LR Growth & Decay

Go round a loop containing a source of emf ℰ, a resistor R and an ideal inductor L, in the direction of the current. The resistor drops iR. The inductor is not dropping a potential because of resistance — it is generating an emf that opposes the change in its own flux, and while the current is rising that emf points backwards, so it enters as −L di/dt. The loop rule gives ℰ − iR − L di/dt = 0. This is not an equation you can solve for a number: it contains i and its own rate of change, so i must be a function of time. Two facts fall out before any integration. An instant after the switch closes the current is still zero, because a finite emf cannot change a current discontinuously, so the resistor drops nothing and the whole ℰ sits across the coil: di/dt = ℰ/L, which for 12.0 V and 0.40 H is 30 A s⁻¹. Long after, the current has stopped changing, L di/dt vanishes, and i = ℰ/R = 0.600 A.

Rearrange ℰ − iR = L di/dt into di/(ℰ − iR) = dt/L, with i on one side and t on the other. Integrating gives −(1/R) ln(ℰ − iR) = t/L + C. The initial condition i = 0 at t = 0 fixes C = −(1/R) ln ℰ, and subtracting leaves ln[(ℰ − iR)/ℰ] = −Rt/L. Exponentiate and rearrange: i(t) = (ℰ/R)(1 − e^(−Rt/L)). Always check a solution by putting it back. Differentiating gives di/dt = (ℰ/L) e^(−Rt/L), so iR + L di/dt = ℰ(1 − e^(−Rt/L)) + ℰ e^(−Rt/L) = ℰ for every t, and i(0) = 0 as required. Notice what the algebra did with L and R: they never appear separately in the exponent, only as the combination Rt/L. So 0.40 H with 20 Ω and 0.80 H with 40 Ω climb on exactly the same timetable — but not to the same place, since on 12.0 V one ends at 0.600 A and the other at 0.300 A. The exponent is blind to the difference; the destination is not.

Write the exponent as −t/τ with τ = L/R. It is a time: H ÷ Ω = (V s A⁻¹) ÷ (V A⁻¹) = s. Its meaning is geometric. The current starts with slope ℰ/L, and a straight line of that slope would reach the final value ℰ/R after a time (ℰ/R) ÷ (ℰ/L) = L/R — so τ is where the initial tangent crosses the asymptote, which is how you measure it off an oscilloscope trace. The curve itself is behind that tangent: at t = τ it has covered 1 − e⁻¹ = 63.2% of the gap, at 2τ 86.5%, at 3τ 95.0%, at 5τ 99.3%. Half the gap is closed at τ ln 2 = 0.693τ. For ℰ = 12.0 V, L = 0.40 H, R = 20 Ω this is τ = 20 ms, final current 0.600 A, 0.379 A at 20 ms, and 0.300 A at 13.9 ms. Doubling L doubles τ and leaves the destination alone; doubling R halves τ and halves the destination too.

A real coil is a length of wire, so the honest circuit model is an ideal inductance L in series with the coil's own resistance, and the battery brings an internal resistance of its own. All of them sit in the same loop and carry the same current, so all of them belong in the R that fixes both the final current and the time constant. Take the 0.40 H coil again, but with 4.0 Ω of winding resistance, an external 20 Ω and a battery of internal resistance 0.5 Ω. The total is 24.5 Ω, so the final current is 12.0 V ÷ 24.5 Ω = 0.490 A, not 0.600 A, and τ = 0.40 H ÷ 24.5 Ω = 0.0163 s = 16.3 ms, not 20 ms. The winding resistance is easy to get: put a DC ohmmeter across the coil, because once the current is steady the inductance contributes nothing and the coil is only its wire. That same fact is the standard long-time check — in the steady state, redraw the circuit with every inductor replaced by its own resistance.

Take the source away and close the loop through whatever resistance R remains. With ℰ = 0 the equation is L di/dt + iR = 0, whose solution is i(t) = I₀ e^(−t/τ) with τ = L/R for the new loop. The current is continuous through the switching instant: it starts the decay at exactly the value it had reached. That is the whole danger. Divert 0.600 A from a 0.40 H coil into a 2.0 kΩ resistor and at the first instant that resistor carries the full 0.600 A, so 1200 V appears across it, a hundred times the 12.0 V that drove the circuit. With the original 20 Ω still in the loop the decay runs on τ = L/R = 0.40 H ÷ 2.02 kΩ = 0.198 ms, so the surge is over inside a millisecond. Open the switch into air instead and R is effectively enormous: the emf climbs until the gap breaks down and the current continues as a spark. This is why inductive loads are switched with a freewheel diode or a shunt resistor across the coil — not to protect the coil, but to give its current somewhere to go.

Multiply the loop equation by i: ℰi = i²R + L i di/dt. The first term is the rate the source supplies energy, the second the rate the resistance turns it into heat, and the third is d/dt(½Li²), the rate the magnetic field is filling. The split is not fixed. The field's share, i(ℰ − iR), is zero at both ends of the growth and largest in the middle, at i = ℰ/(2R), where it equals ℰ²/(4R). For 12.0 V and 20 Ω that is 1.80 W, arriving when the source is delivering 3.60 W and the resistor is taking 1.80 W — heat and field exactly level. On the way back down the field pays for everything: ∫₀^∞ I₀²R e^(−2t/τ) dt = I₀²Rτ/2 = ½LI₀², since τ = L/R. With I₀ = 0.600 A and L = 0.40 H that is 0.072 J of stored energy and 0.072 J of heat, shared between the loop's resistances in proportion to their values.