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{L C}}=\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, $\lambda=\frac{c}{f}=\frac{3 \times 10^{8}}{3.98 \times 10^{5}}=7.54 \times 10^{2} m$