Question

A transmitting station releases waves of wavelength $960\,m.$ 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

Since resonance $\omega _{r}=\frac{1}{\sqrt{LC}}$
$\therefore 2\pi f=\frac{1}{\sqrt{LC}}$
$\therefore 4\pi ^{2}\frac{c^{2}}{\lambda ^{2}}=\frac{1}{LC}$
$\therefore \frac{4 \pi ^{2} \times 9 \times 10^{16}}{960 \times 960}=\frac{1}{ L \times 2 . 56 \times 10^{- 6}}$
$=\frac{375 \times 960}{10^{- 6} \times 4 \times \pi ^{2} \times 9 \times 10^{16}}=\frac{10^{3}}{10^{10}}$
$=10^{- 7}H$
$=10\times 10^{- 8}$