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Solved exercise · level 2 of 5

EMF of an accumulator cell

Skills: Calculate

Statement

An accumulator cell delivers 2 A into a 1 Ω resistor. Given its internal resistance r = 0.05 Ω, compute its EMF.

I = 2 A · R = 1 Ω · r = 0.05 Ω

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

U = R × I = 1 × 2 = 2 V; E = U + r × I = 2 + 0.05 × 2 = 2.1 V; check: E = (R + r) × I.

Course sheet of the chapter : Real generators & loads

This chapter is covered by the paths: General electricity, Measurement prep, ISL-oriented exercises

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.

Shortcuts

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R
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C
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