Question:-
The impedance of a basic ac, bridge arms are
ZAb = 250 ohm ZBC = 100 ohm Ð60°
ZCD = 400 ohm Ð30°, ZDA = Unknown
if w = 1000 Rad/sec, then components of unknown arm has the resistance 500Ö3 in
Option (A)
Series with a capacitor of 2 mF
Option (B)
Parallel with a capacitor of 2 mF
Option(C)
Series with an inductor of 2 mF
Option(D)
Parallel with an inductor of 2 mF
Correct Option:
(A)
Question Solution:
ZDA =(ZAB.ZCD/ZBC)
= ((250 x 400 Ð30°)/100 Ð60°)
ZDA=1000 Ð-30°
Now ZDA =500 √3 –J500
Here R = 500 √3 , Now Xc = 500
So 500 = (1/1000.C)
→ C = 0.2 x 10-5 = 2 x 10-6 F
question-answer-faq-1637