An inductor opposes rapid changes in current because changing current changes magnetic flux and produces an induced voltage. In a resistor-inductor circuit, this creates an exponential current transient.

Consider a series resistor $R$, inductor $L$, and ideal source of EMF $\mathcal E$.

Current growth after connecting the source

Kirchhoff's loop law gives

$$\mathcal E-iR-L\frac{di}{dt}=0.$$

Therefore

$$\boxed{L\frac{di}{dt}+Ri=\mathcal E}.$$

If the initial current is zero,

$$i(0)=0,$$

the solution is

$$\boxed{i(t)=\frac{\mathcal E}{R}\left(1-e^{-tR/L}\right)}.$$

The resistor voltage is

$$v_R(t)=Ri(t) =\mathcal E\left(1-e^{-tR/L}\right),$$

while the inductor voltage is

$$\boxed{v_L(t)=\mathcal E e^{-tR/L}}.$$

Immediately after connection, the current is still zero, so the resistor drop is zero and nearly the full source voltage appears across the inductor.

At long times,

$$\frac{di}{dt}\to0,$$

so

$$v_L\to0$$

and

$$i\to\frac{\mathcal E}{R}.$$

An ideal inductor therefore approaches a short-circuit voltage condition in steady DC while carrying a constant current.

The RL time constant

Define

$$\boxed{\tau=\frac{L}{R}}.$$

Then

$$i(t)=I_\infty\left(1-e^{-t/\tau}\right),$$

where

$$I_\infty=\frac{\mathcal E}{R}.$$

After one time constant,

$$i(\tau)=I_\infty(1-e^{-1})\approx0.632I_\infty.$$

Larger inductance makes current change more slowly; larger resistance reduces the time constant because it dissipates magnetic-field energy more rapidly.

Current decay after removing the source

Suppose the inductor initially carries current $I_0$ and the source is removed while $R$ and $L$ remain in a closed loop.

Kirchhoff's law gives

$$L\frac{di}{dt}+Ri=0.$$

The current decays as

$$\boxed{i(t)=I_0e^{-tR/L}}.$$

The magnetic energy

$$U_B=\frac12Li^2$$

is gradually transferred to the resistor and dissipated as internal thermal energy.

Inductor current cannot jump in the ideal finite-voltage model

Because

$$v_L=L\frac{di}{dt},$$

an instantaneous finite jump in current would require an unbounded voltage impulse. In ordinary ideal-circuit problems with finite voltages,

$$\boxed{i_L(0^+)=i_L(0^-)}.$$

This makes inductor current the magnetic counterpart of capacitor voltage: both are state variables associated with stored field energy that remain continuous across ordinary switching events.

Worked example

Let

$$L=2.0,\mathrm H,\qquad R=4.0,\Omega,$$

so

$$\tau=\frac{L}{R}=0.50,\mathrm s.$$

With source

$$\mathcal E=12,\mathrm V,$$

the final current is

$$I_\infty=\frac{12}{4.0}=3.0,\mathrm A.$$

After

$$t=1.0,\mathrm s=2\tau,$$

$$i=3.0(1-e^{-2})\approx2.59,\mathrm A.$$

At that instant the inductor voltage is

$$v_L=12e^{-2}\approx1.62,\mathrm V.$$

The remaining source voltage appears across the resistor.

RL transients are first-order dynamics produced by competition between magnetic energy storage and resistance. They provide the time-domain foundation for understanding inductors before sinusoidal impedance methods are introduced.