Question

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

Short answer

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

Step-by-step solution

Step 1: Given

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 .$f=\text{3.50\hspace{0.17em}}\text{kHz}={\text{3.50×10}}^{\text{3}}\text{Hz}$

Determining the concept

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,

Determining the value of capacitance of the  LC circuit

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 .$C=\text{1.59\hspace{0.17em}}\text{μF}$