Labo Élec

electrical lab simulator
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Exercise from the chapter "AC power"

Power factor correction

Level 4 of 5 · 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-

Workbench

Circuit diagram CEI 60617

How does it work?

The perfboard: each big hole is a circuit node. You push a single leg of a component (resistor, lamp...) into it. The 4 small holes around it are connected to it: that is where chips and leads plug in.

  1. Drag a component onto the board: it snaps onto the grid. R rotates it.
  2. To join neighbouring big holes, click Jumper chip in the palette (or C) then drag on the board: a line of chips is laid at once. Bridged holes become a single node.
  3. For long links and for the power supply (placed next to the board), drag a lead from one big hole to another (or to a terminal).
  4. As soon as a loop is closed, the circuit is solved in real time (Ohm + Kirchhoff) and the diagram follows.
  5. Zoom with the wheel (or + / −, two-finger pinch); drag the background to pan, 0 fits the board. The + buttons around the board add a row or a column.

Del delete · R rotate · C chip brush · Esc cancel · wheel zoom · drag the background to pan

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