The natural frequency of an L-C circuit is 1, | vedclass

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(C) The natural frequency of an $L-C$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.This implies $f \propto \frac{1}{\sqrt{C}}$.When the capacit

Vedclass · Accepted Answer

$1.56$

Vedclass · Answer

(C) The natural frequency of an $L-C$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.This implies $f \propto \frac{1}{\sqrt{C}}$.When the capacitor $C$ is replaced by a capacitor with a dielectric medium of constant $K$,the new capacitance becomes $C' = KC$.The new frequency $f'$ is $f - 25 	ext{ kHz} = 125 	ext{ kHz} - 25 	ext{ kHz} = 100 	ext{ kHz}$.Using the ratio $\frac{f'}{f} = \sqrt{\frac{C}{C'}} = \sqrt{\frac{C}{KC}} = \frac{1}{\sqrt{K}}$.Substituting the values: $\frac{100}{125} = \frac{1}{\sqrt{K}}$.$\frac{4}{5} = \frac{1}{\sqrt{K}} \implies \sqrt{K} = \frac{5}{4} = 1.25$.Therefore,$K = (1.25)^2 = 1.5625 \approx 1.56$.

The natural frequency of an $L-C$ circuit is $1,25,000 \text{ cycles/s}$. Then the capacitor $C$ is replaced by another capacitor with a dielectric medium of dielectric constant $K$. In this case,the frequency decreases by $25 \text{ kHz}$. The value of $K$ is

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