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Solved exercise · level 3 of 5

Internal resistance of an accumulator

Skills: Calculate

Statement

What is the internal resistance r of an accumulator cell of EMF 1.5 V if the current it delivers into R = 10 Ω is 0.1429 A?

E = 1.5 V · R = 10 Ω · I = 0.1429 A

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

r = E / I − R = 1.5 / 0.1429 − 10 = 10.5 − 10 = 0.5 Ω: the internal drop u = r · I is what U lacks to equal E.

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.

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