(t)⟩=12L1I12+MI1I2+12L2I22, which is the displayed W(t). Differentiating termwise with the product and scalar rules of [F6] gives W˙=L1I1I˙1+M(I˙1I2+I1I˙2)+L2I2I˙2=(L1I˙1+MI˙2)I1+(MI˙1+L2I˙2)I2=V1I1+V2I2, since [LI˙]α=Vα; the sign reading of the mutual term is immediate from W=12L1I12+12L2I22+MI1I2 for M>0.

1.2A1F3F4algebracases

Clause (b): under the realization hypothesis, W(t)=12∑α,βLαβIα(t)Iβ(t)=U(I(t))≥0 at each t by [F3], the block L being by hypothesis the linkage matrix of the two profiles. Substituting I=(L2,−L1) gives W=L1L2−ML1L2≥0, that is M≤L1L2; substituting I=(L2,+L1) gives W=L1L2+ML1L2≥0, that is M≥−L1L2; together, ∣M∣≤L1L2. If M=sL1L2 and I=(L2,−sL1), the same expansion gives W=L1L2−sML1L2=L1L2−L1L2=0, so by the vanishing clause of [F3] the associated current JI vanishes pointwise.

1.3F5F6algebra

Clause (c): with N1=L1/Λ and N2=L2/Λ one has L1=ΛN12, L2=ΛN22 and M=L1L2=ΛN1N2 because N1N2=L1L2/Λ. Hence, using Vα=∑βLαβI˙β, V1=L1I˙1+MI˙2=ΛN1(N1I˙1+N2I˙2)=N1Φ˙c and V2=MI˙1+L2I˙2=ΛN2(N1I˙1+N2I˙2)=N2Φ˙c with Φc=Λ(N1I1+N2I2); dividing by N1,N2>0 gives the common ratio, and eα=−Vα reverses the common sign. For the energy, W=12L1I12+MI1I2+12L2I22=12Λ(N1I1+N2I2)2=Φc2/(2Λ) by the displayed relations and the binomial expansion.

1.4F5F6algebracases

Clause (d), strict-linkage non-identity: suppose ∣M∣<L1L2 and V2=kV1 held for all C1 currents on J; then ∑βL2βI˙β=k∑βL1βI˙β for every such current. Fix any point t0∈J (any point when J is a singleton) and, for each vector a=(a1,a2)∈R2, take the affine history I(t)=c+(t−t0)a (any fixed c∈R2) on J, which is the restriction of the affine map of t on R and hence carries the stated C1 open-neighbourhood extension with I˙≡a; substituting gives (L21−kL11)a1+(L22−kL12)a2=0 for all a, hence L21=kL11 and L22=kL12, i.e. row⁡2L=krow⁡1L. Then L1k=M and L2=kM=M2/L1, so L1L2−M2=0, contradicting L1L2−M2>0 from ∣M∣<L1L2; hence no such k makes the ratio a universal identity. Accidental satisfaction occurs: any history with I˙ in the kernel of the nonzero functional a↦(L21−kL11)a1+(L22−kL12)a2 satisfies V2=kV1 at every time, that kernel containing the nonzero vector (L22−kL12,−(L21−kL11)) because the row difference is nonzero; in particular the constant currents give V1=V2=0. For the reversed winding with M=sL1L2, put Φc=Λ(N1I1+sN2I2); then V1=L1I˙1+MI˙2=ΛN1(N1I˙1+sN2I˙2)=N1Φ˙c and V2=MI˙1+L2I˙2=ΛsN2(N1I˙1+sN2I˙2)=sN2Φ˙c because s2=1, so V1/N1=V2/(sN2)=Φ˙c, and W=12Λ(N1I1+sN2I2)2=Φc2/(2Λ).

