Question

A radio can tune over the frequency range of a portion of $MW$ broadcast band: ( $800\text{kHz}$ to $1200\text{kHz}$ ). If its $\text{LC}$ circuit has an effective inductance of $200\mu \text{H}$, what must be the range of its variable capacitor?

Answer 1

Given,
$L=200\mu H=200\times {10}^{-6}H$
Frequency range of a portion of $MW$ broadcast band $800\text{kHz}(\text{say}{f}_{mrn})$ to $1200\text{kHz}\left(\text{say}{f}_{\text{max}}\right)$.
So, frequency of free oscillations $f$ should lie between this range.
$\Rightarrow f_{min}<f<f_{max}$
As frequency of free oscillations
$f=\frac{1}{2\pi \sqrt{LC}}$
$\Rightarrow C=\frac{1}{4{\pi }^2{f}^{2}L}$
The range of $C$ for given value of $L$
$\Rightarrow C\propto \frac{1}{f^x}$
$\text{If}f=f_{man}=800\text{kHz}$
$C_{max}=\frac{1}{4{\pi }^2\times {(800\times {10}^7)}^2\times 200\times {10}^{-6}}$
$C_{max}=197.7\times {10}^{-12}F=197.7\text{pF}$
$\text{If}f=f_{max}=1200\text{kHz}$
$C_{min}=\frac{1}{4{\pi }^2\times {(1200\times {10}^{-{}^2})}^2\times 200\times {10}^{-6}}$
$C_{min}=87.8\times {10}^{-12}F=87.8pF$
$\text{Hence,}{C}_{min}<C<C_{max}$
$\text{Final Answer: The variable capacitor should have a range of about}88\text{pF}\text{to}198\text{pF}\text{.}$