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In an LCR circuit, the resonating freque...

In an LCR circuit, the resonating frequency is 500 kHz. If the value of L is increased two times and value of C is decreased `(1)/(8)` times, then the new resonating frequency in kHz will be -

A

1500

B

1000

C

500

D

250

Text Solution

Verified by Experts

The correct Answer is:
B
(b) : Frequency for resonance in a circuit is given by
`nu = (1)/(2pi) (1)/(sqrt(LC))`
New frequency can be written as follows
`nu= (1)/(2pi)(1)/(sqrt((2L)((C)/(8))))=2((1)/(2pi)(1)/(sqrt(LC)))=2xx500=1000 kHz`
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