Question

What capacitance is needed in series with an$\mathbf{800}\mathbf{\mu H}$ inductor to form a circuit that radiates a wavelength of $\mathbf{196}\mathbf{}\mathbf{m}$?

Short answer

The capacitance is$13.53\mathrm{p}.\mathrm{F}.$

Step-by-step solution

Define wavelength

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

Evaluate capacitance

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

$\mathrm{f}=\frac{\mathrm{c}}{\mathrm{\lambda }}$

Where is frequency, is the speed of light, is the wavelength.

Substitute the values in the above equation,

$$\begin{gathered} \mathrm{f}=\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{196\mathrm{m}} \\ =1.53\times {10}^6\mathrm{Hz} \end{gathered}$$

The formula for the relationship between inductance and capacitance and the resonant frequency is

$$\begin{gathered} \mathrm{f}=\frac{1}{2\mathrm{\pi }\sqrt{\mathrm{LC}}} \\ \mathrm{C}=\frac{1}{4{\mathrm{\pi }}^2{\mathrm{f}}^2\mathrm{L}} \end{gathered}$$

Where L is inductance C is capacitance f is frequency.

Substitute the values in the above equation,

$$\begin{gathered} \mathrm{C}=\frac{1}{4{\mathrm{\pi }}^2\times {\left(1.53\times {10}^6\mathrm{Hz}\right)}^2\times \left(8.00\times {10}^{-4}\mathrm{H}\right)} \\ =1.353\times {10}^{-11}\mathrm{F}\left(\frac{1\mathrm{pF}}{{10}^{-12}\mathrm{F}}\right) \\ =13.53\mathrm{pF} \end{gathered}$$

Therefore, the capacitance is$13.53\mathrm{pF}.$