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Solved exercise · level 2 of 5

Series: from partial to total voltage

Skills: Calculate

Statement

A circuit has R₁ = 5.6 Ω, R₂ = 27 Ω and R₃ = 1 Ω in series. Across R₁ there is 1.12 V. What is the supply voltage?

R₁ = 5.6 Ω · R₂ = 27 Ω · R₃ = 1 Ω · u₁ = 1.12 V

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

I = u₁ / R₁ = 1.12 / 5.6 = 0.2 A (the same everywhere in series); U = I × (R₁ + R₂ + R₃) = 6.72 V.

Course sheet of the chapter : Resistor networks

This chapter is covered by the paths: 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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