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Solved exercise · level 3 of 5

Impedance of a series RLC circuit

Skills: Calculate

Statement

A circuit contains in series a coil (R = 50 Ω, L = 0.2 H) and a 5 µF capacitor. What is its impedance at 60 Hz?

R = 50 Ω · L = 0.2 H · C = 5 µF · f = 60 Hz

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

ω = 377 rad/s; L·ω = 75.4 Ω; 1/(C·ω) = 530.5 Ω; Z = √(50² + (75.4 − 530.5)²) = 458 Ω.

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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