JEE Mains · Physics · STD 12 -7. Alternating current
In an \(LC\) oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes \(x\) times its initial resonant frequency \(\omega_0\). The value of \(x\) is:
- A \(\frac{1}{4}\)
- B \(16\)
- C \(\frac{1}{16}\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
The resonance frequency of LC oscillations circuit is \(\omega_0=\frac{1}{\sqrt{ LC }}\) \(L \rightarrow 2 L\) \(C \rightarrow 8 C\) \(\omega=\frac{1}{\sqrt{2 L \times 8 C }}=\frac{1}{4 \sqrt{ LC }}\) \(\omega=\frac{\omega_0}{4}\) So \(x =\frac{1}{4}\)
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