LCR circuit, the resonance frequency of circuit increases two times of the initial circuit by changing C and C' and R from `100 Omega " to " 400 Omega` while the inductance was kept the same. The ratio C/C', is

Options

A. 2
B. 8
C. 16
D. 4

Answer

D

Explanation

Given,
In siries `L -C-R` circuite,
`f _(2) = 2 f_(1)`
Initially,
Capacitance, `C _(1) =C`
Resistance, `R_(1) = 100 Omega`
Inductance, `L _(1) =L`
The equation for resonance frequency of a series LCR circuit is given by,
`f _(1) = (1)/(2pi sqrt(L_(1) C_(1)))`
`= (1)/(2pi sqrt(LC))`
Capacitance, `C_(2) =C`
Resistance, `R_(2) =400 Omega`
Inductance, `L _(2) = L_(1) =L`
Then,
`f _(2) = (1)/( 2pi sqrt( L _(2) C_(2)))`
` = (1)/(2pi sqrt(LC))`
Since,
`f _(2) = 2f_(1)`
` (1)/(2pi sqrt(LC))=2 ((1)/(2pi sqrtLC))`
`(1)/(sqrt(LC))= (2)/( sqrt(LC))`
` (C)/(sqrtC) = (2) /(sqrtC)`
`sqrt((C)/(C)) =2`
` C/C = (2)^(L)`
`therefore C/C =4`