Current delay in an inductor (two lamps)

Curricula: APC, Sek II

Open the full lab ↗

Aim

Observe how an inductor delays the rise of the current when the circuit is switched on.

The bench

A 6 V source and a switch (open). Two parallel branches with 6 V / 0.5 W lamps: bottom — a lamp and a 20 Ω resistor, top — a lamp and a 100 H inductor with a winding resistance of 20 Ω.

Procedure

1. Close the switch and watch both lamps.
2. Open the switch.
3. In the inductor's settings, reduce the inductance to 10 H and repeat.

Expected result

The lamp with the resistor lights up at once; the lamp with the inductor lights up about one to two seconds later, and after that their brightness is the same: the resistances of the branches are equal. After the switch is opened, both lamps go out together, in about half a second: the inductor's current keeps flowing round the loop made of the two branches and passes through both lamps. The current rise time is about \(L / R\). At 10 H the delay is ten times shorter — about a tenth of a second, but the eye can still notice it. A 100 H inductor with a 20 Ω winding is very large: for school inductors with an iron core \(L / R\) is a fraction of a second, and the delay on a real bench is just as short.

Questions

1. Why does an inductor prevent the current from rising instantly?
2. Why do the lamps glow equally bright in the steady state?

Answers

1. When the current changes, the magnetic field of the coil changes and induces an EMF of self-induction, proportional to the rate of change, \(L \cdot \Delta I / \Delta t\), and directed against the change. A sudden jump would need an infinitely large EMF, so the current grows gradually, over a time of about \(L / R\): one to two seconds with 100 H on this bench.
2. In the steady state the current no longer changes, so there is no self-induced EMF, and the inductor acts as its 20 Ω winding. Both branches then have a lamp in series with 20 Ω across the same voltage, so the currents and the brightness are equal.