An LC resonant circuit contains a 400 , pF ca | vedclass

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Question

(B) The resonant frequency $
u$ of an $LC$ circuit is given by $
u = \frac{1}{2\pi \sqrt{LC}}$.Given $L = 100 	imes 10^{-6} \, H$ and $C = 400 	im

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$377 \, m$

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(B) The resonant frequency $
u$ of an $LC$ circuit is given by $
u = \frac{1}{2\pi \sqrt{LC}}$.Given $L = 100 	imes 10^{-6} \, H$ and $C = 400 	imes 10^{-12} \, F$.$
u = \frac{1}{2\pi \sqrt{100 	imes 10^{-6} 	imes 400 	imes 10^{-12}}} = \frac{1}{2\pi \sqrt{4 	imes 10^{-14}}} = \frac{1}{2\pi 	imes 2 	imes 10^{-7}} = \frac{10^7}{4\pi} \, Hz$.The wavelength $\lambda$ is given by $\lambda = \frac{c}{
u}$,where $c = 3 	imes 10^8 \, m/s$.$\lambda = \frac{3 	imes 10^8}{10^7 / 4\pi} = \frac{3 	imes 10^8 	imes 4\pi}{10^7} = 120\pi \, m$.Using $\pi \approx 3.14159$,$\lambda \approx 120 	imes 3.14159 \approx 377 \, m$.

An $LC$ resonant circuit contains a $400 \, pF$ capacitor and a $100 \, \mu H$ inductor. It is set into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves is

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