Question

A transmitting station releases waves of wavelength $960m$. A capacitor of $256\mu F$ is used in the resonant circuit. The self inductance of coil necessary for resonance is _____ $\times {10}^{-8}H$.

Answer 1

Step1: Given data and assumptions.

Wavelength, $\lambda =960m$

Capacitance, $C=256\mu F=256\times {10}^{-6}F$

Speed of light, $c=3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step2: Find the self-inductance of the coil.

We know that,

The resonance frequency of the $LC$ circuit is,

${\omega }_r=\frac{1}{\sqrt{LC}}$

Where, ${\omega }_r$ is the resonance frequency, $L$ is the inductance, and $C$ is the capacitance.

Also,

$\Rightarrow$$2\pi f_0=\frac{1}{\sqrt{LC}}$

$\Rightarrow$$2\pi \frac{c}{\lambda }=\frac{1}{\sqrt{LC}}$ $\left[\because f_0=\frac{c}{\lambda }\right]$ …..$\left(i\right)$

Squaring both sides in the equation $\left(i\right)$ we get.

$\Rightarrow$$4{\pi }^2\frac{c^2}{{\lambda }^2}=\frac{1}{LC}$

$\Rightarrow$$L=\frac{{\lambda }^2}{4{\pi }^2{c}^{2}C}$

$\Rightarrow$$L=\frac{{960}^2}{4{\left({\displaystyle \frac{22}{7}}\right)}^2\times \left(3\times {10}^8\right)\times 256\times {10}^{-6}}$

$\Rightarrow$$L=10\times {10}^{-8}H$

Hence, as compared to the given self-inductance of the coil necessary for resonance is $10$$\times {10}^{-8}H$.