2.1A1F1F7step 1.1step 1.2step 1.3step 1.4∎

Steps 1.1-1.4 establish (a)-(d): steps 1.1-1.2 give (a) and the bound and null-direction clauses of (b), and steps 1.3-1.4 give the common-flux model (c) and the limitations (d). The computation is conditional on the adopted lumped model [F1], its units are [F7], and no leakage, loss, material or measurement claim is added; the terminal-voltage jets on a singleton interval are those of the specified extension. Countable Choice is used only through the realization clause (b), via [F3]; the algebraic clauses (a), (c) and (d) are choice-free.

Source notes

Fitzpatrick §7.4, printed pp. 222-225, computes the two-solenoid mutual inductance and records perfect linkage M=L1L2 with the leakage warning; Griffiths §7.2.3, printed pp. 310-317, derives reciprocity and the Neumann formula; Tong §4.1.2, printed pp. 71-73, gives the single-coil energy and its field-energy representation; the mutual term here is obtained by local expansion of the linkage quadratic form. The example separates what is proved for the declared smooth compact profiles (the PSD bound and the null direction) from what is adopted as the common-flux ideal-transformer model, and it proves the failure of a universal voltage ratio under strict linkage by exhibiting the forced row proportionality and its contradiction; no physical transformer behaviour, loss mechanism or measurement is asserted.

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Exponential RL response under a constant terminal voltage

Physical scope

The exponential transient of the scalar RL circuit of the adopted lumped model under a constant source voltage and constant positive L and R, with time constant L/R. It carries the model idealisations (fixed geometry, ideal elements, no radiation or distributed effects) and contains no measurement or empirical claim.

Example

In the RL branch of Restricted RL decay and LC oscillation under the adopted circuit model with the declared units of Framework-local SI units, the vacuum constants and the exact/measured status of their values, let the source voltage be the constant V0∈R, let R>0 and L>0, let t0∈J and I0∈R, with J an interval and derivatives read in the model's domain-relative convention, through the stated C1 open-neighbourhood extension for a non-open J. For nondegenerate J the unique C1 solution of LI˙+RI=V0, I(t0)=I0, is I(t)=V0R+(I0−V0R)exp⁡ ⁣(−R(t−t0)L),t∈J. In particular I(t)−V0/R=(I0−V0/R)exp⁡(−R(t−t0)/L) and hence ∣I(t)−V0/R∣=∣I0−V0/R∣exp⁡(−R(t−t0)/L) for every t∈J, with time constant τ=L/R; I≡V0/R is the unique stationary solution (an extension with zero derivative, also in the singleton convention); if J is unbounded above, then I(t)→V0/R as t→+∞ along J; and the transient part is the free decay exp⁡(−R(t−t0)/L) of the initial deviation I0−V0/R. For J={t0} the restriction is I(t0)=I0 and the jets are those of the same formula read through the stated extension, giving I˙(t0)=(V0−RI0)/L. The computation is conditional on the adopted lumped model, constant elements and fixed geometry; no measurement is involved.

Verification

technique · explicit differentiation of the candidate, with uniqueness inherited from the RL theorem

Given: The RL branch data of the adopted lumped model with constants V0∈R, R>0, L>0, an interval J with t0∈J and initial value I0∈R.

[F1] In the RL branch let L>0, R≥0 and let V:J→R be continuous. Then the initial value problem LI˙+RI=V, I(t0)=I0, has exactly one C1 solution on every nondegenerate interval J, namely I(t)=exp⁡(−R(t−t0)/L)[I0+L−1∫t0texp⁡(R(s−t0)/L)V(s) ds]; for J={t0} the restriction is I(t0)=I0 and the canonical extension for the constant source V(t0) has jet I˙(t0)=(V(t0)−RI0)/L; the RL solution uses no choice principle. (Restricted RL decay and LC oscillation under the adopted circuit model).

[F2] The real exponential is C∞ and exp⁡′=exp⁡. (The exponential function is smooth and (exp⁡)′=exp⁡).

[F3] Sums, scalar multiples and products of differentiable functions are differentiable, with the sum, scalar and product rules; the derivative of a constant is 0 and the derivative of the identity is 1, since their difference quotients are respectively 0 and 1. (Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0).

