Inductor

An inductor stores energy in a magnetic field and opposes changes in current. V1 switches from $0$ to $5\,\mathrm{V}$ and back. Current through L1 builds gradually, then continues in the same direction as it decays. Watch the inductor voltage reverse when the source falls to zero.

The response speed is set by the time constant:

$$\begin{aligned} \tau &= \frac{L}{R} \\ &= \frac{1\,\mathrm{H}}{100\,\Omega} \\ &= 10\,\mathrm{ms} \end{aligned}$$

Read the circuit and chart

• VL — green: voltage across L1, from its upper terminal to ground. A negative value means the upper terminal is below ground.
• Vs — red: source voltage, showing the rising and falling edges.
• IL — blue: current toward L1, flowing clockwise through the loop and downward through the inductor.

R1 limits the final current. The bottom wire is the shared return, and ground defines $0\,\mathrm{V}$. Voltage traces share the upper chart group; current has its own scale below. Both groups use the same time axis. The probes do not materially change the circuit.

Follow one pulse

The circuit starts with zero current because V1 is initially at $0\,\mathrm{V}$. At about $5\,\mathrm{ms}$, Vs rises to $5\,\mathrm{V}$. IL cannot jump when the applied voltage is finite: $v_L=L\,di_L/dt$. Initially, almost all the source voltage appears across L1. As current builds, the resistor's voltage drop grows and VL falls toward zero.

For an ideal voltage step, with $t$ measured from the rising edge:

$$\begin{aligned} i_L(t) &= 50(1-e^{-t/\tau})\,\mathrm{mA} \\ v_L(t) &= 5e^{-t/\tau}\,\mathrm{V} \end{aligned}$$

The eventual current is $I_\infty=V/R=5\,\mathrm{V}/100\,\Omega=50\,\mathrm{mA}$. After one time constant, IL reaches about 63.2% of that value; after five, about 99.3%. The source uses short $10\,\mu\mathrm{s}$ edges, so these chart landmarks are approximate:

Chart time	What to look for
Just after $5\,\mathrm{ms}$	VL is nearly $+5\,\mathrm{V}$ while IL is still near zero.
$15\,\mathrm{ms}$	One time constant later: IL is about $31.6\,\mathrm{mA}$ and VL about $+1.84\,\mathrm{V}$.
Just before $55\,\mathrm{ms}$	IL is about $49.7\,\mathrm{mA}$; VL is nearly zero.
Just after $55\,\mathrm{ms}$	Vs returns to zero. VL reverses to nearly $-4.97\,\mathrm{V}$, while IL remains continuous and positive.
$65\,\mathrm{ms}$	One decay time constant later: IL is about $18.3\,\mathrm{mA}$ and VL about $-1.83\,\mathrm{V}$.

During decay, $i_L(t_d)\approx49.7e^{-t_d/\tau}\,\mathrm{mA}$, where $t_d$ is time since the falling edge. The negative VL means current is decreasing, not reversing. L1 releases stored energy into R1, which dissipates it as heat.

Setting the ideal source to zero keeps the loop closed through R1. The source is not disconnected, so this circuit demonstrates a controlled current decay. At steady DC, an ideal inductor has zero voltage across it while carrying current.

Stored energy is $E=\tfrac12 LI^2$. At $50\,\mathrm{mA}$, L1 stores about $1.25\,\mathrm{mJ}$. This is a calculated value, not an additional chart trace.

Use CircuitSim

This example runs Transient analysis automatically from 0 to 110 ms. The target and maximum time steps are 50 µs, small compared with the time constant. Normal operating-point initialization is enabled; Use initial conditions is off.

Hover over the chart to inspect values. Open Table for numeric samples in the order time, VL, Vs, IL. Select a component and edit the named field in Properties, then use Run simulation. Leave advanced parameters at their defaults.

Try it

1. Double inductance. Change L1's Inductance from 1 to 2. The time constant becomes $20\,\mathrm{ms}$, so IL reaches about $31.6\,\mathrm{mA}$ at chart time $25\,\mathrm{ms}$. The eventual current remains $50\,\mathrm{mA}$, but this pulse ends before it is fully reached.
2. Double resistance. Restore L1 to 1, then change R1's Resistance from 100 to 200. The time constant becomes $5\,\mathrm{ms}$ and the eventual current is $25\,\mathrm{mA}$. At chart time $10\,\mathrm{ms}$, IL is about $15.8\,\mathrm{mA}$: it rises faster toward a lower final value.

Restore L1 = 1 and R1 = 100, then rerun. Inductance is in henries and resistance in ohms. V1 uses VI = 0, VP = 5, TD = 5m, TR = TF = 10u, PW = 50m, and Per = 120m. The displayed interval includes one rise-and-decay cycle.

Model assumptions

L1 is an ideal linear inductor with no winding resistance, magnetic saturation, or core losses. Real inductors have these effects and a current rating. R1 is constant, and V1 is an ideal source. This example isolates magnetic energy storage and the $L/R$ time constant.