Question

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

• A. 750m
• B. 751m
• C. 752m
• D. 753m

Answer 1

Answer: (D)

$L=0.4\, mH =0.4 \times 10^{-3} H$
$C=400\, pF =4 \times 10^{-10} F$
$\lambda=?$
$v=\frac{1}{2 \pi \sqrt{L C}}$
$v=\frac{1}{2 \times 3.14 \times \sqrt{0.4 \times 10^{-3} \times 4 \times 10^{-10}}}$
$v=\frac{10^{7}}{6.28 \times 4}\, Hz$
If $c$ is velocity of em waves, then
$\lambda=\frac{c}{v}=\frac{3 \times 10^{8}}{10^{7} / 6.28 \times 4}$
$\lambda=6.28 \times 10 \times 4 \times 3$
$\lambda=753.6\, m$