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LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
Knowledge / Understand components
LNGR · Learning workshop

The inductor

A capacitor opposes a rapid voltage change. An inductor opposes a rapid current change.

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Turns store magnetic-field energy

Start without prior knowledgeUnderstand the model limitsDE / EN
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Circuit 08 · 5 small steps

An inductor builds up current

An inductor builds up currentAt t = 0 the source steps from 0 to 5 V. Initial current is zero. L1 is ideal with no winding resistance. R1 limits the eventual DC current.+−5 VAB · 0 VR1 = 100 ΩL1 = 100 mHM
At t = 0 the source steps from 0 to 5 V. Initial current is zero. L1 is ideal with no winding resistance. R1 limits the eventual DC current.

  1. What is I just after switching on?

    Hint

    Inductor current cannot jump with finite voltage.

    Answer and working
    Calculation notes
    {i_L(0^+)=0\,\mathrm{A}}
    u_R(0^+)=0\,\mathrm{V}
    u_L(0^+)=5\,\mathrm{V}

    Inductor current cannot jump with finite voltage.

  2. What current is reached eventually?

    Hint

    An ideal inductor has no voltage drop at steady DC.

    Answer and working
    Calculation notes
    I_{\infty}=\frac{U}{R}
    I_{\infty}=\frac{5\,\mathrm{V}}{100\,\Omega}
    {I_{\infty}=0.05\,\mathrm{A}=50\,\mathrm{mA}}

  3. Calculate τ.

    Hint

    100 mH = 0.1 H; τ = L/R.

    Answer and working
    Calculation notes
    L=100\,\mathrm{mH}=0.1\,\mathrm{H}
    \tau=\frac{L}{R}
    \tau=\frac{0.1\,\mathrm{H}}{100\,\Omega}
    {\tau=0.001\,\mathrm{s}=1\,\mathrm{ms}}

  4. What is I at t = τ?

    Hint

    I = I∞ × (1 − e−t/τ).

    Answer and working
    Calculation notes
    i(t)=I_{\infty}\left(1-e^{-\frac{t}{\tau}}\right)
    i(1\,\mathrm{ms})=50\,\mathrm{mA}\left(1-e^{-1}\right)
    {i(1\,\mathrm{ms})\approx31.6\,\mathrm{mA}}
    u_R\approx100\,\Omega\cdot0.0316\,\mathrm{A}=3.16\,\mathrm{V}
    u_L\approx5\,\mathrm{V}-3.16\,\mathrm{V}=1.84\,\mathrm{V}

  5. Double L. What changes, and why is sudden disconnection different?

    Hint

    τ depends on L; I∞ does not.

    Answer and working
    Calculation notes
    \tau_{\mathrm{new}}=\frac{0.2\,\mathrm{H}}{100\,\Omega}=2\,\mathrm{ms}
    {I_{\infty}=\frac{5\,\mathrm{V}}{100\,\Omega}=50\,\mathrm{mA}}

    The larger inductor slows current rise. Sudden interruption can create high voltage; the turn-on model does not simulate this turn-off case.

Think it through before opening the answer. Always write the formula, substituted values and units.

01

The mental picture

A heavy flywheel cannot speed up or stop instantly. An inductor similarly reacts to changes in current. Its energy is stored in a magnetic field rather than a rotating mass. The analogy explains inertia, not all the electromagnetic physics.

02

The principle

Wire turns concentrate the magnetic effect of current. Inductance L is measured in henries; speaker crossovers often use millihenries. The wire also has DC resistance. Air-core and magnetic-core inductors differ in size, losses and possible core saturation.

03

Try it yourself

Interactive experiment

The inductor

Double L at fixed R and U. Final current stays unchanged, but the rise takes longer. Then increase R and watch both final current and time constant.

Model limits: Linear ideal inductor plus the selected series resistance; no core saturation. Time axis automatically spans 0 to 5τ.

04

A worked example

10 mH with 10 Ω gives τ = 1 ms. At 5 V, current approaches 0.5 A. After 1 ms it reaches about 0.316 A. At 1 kHz, the ideal inductor has a reactance of about 62.8 Ω.

A closer look · equations and assumptions
Formula notes
u_L=L\frac{\mathrm{d}i}{\mathrm{d}t}
\tau=\frac{L}{R}
i(t)=\frac{U}{R}\left(1-e^{-\frac{t}{\tau}}\right)
X_L=2\pi fL
E=\frac{1}{2}LI^2

For sinusoidal current, voltage across an ideal L leads current by 90°. A magnetic core can saturate at high current, reducing effective inductance. A real coil also has parasitic capacitance, so XL = 2πfL does not remain valid to arbitrarily high frequencies.

A common pitfall. Abruptly interrupting inductor current can produce a high voltage. A transistor driving an inductive load needs suitable clamping, such as a flyback diode for a DC relay.

05

Workshop takeaway

Compare more than millihenries: include winding resistance, current and mounting. Adjacent crossover coils can couple magnetically.

Sources & transparency

Original illustrations and educational calculations. No measurements of a specific product. Edited 11 October 2026.

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

17 · An ideal 10 mH inductor carries 0.2 A. Calculate stored magnetic energy.

Show worked answer
  1. L = 0.01 H.
  2. Square current and multiply by L/2.
  3. E = 0.2 mJ.
Calculation
E_L=\frac{1}{2} LI^2=0.0002\,J

18 · 10 mH and 100 Ω are in series across 5 V. Find τ and steady current; ideal inductor.

Show worked answer
  1. τ = L/R = 0.1 ms.
  2. The ideal inductor has zero steady DC voltage.
  3. Final I = 5/100 = 50 mA.
Calculation
\tau=\frac{L}{R}=0.1\,ms

19 · Calculate inductor current in task 18 after 0.1 ms, starting at zero.

Show worked answer
  1. Time equals τ.
  2. Current rises to 63.21% of final value.
  3. I = 31.61 mA.
Calculation
I(\tau)=50\,mA(1-e^{-1})=31.61\,mA

20 · A 10 mH inductor changes current by 0.2 A in 10 µs. What ideal voltage magnitude is needed? What does this imply at turn-off?

Show worked answer
  1. Slope: ΔI/Δt = 20000 A/s.
  2. U = L · ΔI/Δt = 200 V.
  3. A limiting current path, such as a suitable flyback diode, prevents uncontrolled spikes.
Calculation
|U_L|=L\left|\frac{\Delta I}{\Delta t}\right|=200\,V

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