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Solved exercise · level 5 of 5

Workshop: capacitor, heater and cos φ

Skills: Calculate, Analyse

Statement

A workshop at 230 V, 50 Hz has a machine running continuously with a motor of total power 2400 W and cos φ = 0.75. What capacitance (in µF) is needed to raise this cos φ to 0.9?

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

1 hint from Bastaya available in the lab
Solution

Q = P · tan φ = 2120 var; with cos φ′ = 0.9, Q′ = 1160 var; C = (Q − Q′) / (U²ω) = 57.4 µF. With the heater: P = 2400 + 1500 W, cos φ = 0.958. Without capacitor: cos φ = 0.879.

Course sheet of the chapter : AC power

This chapter is covered by the paths: General electricity, 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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