[F4] If f is totally differentiable at a and g is totally differentiable at f(a), then g∘f is totally differentiable at a with D(g∘f)(a)=Dg(f(a))∘Df(a). (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

[F5] exp⁡(x)→+∞ as x→+∞ and exp⁡(x)→0 as x→−∞. (The exponential tends to +∞ at +∞ and to 0 at −∞).

[F6] For every real x, exp⁡(x)>0 and exp⁡(−x)=1/exp⁡(x). (The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x)).

[F7] The SI units are those of the framework: V in volts, I in amperes, t in seconds, R in ohms and L in henries. (Framework-local SI units, the vacuum constants and the exact/measured status of their values).

[F8] A subset of R is an interval when it has one of the nine order-convex forms of the definition; an interval is unbounded above exactly when it contains a ray [r,+∞). (Intervals of R: the nine order-convex forms, nondegeneracy, and length).

1.1F2F3F4algebra

Define the candidate φ(t):=V0R+(I0−V0R)exp⁡(−R(t−t0)L) on J. The function t↦−R(t−t0)L is affine with derivative −RL by [F3], and exp⁡′=exp⁡ by [F2], so the chain rule [F4] gives φ′(t)=(I0−V0R)(−RL)exp⁡(−R(t−t0)L)=−RL(φ(t)−V0R). Therefore Lφ′(t)+Rφ(t)=−R(φ(t)−V0R)+Rφ(t)=V0 identically, and φ(t0)=V0R+(I0−V0R)=I0.

1.2F3algebra

Stationary solutions: if I≡c is constant, then LI˙+RI=Rc, so the equation LI˙+RI=V0 holds exactly when c=V0/R; since R>0 there is exactly one such constant, namely I≡V0/R.

1.3F5F8algebra

Limit statement: if J is unbounded above, then for t∈J∩[t0,+∞) the argument −R(t−t0)L satisfies −R(t−t0)L→−∞ as t→+∞ because R/L>0, so [F5] gives exp⁡(−R(t−t0)L)→0 and hence I(t)→V0/R along that ray; the time constant is τ:=L/R, so the decay factor is exp⁡(−(t−t0)/τ).

1.4F6algebra

Deviation identity: subtracting V0/R from the displayed formula gives I(t)−V0/R=(I0−V0R)exp⁡(−R(t−t0)L); since exp⁡(x)>0 for real x by [F6], taking absolute values preserves the exponential factor, so ∣I(t)−V0/R∣=∣I0−V0R∣exp⁡(−R(t−t0)L) for every t∈J.

2.1F1step 1.1algebracases

Identification of the solution: by step 1.1 the candidate φ is a C1 solution of the initial value problem (its derivative exists on all of J through the stated extension convention) and by [F1] the initial value problem with the constant source V≡V0 has exactly one C1 solution on every nondegenerate J; hence the displayed I equals φ there and is the unique solution, and for J={t0} the restriction and jets are those of [F1] with V(t0)=V0, namely I(t0)=I0 and I˙(t0)=(V0−RI0)/L.

3.1F1F7step 1.1step 1.2step 1.3step 1.4step 2.1∎

Steps 1.1-1.4 and 2.1 establish the formula, its uniqueness, the deviation identity with time constant τ=L/R, the unique stationary solution, the limit behaviour on intervals unbounded above, and the singleton reading. The computation uses only the adopted lumped model and its constant positive elements, with units [F7]; no radiation, distributed-element, material or measurement claim is added, and the RL existence and uniqueness supplier is the choice-free clause (a) of [F1].

Source notes

Griffiths §7.2.3, printed pp. 310-317, records the battery-plus-RL solution I(t)=E0/R+ke−Rt/L with time constant τ=L/R; Fitzpatrick §7.5, printed pp. 225-231, writes the RL equation and its exponential approach to the steady current. The example verifies the constant-source candidate by explicit differentiation and inherits uniqueness from the RL theorem; no other current history, transient measurement or numerical value is asserted.

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