Question:
A series RLC circuit has \(L=0.1\,\mathrm{H}\) and \(C=10\,\mu\mathrm{F}\), then the resonant frequency is
A series RLC circuit has \(L=0.1\,\mathrm{H}\) and \(C=10\,\mu\mathrm{F}\), then the resonant frequency is
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For a series RLC circuit,
\[
\boxed{
f_r=\frac{1}{2\pi\sqrt{LC}}
}
\]
where \(L\) is in henry and \(C\) is in farad.
Updated On: Jul 14, 2026
- \(159\,\mathrm{Hz}\)
- \(318\,\mathrm{Hz}\)
- \(503\,\mathrm{Hz}\)
- \(1000\,\mathrm{Hz}\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the resonance formula.
The resonant frequency is
\[
f_r=\frac{1}{2\pi\sqrt{LC}}.
\]
Given,
\[
L=0.1\,\mathrm{H},\qquad
C=10\,\mu\mathrm{F}=10\times10^{-6}\,\mathrm{F}.
\]
Step 2: Calculate the frequency. \[ LC = 0.1\times10^{-5} = 10^{-6}, \] \[ \sqrt{LC}=10^{-3}. \] Hence, \[ f_r = \frac{1}{2\pi\times10^{-3}} \approx159\,\mathrm{Hz}. \] Therefore, \[ \boxed{159\,\mathrm{Hz}} \] is the correct answer. Hence, \[ \boxed{(A)} \] is the correct answer.
Step 2: Calculate the frequency. \[ LC = 0.1\times10^{-5} = 10^{-6}, \] \[ \sqrt{LC}=10^{-3}. \] Hence, \[ f_r = \frac{1}{2\pi\times10^{-3}} \approx159\,\mathrm{Hz}. \] Therefore, \[ \boxed{159\,\mathrm{Hz}} \] is the correct answer. Hence, \[ \boxed{(A)} \] is the correct answer.
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