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Solved exercise · level 5 of 5

Coil resistance from cos φ

Skills: Calculate

Statement

A circuit contains in series a coil of inductance L = 0.1 H and a capacitor C = 3.3 µF; a voltmeter across the capacitor reads 60.3 V, the frequency is 400 Hz and the phase angle between U and I has cosine 0.0763. What is the coil resistance R? (From cos φ get sin φ; I = U_C·C·ω; U = (U_L − U_C)/sin φ; Z = U/I; R = Z·cos φ.)

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

2 hints from Bastaya available in the lab
Solution

I = U_C·C·ω = 0.5 A; L·ω − 1/(C·ω) = 130.8 Ω; Z = 131.1 Ω; R = Z · cos φ = 131.1 × 0.0763 = 10 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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