wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An electronic test circuit produced a resonant curve of half power frequency points at 414Hz and 436Hz. If Q factor be 10, the resonant frequency of the circuit is:

A
220Hz
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
22Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2.2Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.22Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 220Hz

We have

Q=frBW

Where

Q= quality factor

fr= resonant frequency

BW= Bandwidth

Also, bandwidth is the difference is half power frequencies.

So,

Q=frf2f1

Given

f1=414Hz,f2=436Hz,Q=10

So,

fr=Q(f2f1)

fr=10(436414)=10×22=220Hz

So the resonant frequency is 220Hz


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Electrical Power in an AC Circuit
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon