The frequency of the output signal of an LC oscilliator circuit is 100 Hz, with a capacitance of `0.1 mu F`. If the value of the capacitor is increased to `0.2 mu F`, then the frequency of the output signal will

Options

A. be doubled
B. be half
C. increase by `1/(sqrt2)`
D. decrease by `1/(sqrt2)`

Answer

D

Explanation

`f = 1/(2pisqrt(LC)) :. f prop 1/(sqrt(C)) as 1/(2pisqrt(L))` is consant.
In this case `C_(1) = 0.1 mu F and C_(2) = o.2 mu F i.e. C_(2) = 2 C_(1)`
`:. (f_2)/(f_1) = (sqrt(C_1))/(sqrt(C_2)) = 1/(sqrt(2))`. Thus `f_(2) = (f_1)/(sqrt(2))`
Thus the frequency decreases by `1/(sqrt(2))`.