KCET · Physics · Alternating Current
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
- A \(10 \mathrm{~Hz}\)
- B \(\frac{10^5}{2 \pi} \mathrm{Hz}\)
- C \(\frac{10}{2 \pi} \mathrm{Hz}\)
- D \(10^5 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(B) \(\frac{10^5}{2 \pi} \mathrm{Hz}\)
Step-by-step Solution
Detailed explanation
Given, capacitance, \(C=10^{-6} \mathrm{~F}\)
Inductance, \(L=10^{-4} \mathrm{H}\)
Frequency of oscillation of \(L-C\) series resonant . circuit,
\(\begin{aligned}f & =\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{10^{-4} \times 10^{-6}}} \\& =\frac{1}{2 \pi \times 10^{-5}}=\frac{10^5}{2 \pi} \mathrm{Hz}\end{aligned}\)
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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