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Solved exercise · level 3 of 5

Self-induction

Skills: Calculate

Statement

The current in a coil of inductance L = 0.2 H rises from 0 to 3 A in 0.01 s. What self-induced EMF appears?

L = 0.2 H · ΔI = 3 A · Δt = 0.01 s

Parametric exercise: the values change with every draw; these are one example.

Guided problem in 4 steps: unknown, formula, units, data, result.

1 hint from Bastaya available in the lab
Solution

E = L × ΔI / Δt = 0.2 × 3 / 0.01 = 60 V. On sudden opening of an inductive circuit Δt is very short: overvoltage (arc).

Course sheet of the chapter : Induction & self-induction

This chapter is covered by the paths: General electricity

Other exercises of the chapter

Simulated time00:00:00
Q = I·t-
W = P·t-

Circuit diagram CEI 60617

How does it work?

Each big hole is a circuit node: you push one leg of a component into it. The 4 small holes around it are connected to it: chips and leads plug in there.

  1. Place a component Drag it from the palette: it snaps onto the grid. R rotates it.
  2. Bridge neighbouring holes Pick the jumper chip (or C) and drag on the board: bridged holes become a single node.
  3. Draw a lead For long links and the power supply: from one big hole to another, or to a terminal.
  4. Close the loop The circuit is solved live (Ohm, Kirchhoff) and the diagram follows.

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