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Solved exercise · level 3 of 5

Junction law in the lab

Skills: Analyse

Statement

In this circuit, E1's branch brings I₁ = 2.82 A to the common junction and E2's branch brings I₂ = 0.706 A. What current flows through R3?

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

2 hints from Bastaya available in the lab
Solution

I₃ = I₁ + I₂ = 2.82 + 0.706 = 3.53 A; check with the left loop: E₁ = R₁ I₁ + R₃ I₃.

Course sheet of the chapter : Kirchhoff's laws

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

Del
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R
rotate
C
chip brush
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