Solve an LR circuit without erasing its initial current

Define the passive voltage and initial state

An ideal source E, resistor R and constant inductor L form one series loop. Choose current in the direction driven by the source. The passive inductor voltage drop is vL = L dI/dt, while its induced emf around the current direction is −L dI/dt. The loop equation is E = R I + L dI/dt. An initial current may be positive, negative or above the eventual source-set current; it cannot be discarded merely because a new interval begins.

Derive the exponential offset

Let Iinf = E/R and D = I0 − Iinf. Then d(I − Iinf)/dt = −(R/L)(I − Iinf), giving I(t) = Iinf + D exp(−t/tau), tau = L/R. Differentiation verifies the equation and substitution at zero verifies the initial condition. The tangent at zero reaches the final asymptote after one tau if extended linearly; the actual curve only approaches it. If D = 0, the state is already steady and there is no transient.

Use the 63 and 37 percent statements conditionally

From I0 = 0 with a constant source, I(tau) = Iinf(1 − exp(−1)), about 63 percent of the final current. In a separate source-free discharge, I(tau) = I0 exp(−1), about 37 percent of the initial current. An arbitrary nonzero-current driven case instead reduces its difference from the asymptote to 37 percent. At steady DC the ideal inductor has zero voltage drop, but a real winding resistance is not thereby erased.

Integrate a signed energy account

Source work is integral E I dt; resistor heat is integral R I² dt; magnetic-energy change is L[I(t)² − I0²]/2. Multiplying the differential equation by I establishes source power = heat rate + stored-energy rate at every instant. The model integrates all three terms. For an initially overdriven current, the inductor can release energy while the source continues to supply positive work. Stored energy is the square of an offset exponential, not generally a single exponential with the same time constant.

Qualify the initial back-emf shortcut

When I0 = 0, the initial resistor drop is zero and the passive inductor voltage equals the applied source voltage. For a nonzero initial current, vL(0) = E − R I0 instead. A blanket claim that the inductor always takes the entire source voltage at a switching instant would contradict the actual branch current and loop equation.

Worked original case

E = 6 V, R = 2 ohm, L = 2 H and I0 = 4 A give tau = 1 s and Iinf = 3 A. At two tau, I ≈ 3.135335 A and vL ≈ −0.270671 V. Source work ≈ 41.187988 J and resistor heat ≈ 47.357661 J; the difference is the stored-energy change ≈ −6.169673 J, not a conservation failure.

Investigation

Derive and differentiate the general solution, then compare initial currents below, at and above E/R. Retain source work, heat and stored-energy change over a finite interval. Explain why the initial-voltage and 63-percent shortcuts apply only to their stated special cases.

Changed-case practice

Sources and boundaries

Original teaching checked 31 October 2026 against current official Unit 13 and the relevant OpenStax discussion. No source exercises or diagrams are reproduced. All records and apparatus are virtual, supplied or generated; no real wiring, switching, magnetic braking or electrical measurement is requested. Public feedback and retained work do not establish independent mastery, supervised practical participation or AP authorization.