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Exercise from the chapter "Real generators & loads"

Internal resistance of an accumulator

Level 3 of 5 · 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 = 22 Ω is 0.06757 A?

E = 1.5 V · R = 22 Ω · I = 0.06757 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 / 0.06757 − 22 = 0.2 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-

Workbench

Circuit diagram CEI 60617

How does it work?

The perfboard: each big hole is a circuit node. You push a single leg of a component (resistor, lamp...) into it. The 4 small holes around it are connected to it: that is where chips and leads plug in.

  1. Drag a component onto the board: it snaps onto the grid. R rotates it.
  2. To join neighbouring big holes, click Jumper chip in the palette (or C) then drag on the board: a line of chips is laid at once. Bridged holes become a single node.
  3. For long links and for the power supply (placed next to the board), drag a lead from one big hole to another (or to a terminal).
  4. As soon as a loop is closed, the circuit is solved in real time (Ohm + Kirchhoff) and the diagram follows.
  5. Zoom with the wheel (or + / −, two-finger pinch); drag the background to pan, 0 fits the board. The + buttons around the board add a row or a column.

Del delete · R rotate · C chip brush · Esc cancel · wheel zoom · drag the background to pan

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