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Solved exercise · level 4 of 5

Shunt calculation

Skills: Calculate

Statement

An ammeter with internal resistance r_A = 2 Ω reads full scale at i = 0.01 A. We want to measure up to I = 5 A. Compute the resistance of the shunt to place in parallel.

rA = 2 Ω · i = 0.01 A · I = 5 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

Same voltage across the meter and the shunt: r_A·i = R_s·(I − i) → R_s = 2 × 0.01 / (5 − 0.01) = 4.01 mΩ.

Course sheet of the chapter : Measurement & instruments

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.

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