What capacitance would you connect across an$\mathbf{\text{1.30mH}}$ inductor to make the resulting oscillator resonate $\mathbf{\text{3.50kHz}}$?

i) The inductance of an inductor is .$L=\text{1.30\hspace{0.17em}}\text{mH}={\text{1.30×10}}^{\text{-3}}\text{H}$

ii) The resonant frequency of the oscillator is .$f=\text{3.50\hspace{0.17em}}\text{kHz}={\text{3.50×10}}^{\text{3}}\text{Hz}$

If the frequency of oscillations of an oscillator becomes equal to the natural frequency of the oscillator, the intensity of oscillations goes on increasing. This phenomenon is known as resonance.

The formula is as follows:

.$\omega =\frac{\text{1}}{\sqrt{LC}}$

$\omega =\text{2}\pi f$

Where,

$\omega$= angular frequency,

L= inductance,

C = capacitance,

The value of capacitance:

The expression of the angular frequency oscillation is,

$\omega =\frac{\text{1}}{\sqrt{LC}}$,

${\omega }^2=\frac{\text{1}}{LC}$,

$C=\frac{\text{1}}{L{\omega }^2}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \mathrm{..}\left(1\right)$,

The expression of the angular frequency in terms of the frequency of oscillation is,

$\omega =\text{2}\pi f$.

The equation (1) becomes as,

$C=\frac{\text{1}}{L{\left(\text{2π}f\right)}^2}$.

$C=\frac{\text{1}}{{\text{1.30×10}}^{\text{-3}}\text{H×}{\left({\text{2×3.14×3.50×10}}^{\text{3}}\text{Hz}\right)}^{\text{2}}}$.

$C={\text{1.59×10}}^{\text{-6}}\text{F}$.

$C=\text{1.59μF}$.

Hence, the value of capacitance of the $LC$ circuit is .$C=\text{1.59\hspace{0.17em}}\text{μF}$