Elec Class

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Solved exercise · level 2 of 5

A cell and its load

Skills: Build, Analyse

Statement

No bench supply this time: snap a cell (E = 9 V, r = 0.5 Ω) and a resistor R1 = 10 Ω onto the board, join them in a loop (chips and leads), then compute the current.

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

2 hints from Bastaya available in the lab
Solution

I = E / (r + R1) = 9 / (0.5 + 10) = 857 mA.

Course sheet of the chapter : Real generators & loads

This chapter is covered by the paths: General electricity, Measurement prep

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

Del
delete
R
rotate
C
chip brush
Esc
cancel
wheel
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fit the board
drag background
pan

Confirmer

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