A) \[\frac{{{f}_{0}}}{4}\]
B) \[2\,{{f}_{0}}\]
C) \[{{f}_{0}}\]
D) \[\frac{{{f}_{0}}}{2}\]
Correct Answer: D
Solution :
We know that \[{{f}_{0}}= resonance frequency\] \[=\,\,\frac{1}{2\pi }\times \frac{1}{\sqrt{LC}}\] \[\Rightarrow \,\,{{f}_{0}}\propto \frac{1}{\sqrt{C}}\] when capacitance of the circuit is made 4 times let new resonance frequency is \[{{\operatorname{f}}_{0}}'\] \[\Rightarrow \,\,\,\,\,\,\frac{{{f}_{0}}'}{{{f}_{0}}}=\frac{\sqrt{C}}{\sqrt{4C}}=\frac{1}{2}\,\,\,\,\,\,\,\Rightarrow \,\,\,{{f}_{0}}'\,\,=\,\,\frac{{{f}_{0}}}{2}\]
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