Machine-translated by Google. Questions are authored in English.
Top

Discussion

Three arms of an alternating bridge have impedance magnitudes of 200, 400 and 100 ohms, the last two being opposite each other. For magnitude balance the fourth arm, opposite the first, must be

  • A.200 \( \Omega \)
  • B.800 \( \Omega \)
  • C.50 \( \Omega \)
  • D.100 \( \Omega \)

Answer: A

Magnitude balance requires the products of opposite pairs to be equal, so \( 200 \times Z = 400 \times 100 \) and \( Z = 200\ \Omega \). The phase condition is a separate equation and must be satisfied at the same time before the detector goes silent.

No comment is present. Be the first to comment.
Loading…

Post your comment

Please log in or register to post a comment.