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Solved exercise · level 3 of 5

Pure inductor and capacitor in parallel

Skills: Calculate

Statement

A pure inductor L = 0.9 H is connected in parallel with a 12 µF capacitor. What current does the circuit draw at 380 V, 50 Hz?

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

2 hints from Bastaya available in the lab
Solution

I_L = U / (L·ω) = 1.34 A and I_C = U·C·ω = 1.43 A are in phase opposition: I = |I_L − I_C| = 0.0886 A.

Course sheet of the chapter : RLC circuits

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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