COMEDK · Physics · 23. Alternating Current
Quality factor of a series \(L-C-R\) circuit decreases from 3 to 2 . Resonant frequency is \(600 \mathrm{~Hz}\). Change in bandwidth is
- A zero
- B \(100 \mathrm{~Hz}\) increase
- C \(100 \mathrm{~Hz}\) decrease
- D \(300 \mathrm{~Hz}\) increase
Answer & Solution
Correct Answer
(B) \(100 \mathrm{~Hz}\) increase
Step-by-step Solution
Detailed explanation
Given, \(f_{0}=600 \mathrm{~Hz}, Q_{1}=3, Q_{2}=2\) The bandwidth in \(L-C-R\) circuit, \(\beta=\frac{f_{0}}{Q}\) As, quality factor decreases, bandwidth increases. This increase in bandwidth is given by…
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