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 (integer)
10
Solution
$\lambda$ = 960 m<br><br>C = 2.56 $\mu$F = 2.56 $\times$ 10<sup>$-$6</sup><sup></sup> F<br><br>c = 3 $\times$ 10<sup>8</sup> m/s<br><br>L = ?<br><br>Now at resonance, ${\omega _0} = {1 \over {\sqrt {LC} }}$<br><br>[Resonant frequency]<br><br>$2\pi {f_0} = {1 \over {\sqrt {LC} }}$<br><br>On substituting ${f_0} = {c \over \lambda }$, we have $2\pi {c \over \lambda } = {1 \over {\sqrt {LC} }}$<br><br>Squaring both sides : $4{\pi ^2}{{{c^2}} \over {{\lambda ^2}}} = {1 \over {LC}}$<br><br>$$ = {{4 \times 10 \times {{(3 \times {{10}^8})}^2}} \over {{{(960)}^2}}} = {1 \over {L \times 2.56 \times {{10}^{ - 6}}}}$$<br><br>$$ \Rightarrow {1 \over L} = {{4 \times 10 \times 9 \times {{10}^{16}} \times 2.56 \times {{10}^{ - 6}}} \over {960 \times 960}}$$<br><br>$\Rightarrow L = 10 \times {10^{ - 8}}$ H
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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