The frequency of an L-C circuit is f_{1}. If | vedclass

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(B) The frequency of an ideal $L-C$ circuit is given by $f_{1} = \frac{1}{2\pi\sqrt{LC}}$.When a resistance $R$ is added in series,the circuit becomes

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$\sqrt{1-\frac{R^{2}C}{4L}}$

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(B) The frequency of an ideal $L-C$ circuit is given by $f_{1} = \frac{1}{2\pi\sqrt{LC}}$.When a resistance $R$ is added in series,the circuit becomes a damped $L-C-R$ circuit. The damped angular frequency is given by $\omega_{d} = \sqrt{\omega_{0}^{2} - \beta^{2}}$,where $\omega_{0} = \frac{1}{\sqrt{LC}}$ and $\beta = \frac{R}{2L}$.Thus,the damped frequency $f_{2}$ is given by:$f_{2} = \frac{1}{2\pi} \sqrt{\left(\frac{1}{\sqrt{LC}}\right)^{2} - \left(\frac{R}{2L}\right)^{2}}$Taking the ratio $\frac{f_{2}}{f_{1}}$:$\frac{f_{2}}{f_{1}} = \frac{\frac{1}{2\pi} \sqrt{\frac{1}{LC} - \frac{R^{2}}{4L^{2}}}}{\frac{1}{2\pi\sqrt{LC}}}$$= \sqrt{\frac{\frac{1}{LC} - \frac{R^{2}}{4L^{2}}}{\frac{1}{LC}}}$$= \sqrt{1 - \frac{R^{2}C}{4L}}$

The frequency of an $L-C$ circuit is $f_{1}$. If a resistance $R$ is also added to it,the frequency becomes $f_{2}$. The ratio $\frac{f_{2}}{f_{1}}$ will be:

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