## Related questions with answers

The AM band extends from approximately 500 kHz to 1 600 kHz. If a 2.0-$\mu \mathrm { H }$ inductor is used in a tuning circuit for a radio, what are the extremes that a capacitor must reach to cover the complete band of frequencies?

Solutions

VerifiedIn this problem, the frequency of an AM band ranges from $f_\text{min} = 500~\mathrm{kHz} = 5.0 \times 10^{5}~\mathrm{Hz}$ to $f_\text{max} = 1600~\mathrm{kHz} = 1.6 \times 10^{6}~\mathrm{Hz}$. The inductor in the radio has inductance $L = 2.0~\mathrm{\mu H} = 2.0 \times 10^{-6}~\mathrm{H}$. We calculate the range of the capacitor to cover the frequency band.

The frequency at which the current has its maximum value when the impedance has its minimum value is known as the resonance frequency and is given by

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

where

$L$ is the inductance

$C$ is the capacitance.

The resonance frequency is given as:

$\begin{align*} f &= \dfrac{1}{{2\pi \sqrt {LC} }}\\ f^2 &= \dfrac{1}{{{{(2\pi )}^2}LC}}\\ \end{align*}$

If we make C as the subject of the equation, we get:

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

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