Physics - EMI/AC Question with Solution | TestHub

$\lambda =960 \mathrm{m}$

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

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

$L=?$

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

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

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

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

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

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

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

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