Question

A L - C resonant circuit contains a 200 pF capacitor and a $100\mu$ H inductor it is set into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves is

• A. 377 mm
• B. 266 m
• C. 377 cm
• D. 3.77 cm

Answer 1

The correct option is B 266 m

Given that,

$C=200pF$

$C=200\times {10}^{-12}F$

$L=100\mu H$

$L=100\times {10}^{-6}H$

We know that,

Frequency of wave = $\nu =\frac{1}{2\pi \sqrt{LC}}$

$\nu =\frac{1}{2\times 3.14\sqrt{200\times {10}^{-12}\times 100\times {10}^{-6}}}$

$\nu =\frac{1}{2\times 3.14\times 1.4142\times {10}^{-7}}$

Now, the wavelength is

$\lambda =\frac{3\times {10}^8}{\nu }$

$\lambda =\frac{3\times {10}^8}{\frac{1}{2\times 3.14\times 1.4142\times {10}^{-7}}}$

$\lambda =3\times {10}^8\times 2\times 3.14\times 1.4142\times {10}^{-7}$

$\lambda =26.643\times 10$

$\lambda =266.43m$

Hence, the wavelength of radiated electromagnetic wave is $266$ m.