The current that cannot jump

Two snapshots settle most LR problems before any exponential is written, and both come from $V_L = L\,dI/dt$.

1. Just after switching, $t = 0^+$. The inductor's current is whatever it was just before. In a fresh circuit that is zero, so the coil behaves as an open branch: no current through it, and the full voltage the loop demands sits across it. Redraw the circuit with the inductor removed and solve.
2. Long after, $t \to \infty$. Every current has settled, $dI/dt = 0$, and the ideal inductor drops zero volts. Redraw it as a plain wire and solve again.

The capacitor did exactly the reverse: wire at $t = 0$, open at $t \to \infty$. Students who remember only that "there is a pair of endpoints" import the capacitor's pair unmirrored and get both snapshots wrong. Say the mirror out loud once: capacitor (wire, open), inductor (open, wire).

Worked once with numbers: a $12$ V battery, a $4.0\ \Omega$ resistor in series, then a $4.0\ \Omega$ resistor and an ideal inductor in parallel with each other. At $t = 0^+$ the inductor branch is open, so the battery sees $8.0\ \Omega$ and drives $1.5$ A, all through the parallel resistor. Long after, the inductor is a wire that shorts the parallel resistor, the battery sees $4.0\ \Omega$ and drives $3.0$ A, all of it through the coil.

Series $\varepsilon$, $R$, $L$, switch closed at $t = 0$. Walk the loop in the direction of the current:

$$\varepsilon - IR - L\frac{dI}{dt} = 0 \quad\Longrightarrow\quad \frac{dI}{dt} = \frac{\varepsilon - IR}{L}.$$

Separate and integrate from $I = 0$ at $t = 0$:

$$I(t) = \frac{\varepsilon}{R}\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{L}{R}.$$

Check the endpoints against the snapshots: $I(0) = 0$ and $I(\infty) = \varepsilon/R$. Check the start against the loop rule: at $t = 0$ the whole $\varepsilon$ is across the coil, so the initial slope is $\varepsilon/L$, and the tangent there reaches the final current at exactly $t = \tau$. Differentiate to get the inductor voltage, $V_L = \varepsilon e^{-t/\tau}$, which starts at the full battery voltage and decays.

The time constant is a quotient. Henry per ohm is a second; henry times ohm is not any unit of time. So doubling $R$ halves $\tau$: the current heads toward a smaller final value and gets there sooner. This is opposite to $RC$, where more resistance slows the charging. Park the two side by side once, $\tau_{RC} = RC$ and $\tau_{LR} = L/R$, and let the units referee.

Decay is the same equation with $\varepsilon$ removed. Disconnect the battery and let the coil drive its current $I_0$ around a loop of resistance $R$: $-IR - L\,dI/dt = 0$ gives $I(t) = I_0 e^{-t/\tau}$ with the same $\tau = L/R$.

The state variable of an inductor is its current, the way a capacitor's is its voltage. Across any switching instant,

$$I_L(0^+) = I_L(0^-).$$

A jump would mean $dI/dt \to \infty$ and so $V_L \to \infty$, and no real circuit supplies that. So freeze the coil's current across the switching moment, then let the new circuit evolve it.

The consequences are not small. Take the circuit above at its long-time state, $3.0$ A through the coil, and open the battery switch. If a $20\ \Omega$ resistor sits in the loop the coil can now drive, the $3.0$ A continues around that loop at the first instant, and the resistor drops $3.0 \times 20 = 60$ V, five times the battery voltage that built the current. If instead the only path is an air gap at the switch, the coil raises its EMF until the gap breaks down, and the arc is the current continuing. That is why an inductive load sparks its switch on opening and not on closing.

The error to watch is letting the current teleport: zeroing it the instant the switch opens, or jumping it straight to the new steady value. Both throw away the one quantity the physics promises will not change.

Multiply the growth loop equation by $I$:

$$\varepsilon I = I^2R + LI\frac{dI}{dt}.$$

Battery power on the left; on the right, heat in the resistor plus the rate at which energy is banked in the field. Integrate over the whole build-up and the last term gives $\tfrac12 LI_f^2$. So the battery delivers more than the field ends up holding: the field's $\tfrac12 LI_f^2$ plus every joule the resistor turned to heat along the way. Nothing is double-booked, and the inductor is not free to start.

At shutdown the ledger runs the other way. Let $I_0$ decay through resistance $R$:

$$\int_0^{\infty} I^2R\,dt = I_0^2R\int_0^{\infty} e^{-2t/\tau}\,dt = I_0^2R\,\frac{\tau}{2} = \tfrac12 LI_0^2.$$

The resistor receives exactly the field's energy, to the joule. If the problem offers no resistor, the withdrawal is made somewhere else: the arc across the switch, or the capacitor of an LC loop. Energy in an inductor's field never simply ends when a switch is thrown.