# Resonance and its applications

### Explanation of question 16

A parallel circuit is resonating at 2MHz, and has a Q factor of 200. The bandwidth will be

1. 5kHz
2. 10kHz
3. 20kHz
4. 125kHz

The answer is :- 2

Bandwidth = f / Q = 2,000,000 / 200 = 10,000 Hz = 10kHz

BACK

page (tq1016ex)
Some material Copyright John Bowyer G4KGS.
Click here for important Copyright information
Web Space provided by Hostroute.com Ltd
Contact details