Search any question & find its solution
Question:
Answered & Verified by Expert
A radio can tune over the frequency range of a portion of MW broadcast band: $(800 \mathrm{kHz}$ to $1200 \mathrm{kHz})$. If its $\mathrm{LC}$ circuit has an effective inductance of $200 \mu \mathrm{H}$, what must be the range of its variable capacitor?
[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.]
[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.]
Solution:
1188 Upvotes
Verified Answer
Given: Max. frequency $v_1=1200 \mathrm{kHz}$
$\min$. frequency $v_2=800 \mathrm{kHz}$
inductance $\mathrm{L}=200 \mu \mathrm{F}$
To find: Range of capacitor $\mathrm{C}_1$ and $\mathrm{C}_2$
Formula: $v=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
Tuning of radio means natural frequency of LC circuit matches broadcast frequency.
For lower limit $v_2=800 \mathrm{kHz}$
$$
\begin{aligned}
&=800 \times 10^3 \mathrm{~Hz}=8 \times 10^5 \mathrm{~Hz} \\
&\Rightarrow \mathrm{C}_2=\frac{1}{4 \pi^2 \mathrm{Lv}_2^2} \\
&=\frac{1}{4 \times(3.14)^2 \times 200 \times 10^{-6} \times\left(8 \times 10^5\right)^2} \\
&=\frac{1}{4 \times(3.14)^2 \times 2 \times 64 \times 10^{-4+10}} \\
&=\frac{10^{-6}}{8 \times 64 \times 3.14}=197.9 \times 10^{-12}=198 \mathrm{pF}
\end{aligned}
$$
For upper limit $v_1=1200 \mathrm{kHz}=12 \times 10^5 \mathrm{~Hz}$
$$
\begin{aligned}
\mathrm{C}_1 &=\frac{1}{4 \pi^2 v_1^2 \mathrm{~L}} \\
=& \frac{1}{4 \times(3.14)^2 \times\left(12 \times 10^5\right)^2 \times 200 \times 10^{-6}} \\
=& \frac{1}{8 \times(3.14)^2 \times 144 \times 200 \times 10^{10-4}} \\
=& \frac{10^{-6}}{8 \times(3.14)^2 \times 144}=87.9 \times 10^{-12} \mathrm{~F} \\
\therefore \quad \mathrm{C}_1 \approx 88 \mathrm{pF} \\
\text { Range of capacitor is } 88 \text { to } 198 \mathrm{pF}
\end{aligned}
$$
$\min$. frequency $v_2=800 \mathrm{kHz}$
inductance $\mathrm{L}=200 \mu \mathrm{F}$
To find: Range of capacitor $\mathrm{C}_1$ and $\mathrm{C}_2$
Formula: $v=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
Tuning of radio means natural frequency of LC circuit matches broadcast frequency.
For lower limit $v_2=800 \mathrm{kHz}$
$$
\begin{aligned}
&=800 \times 10^3 \mathrm{~Hz}=8 \times 10^5 \mathrm{~Hz} \\
&\Rightarrow \mathrm{C}_2=\frac{1}{4 \pi^2 \mathrm{Lv}_2^2} \\
&=\frac{1}{4 \times(3.14)^2 \times 200 \times 10^{-6} \times\left(8 \times 10^5\right)^2} \\
&=\frac{1}{4 \times(3.14)^2 \times 2 \times 64 \times 10^{-4+10}} \\
&=\frac{10^{-6}}{8 \times 64 \times 3.14}=197.9 \times 10^{-12}=198 \mathrm{pF}
\end{aligned}
$$
For upper limit $v_1=1200 \mathrm{kHz}=12 \times 10^5 \mathrm{~Hz}$
$$
\begin{aligned}
\mathrm{C}_1 &=\frac{1}{4 \pi^2 v_1^2 \mathrm{~L}} \\
=& \frac{1}{4 \times(3.14)^2 \times\left(12 \times 10^5\right)^2 \times 200 \times 10^{-6}} \\
=& \frac{1}{8 \times(3.14)^2 \times 144 \times 200 \times 10^{10-4}} \\
=& \frac{10^{-6}}{8 \times(3.14)^2 \times 144}=87.9 \times 10^{-12} \mathrm{~F} \\
\therefore \quad \mathrm{C}_1 \approx 88 \mathrm{pF} \\
\text { Range of capacitor is } 88 \text { to } 198 \mathrm{pF}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.