Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Anonymous
Anonymous asked in Science & MathematicsEngineering · 2 months ago

Consider the circuit below, where E = 240<0(degrees) v? r1 = 40 ohms r2 = j 80 ohms r3 = j 120 ohms?

Attachment image

1 Answer

Relevance
  • 2 months ago
    Favorite Answer

    Using phasors

    iR = (240/0°)/(40/0°) = 6A at 0° = 6 + j0

    iL = (240/0°)/(80/90°) = 3/-90° = 0 -3j

    ic = (240/0°)/(120/-90°) = 2/+90° = 0+2j

    Current leaving the source = 6-j

    Ztp = 240/(6-j) = 39.5 /9.462°

    iR = 240/40 = 6A /_0°

    iL = 240/80 = 3A /_-90°

    iC = 240/120 = 2A /_90°

    ic(t) = 2sin(377t+π/2)

Still have questions? Get your answers by asking now.