AP EAMCET · PHYSICS · Alternating Current
Resonance frequency of \(L-C-R\) series AC circuit is \(f_0\). Now inductance is reduced to \(\frac{1}{4}\) times and capacitance is increased to 16 times, then the resonance frequency becomes.
- A \(\frac{f_0}{4}\)
- B \(\frac{t_0}{2}\)
- C \(2 f_0\)
- D \(4 f_0\)
Answer & Solution
Correct Answer
(B) \(\frac{t_0}{2}\)
Step-by-step Solution
Detailed explanation
Resonance frequency of series \(L-C-R\) circuit, \[ f_0=\frac{1}{2 \pi \sqrt{L C}}...(i) \] \(\Rightarrow\) When inductance is reduced to \(\frac{L}{4}\) and capacitance is increased to \(16 \mathrm{C}\), then new value of resonance frequency,…
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