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