Inductors, self-inductance and the transformer

Key ideas

• An inductor is a coil. When the current through it changes, the flux through its own turns changes, inducing an EMF in itself that opposes the change — self-inductance:

$\varepsilon = -L\frac{\Delta I}{\Delta t}$

• $L$ — the inductance (henry, H)
• $\Delta I/\Delta t$ — how fast the current is changing (A s⁻¹)
• An inductor resists changes in current (the mirror image of a capacitor resisting changes in voltage):
  • at switch-on, the induced back-EMF is large, so the current starts at zero and grows toward its final value $\varepsilon/R$
  • the growth has time constant $\tau = \dfrac{L}{R}$ — after one $\tau$ the current reaches 63% of its final value
  • at switch-off, the inductor fights the collapse — the rapidly changing flux can induce a large voltage spike (the spark in a switch).
• Energy stored in the inductor's magnetic field:

$E = \tfrac{1}{2}LI^2$

The transformer

• A transformer is two coils sharing an iron core. Alternating current in the primary makes a continually changing flux in the core, which induces an EMF in the secondary (Faraday's law).
• For an ideal transformer, the voltages follow the turns ratio:

$\frac{N_p}{N_s} = \frac{V_p}{V_s}$

• more secondary turns → step-up (higher voltage); fewer → step-down
• An ideal transformer passes power through unchanged: $V_pI_p = V_sI_s$ — step the voltage up and the current steps down in proportion.
• Transformers only work on AC: a steady DC current makes a steady flux, and a steady flux induces nothing.

Tips

• τ = L/R, not LR. A bigger resistance makes an inductive circuit settle faster — the opposite of the capacitor's τ = RC. Check which way your answer should move.
• If a question mentions a spark or voltage spike at switch-off, the explanation is Faraday + Lenz: breaking the circuit makes $\Delta I/\Delta t$ huge, so $\varepsilon = -L\,\Delta I/\Delta t$ is huge.