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Solved exercise · level 4 of 5

Power factor correction

Skills: Calculate

Statement

A single-phase installation (230 V, 50 Hz) draws P = 2000 W at cos φ = 0.6. The power factor is to be corrected to 0.95. Compute the capacitance to connect in parallel (answer in µF).

P = 2000 W · cos φ₁ = 0.6 · U = 230 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

Q_C = P (tan φ₁ − tan φ₂) = 2.01 kvar; C = Q_C / (U²·ω) = 121 µF.

Course sheet of the chapter : AC power

This chapter is covered by the paths: General electricity

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
Esc
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drag background
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