Question

A transmitting station releases waves of wavelength $960 m$. A capacitor of $2.56 \mu F$ is used in the resonant circuit. The self inductance of coil necessary for resonance is____ $\times 10^{-8} H$

Answer 1

$\lambda=960 m$
$C =2.56 \mu F =2.56 \times 10^{-6} F$
$c =3 \times 10^{8} m / s$
$L =?$
Now at resonance, $\omega_{0}=\frac{1}{\sqrt{ LC }}$
[Resoant frequency]
$2 \pi f _{0}=\frac{1}{\sqrt{ LC }}$
On substituting $f _{0}=\frac{ c }{\lambda},$ we have $2 \pi \frac{ c }{\lambda}=\frac{1}{\sqrt{ LC }}$
Squaring both sides : $4 \pi^{2} \frac{ c ^{2}}{\lambda^{2}}=\frac{1}{ LC }$
$=\frac{4 \times 10 \times\left(3 \times 10^{8}\right)^{2}}{(960)^{2}}=\frac{1}{ L \times 2.56 \times 10^{-6}}$
$\Rightarrow \frac{1}{ L }=\frac{4 \times 10 \times 9 \times 10^{16} \times 2.56 \times 10^{-6}}{960 \times 960}$
$\Rightarrow L =10 \times 10^{-8} H$