Put an inductor in series with a resistor and a battery. Because the inductor fights every change in current, the current cannot jump when the switch closes; it grows smoothly, in a curve you have seen before in charging a capacitor.
Why it works
Loop rule, with the inductor's back emf: $\varepsilon - IR - L\dfrac{dI}{dt} = 0$, so
$$\frac{dI}{dt} = \frac{\varepsilon - IR}{L} \quad\Longrightarrow\quad \frac{dI}{\varepsilon/R - I} = \frac{R}{L}\,dt.$$
Integrating from $I = 0$ at $t = 0$ gives $I = \dfrac{\varepsilon}{R}\left(1 - e^{-Rt/L}\right)$. At first $I = 0$, so the whole battery voltage sits across the inductor: the current starts rising at the rate $\varepsilon/L$. As $I$ grows, more voltage goes to the resistor and the rise slows.
Key idea
$$I = \frac{\varepsilon}{R}\left(1 - e^{-t/\tau}\right), \qquad V_L = \varepsilon e^{-t/\tau}, \qquad \tau = \frac{L}{R}.$$
If the battery is removed and the loop stays closed through a resistance $R$, the current dies away as $I = I_0e^{-t/\tau}$.
Worked examples
Worked example: Current after one time constant
A $0.50\ \text{H}$ inductor and a $10\ \Omega$ resistor are switched onto $12\ \text{V}$. Find $\tau$, the final current, the current after one time constant and the starting rate of rise.
- $\tau = L/R = 0.50/10 = 50.0\ \text{ms}$; final current $\varepsilon/R = 1.2\ \text{A}$.
- $I(\tau) = 1.2(1 - e^{-1}) = 0.759\ \text{A}$, 63% of the way.
- At $t = 0$: $\dfrac{dI}{dt} = \dfrac{\varepsilon}{L} = \dfrac{12}{0.50} = 24\ \text{A/s}$.
Answer: $I(\tau) \approx 0.759\ \text{A}$
Worked example: Time to 90%
When does the current in that circuit reach 90% of its final value?
- $1 - e^{-t/\tau} = 0.90$, so $t = \tau\ln 10 = (0.050)(2.303) = 115\ \text{ms}$.
Answer: $t \approx 115\ \text{ms}$
Worked example: Switching off gently
After the current has settled at $1.2\ \text{A}$, the battery is removed and the inductor and resistor are left connected. What is the current $50\ \text{ms}$ later?
- Loop rule without the battery: $L\dfrac{dI}{dt} = -IR$, so $I = I_0e^{-t/\tau}$ with the same $\tau = 50\ \text{ms}$.
- $I = 1.2e^{-1} = 0.441\ \text{A}$. The inductor keeps the current flowing for a while after the battery is gone.
Answer: $I \approx 0.441\ \text{A}$
Common mistake
"The current jumps to $\varepsilon/R$ the instant the switch closes." A sudden jump would need an infinite $dI/dt$, and so an infinite back emf. Instead the inductor takes the whole battery voltage at the start and the current begins at zero. Check with the loop rule at $t = 0$: $I = 0$, so $L\,dI/dt = \varepsilon$. (A capacitor is the opposite: its current jumps at once and its voltage builds slowly.)