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117.An L-C resonant circuit contains a 400 pF capacitor and an inductor of 400 μH. It is coupled to an antenna. Wavelength of
radiated electromagnetic wave is
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radiated electromagnetic wave is
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The correct answer is:
754 m
Given, capacitance, C = 400 pF
Inductance, $L=400 \mu \mathrm{H}$
Frequency of resonating $L C$-circuit, $f=\frac{1}{2 \pi \sqrt{L C}}$
Wavelength $(\lambda)=\frac{\text { Velocity }(c)}{\text { Frequency }(f)}$
$\Rightarrow$ Wavelength, $\lambda=c \times 2 \pi \times \sqrt{L C}$
where, c is velocity of electromagnetic radiation
$\begin{aligned} \lambda & =3 \times 10^8 \times 2 \times 3.14 \times \sqrt{400 \times 10^{-12} \times 400 \times 10^{-6}} \\ & =754 \mathrm{~m}\end{aligned}$
Inductance, $L=400 \mu \mathrm{H}$
Frequency of resonating $L C$-circuit, $f=\frac{1}{2 \pi \sqrt{L C}}$
Wavelength $(\lambda)=\frac{\text { Velocity }(c)}{\text { Frequency }(f)}$
$\Rightarrow$ Wavelength, $\lambda=c \times 2 \pi \times \sqrt{L C}$
where, c is velocity of electromagnetic radiation
$\begin{aligned} \lambda & =3 \times 10^8 \times 2 \times 3.14 \times \sqrt{400 \times 10^{-12} \times 400 \times 10^{-6}} \\ & =754 \mathrm{~m}\end{aligned}$
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