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Solved exercise · level 4 of 5

Two cells, one motor, two resistors

Skills: Analyse

Statement

A single loop: E1 (9 V, r = 1 Ω) and E2 (4.5 V, r = 0.5 Ω) in the same direction, motor M1 (E′ = 3 V, r′ = 2 Ω) and resistors R1 = 3.3 Ω and R2 = 1 Ω.

What current will A1 show?

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

2 hints from Bastaya available in the lab
Solution

I = (ΣE − ΣE′) / ΣR = (9 + 4.5 − 3) / (1 + 0.5 + 2 + 3.3 + 1) = 1.35 A.

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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C
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