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Question: Answered & Verified by Expert
A series resonant $\mathrm{AC}$ circuit contains a capacitance $10^{-6} \mathrm{~F}$ and an inductor of $10^{-4} \mathrm{H}$. The frequency of electrical oscillations will be
PhysicsAlternating CurrentKCETKCET 2022
Options:
  • A $10 \mathrm{~Hz}$
  • B $\frac{10^5}{2 \pi} \mathrm{Hz}$
  • C $\frac{10}{2 \pi} \mathrm{Hz}$
  • D $10^5 \mathrm{~Hz}$
Solution:
2916 Upvotes Verified Answer
The correct answer is: $\frac{10^5}{2 \pi} \mathrm{Hz}$
Given, capacitance, $C=10^{-6} \mathrm{~F}$
Inductance, $L=10^{-4} \mathrm{H}$
Frequency of oscillation of $L-C$ series resonant . circuit,
$$
\begin{aligned}
f & =\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{10^{-4} \times 10^{-6}}} \\
& =\frac{1}{2 \pi \times 10^{-5}}=\frac{10^5}{2 \pi} \mathrm{Hz}
\end{aligned}
$$

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