Question

An LC circuit with a $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{-}\mathit{p}\mathit{F}$capacitor oscillates in such a manner as to radiate at a wavelength of $\mathbf{3}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{m}$. (a) What is the resonant frequency? (b) What inductance is in series with the capacitor?

Short answer

(a) The resonant frequency is $9.09\times {10}^7\mathrm{Hz}$.

(b) The inductance is $6.13\times {10}^{-7}\mathrm{H}$.

Step-by-step solution

Define wavelength

The distance between two successive crests or troughs of a wave is known as its wavelength.

Evaluate resonant frequency

(a)

The frequency of light and its amplitude is known to be connected by the formula

$f=\frac{C}{\lambda }$

Where $f$is frequency, $c$is the speed of light, $\lambda$is the wavelength.

Substitute the values in the above equation,

role="math" localid="1655877235774" $$\begin{gathered} \mathrm{f}=\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{3.3\mathrm{m}} \\ =9.09\times {10}^7\mathrm{Hz} \end{gathered}$$

Therefore, the resonant frequency is $9.09\times {10}^6\mathrm{Hz}$.

Evaluate inductance

(b)

The formula for inductance and capacitance in relation to resonant frequency is

$f=\frac{1}{2\pi \sqrt{LC}}$

$L=\frac{1}{4{\pi }^2{f}^{2}C}$

Where $L$is inductance, $C$is capacitance, and $f$is frequency.

Substitute the values in the above equation,

$\begin{array}{l}\mathrm{L}=\frac{1}{4{\mathrm{\pi }}^2\mathrm{x}(9.09\mathrm{x}{10}^7\mathrm{Hz})\mathrm{x}\left(5\mathrm{x}{10}^{-12}\mathrm{F}\right)}\\ =6.13\times {10}^{-7}\mathrm{H}\end{array}$

Therefore, the inductance is $6.13\times {10}^{-7}\mathrm{H}$.