Physics - EMI/AC Question with Solution | TestHub

$\lambda =960m$

$C=2.56\mu \mathrm{F}=2.56\times 10^{-6}\mathrm{F}$

$c=3\times 10^{8}m{\text{s}}^{-1}$

$L=?$

Now at resonance, ${\omega }_0=\frac{1}{\sqrt{L\mathrm{C}}}$ [Resonant frequency]

$2\pi f_0=\frac{1}{\sqrt{L\mathrm{C}}}$

On substituting $f_0=\frac{c}{\lambda },$ we have $2\pi \frac{c}{\lambda }=\frac{1}{\sqrt{L\mathrm{C}}}$

On substituting $f_0=\frac{c}{\lambda },$ we have $2\pi \frac{c}{\lambda }=\frac{1}{\sqrt{L\mathrm{C}}}$

Squaring both sides: $4{\pi }^2\frac{c^2}{{\lambda }^2}=\frac{1}{L\mathrm{C}}$

$=\frac{4\times 10\times {\left(3\times 10^8\right)}^2}{{\left(960\right)}^2}=\frac{1}{L\times 2.56\times 10^{-6}}$

$\Rightarrow \frac{1}{L}=\frac{4\times 10\times 9\times 10^{16}\times 2.56\times 10^{-6}}{960\times 960}$

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