In an LCR circuit,the resonating frequency is | vedclass

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(C) The resonant frequency of an $LCR$ circuit is given by the formula $f = \frac{1}{2 \pi \sqrt{LC}}$.From this,we can see that $f \propto \frac{1}{\

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(C) The resonant frequency of an $LCR$ circuit is given by the formula $f = \frac{1}{2 \pi \sqrt{LC}}$.From this,we can see that $f \propto \frac{1}{\sqrt{LC}}$.Let the initial frequency be $f_1 = 500 \,kHz$ with inductance $L$ and capacitance $C$.Given the new values: $L' = 2L$ and $C' = \frac{1}{8}C$.The new frequency $f_2$ is given by $f_2 = \frac{1}{2 \pi \sqrt{L'C'}} = \frac{1}{2 \pi \sqrt{(2L)(\frac{1}{8}C)}} = \frac{1}{2 \pi \sqrt{\frac{1}{4}LC}}$.Simplifying this,we get $f_2 = \frac{1}{2 \pi \sqrt{LC} \cdot \sqrt{\frac{1}{4}}} = \frac{1}{2 \pi \sqrt{LC} \cdot \frac{1}{2}} = 2 	imes \frac{1}{2 \pi \sqrt{LC}} = 2f_1$.Substituting the value of $f_1$,we get $f_2 = 2 	imes 500 \,kHz = 1000 \,kHz$.

In an $LCR$ circuit,the resonating frequency is $500 \,kHz$. If the value of $L$ is doubled and the value of $C$ is decreased to $\frac{1}{8}$ times its initial value,then the new resonating frequency in $kHz$ will be .......

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