Question

A coil of inductance $8\mu H$ is connected to a capacitor of capacitance $0.02\mu F$. To what wavelength is this circuit tuned ?

• A. 7.54 × 10³ m
• B. 4.12 × 10² m
• C. 15.90 × 10³ m
• D. 7.54 × 10² m

Answer 1

Answer: (D)

Here, $L=8\mu H=8\times 10^{-6}H$;
$C=0.02\mu F=0.02\times 10^{-6}F$
$\therefore$ Resonant frequency,
$f_r=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{8\times 10^{-6}\times 0.02\times 10^{-6}}}$
$=3.98\times 10^{5}Hz$
If $c\left(=3\times 10^{8}m{s}^{-1}\right)$ is the velocity of the electromagnetic wave, then,
Wavelength, $\lambda =\frac{c}{f}=\frac{3\times 10^8}{3.98\times 10^5}=7.54\times 10^{2}m$