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Solved exercise · level 4 of 5

Correction capacitor for a coil

Skills: Calculate

Statement

An inductive load has a resistance of 20 Ω and an inductance of 0.08 H. The supply voltage is 120 V, 50 Hz. What capacitor must be placed in parallel to raise cos φ to 0.9? (Answer in µF.)

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

2 hints from Bastaya available in the lab
Solution

Z = 32.1 Ω, I_Z = 3.74 A; P = R·I² = 279 W, Q_Z = Lω·I² = 351 var; with cos φ = 0.9: Q = P · tan φ = 135 var; Q_C = Q_Z − Q = 216 var; C = Q_C / (U² · ω) = 47.7 µF.

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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