Question

A coil of inductance 0.4 mH is connected to a capacitor of capacitance 400 pF. To what wavelength is this circuit tuned ?

Answer 1

`L=0.4 mH = 0.4 xx 10^(-3) H, C = 400 p f = 4 xx 10^(-10)F`
Frequency `f=(1)/(2 pi sqrt(LC))=(1)/(2 xx 3.14)xx(1)/(sqrt(0.4xx10^(-3)xx4xx10^(-10)))=(10^(-7))/(6.28xx4)Hz`
If the speed of electromagnetic wave is v, then
`lambda = (v)/(l) = (3xx10^(8) ms^(-1))/((10^(7))/(6.28xx4 Hz))=753.6 m`