Inductors — Inductance, RL Circuits & Energy Storage

Flip a switch in a circuit with an inductor and the current does not snap to its new value — it eases there. An inductor stores energy in a magnetic field and makes current change gradually rather than instantly. In the ideal circuit model, one equation captures it:

The inductor does not oppose current itself; it opposes changes in current. If the current is steady, $dI/dt = 0$ and the ideal inductor has zero voltage across it.

When Inductors Drive The Analysis

Inductors matter most when current is rising, falling, or oscillating — RL switching transients, filters, switching power supplies, electromagnets, transformers, and motor systems. The details differ, but the core pattern is constant: they matter whenever changing current and magnetic energy storage matter. They also connect naturally to electromagnetic induction, since a changing current creates a changing magnetic field, and that changing field produces an induced emf opposing the change.

Two quantities set the stage. Inductance $L$ tells you how much voltage is needed to change current at a given rate; a larger $L$ changes the current more slowly, which is how a coil smooths current changes. And an ideal inductor carrying current $I$ stores magnetic energy

The square matters: double the current and the stored energy becomes four times as large. This is why inductors appear in filters, power supplies, and switching circuits — they store energy briefly and release it as the circuit changes.

The Steps For An RL Problem

1. Check whether current is changing. Inductors matter most when current is rising, falling, or oscillating.
2. Use the ideal inductor law. For an ideal inductor, $V_L = L\,dI/dt$, so voltage depends on the rate of change.
3. Use the RL time constant. In a simple series RL circuit, $\tau = L/R$ sets the time scale for how fast the current responds.
4. Track stored energy. If the current is $I$, the magnetic energy in an ideal inductor is $U = \tfrac{1}{2}LI^2$.

Why can current not jump in step 2? An instantaneous jump would make $dI/dt$ extremely large, and from $V_L = L\,dI/dt$ that would demand an extremely large voltage. In an ordinary finite-voltage circuit that voltage is not available, so current through an ideal inductor changes continuously. Right after a switch opens or closes, the inductor is what stops the current from jumping to its new value.

The Whole Procedure On A 12 V RL Circuit

Consider a DC source of $12\ \mathrm{V}$ in series with a resistor $R = 6\ \Omega$ and an ideal inductor $L = 3\ \mathrm{H}$, with the switch closing at $t = 0$. The current is changing (step 1). The time constant (step 3) is

and the final steady current is

The current does not jump to $2\ \mathrm{A}$. For this step-input case it rises as

so here

After one time constant, $t = 0.5\ \mathrm{s}$:

about $63\%$ of its final value, the standard RL benchmark. At that moment the resistor voltage is

and the rest of the source voltage is across the inductor:

Early on the inductor takes a larger share because the current is changing quickly; later, as the current settles and $dI/dt$ shrinks, the inductor voltage falls toward zero. Tracking stored energy (step 4) after one time constant:

Where Each Step Goes Wrong, And How To Check

Saying an inductor "blocks DC." In steady ideal DC the inductor has zero voltage drop, but during the transient before the current settles it strongly affects the circuit. Watch this at step 1.

Treating $V_L = L\,dI/dt$ as a formula about current alone. Voltage depends on how fast current changes, not on whether it is large or small; a large steady current can coexist with zero ideal-inductor voltage. Self-check at step 2.

Thinking the time constant is the finishing time. $\tau = L/R$ is a time scale, not a hard cutoff — after one time constant the process is still going, just well underway. Keep this in mind at step 3.

Forgetting the ideal-model condition. Real inductors have winding resistance, parasitic capacitance, and core limits. The ideal equations are useful, but still a model.

To deepen the procedure, keep the same $12\ \mathrm{V}$ source and $6\ \Omega$ resistor but change the inductance from $3\ \mathrm{H}$ to $1.5\ \mathrm{H}$: the final current stays $2\ \mathrm{A}$, but $\tau$ halves, so the current rises faster. Predict the new time constant before you recompute the response.