MHT CET · Physics · Alternating Current
The frequency of the output signal of an LC oscillator circuit is ' \(F\) ' \(\mathrm{Hz}\) with a capacitance of \(0.1 \mu \mathrm{F}\). If the value of the capacitor is increased to \(0.2 \mu \mathrm{F}\), then the frequency of the output signal will be
- A \(\frac{F}{\sqrt{2}} \mathrm{~Hz}\)
- B \(\frac{F}{\sqrt{3}} \mathrm{~Hz}\)
- C \(\frac{F}{2} \mathrm{~Hz}\)
- D \(2 F \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(A) \(\frac{F}{\sqrt{2}} \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
The frequency of LC oscillation is given by
\(
\begin{aligned}
& F=\frac{1}{2 \pi \sqrt{\mathrm{LC}}} \\
& \therefore \frac{F_2}{F_1}=\sqrt{\frac{C_1}{C_2}}=\sqrt{\frac{0.1}{0.2}}=\frac{1}{\sqrt{2}} \\
& \therefore F_2=\frac{F_1}{\sqrt{2}}
\end{aligned}
\)
\(
\begin{aligned}
& F=\frac{1}{2 \pi \sqrt{\mathrm{LC}}} \\
& \therefore \frac{F_2}{F_1}=\sqrt{\frac{C_1}{C_2}}=\sqrt{\frac{0.1}{0.2}}=\frac{1}{\sqrt{2}} \\
& \therefore F_2=\frac{F_1}{\sqrt{2}}
\end{aligned}
\)
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