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(C) The natural frequency of an $L-C$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.When the capacitor is filled with a dielectric material of c

Vedclass · Accepted Answer

$1.56$

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(C) The natural frequency of an $L-C$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.When the capacitor is filled with a dielectric material of constant $K$,the new capacitance becomes $C' = KC$.The new frequency is $f' = \frac{1}{2 \pi \sqrt{L(KC)}} = \frac{1}{\sqrt{K}} f$.Given $f = 125 \ kHz$ and the frequency decreases by $25 \ kHz$,the new frequency is $f' = 125 \ kHz - 25 \ kHz = 100 \ kHz$.Substituting these values: $\frac{100}{125} = \frac{1}{\sqrt{K}}$.$\frac{4}{5} = \frac{1}{\sqrt{K}} \Rightarrow \sqrt{K} = \frac{5}{4} = 1.25$.Therefore,$K = (1.25)^2 = 1.5625 \approx 1.56$.

The natural frequency of an $L-C$ circuit is $125 \ kHz$. When the capacitor is totally filled with a dielectric material,the natural frequency decreases by $25 \ kHz$. The dielectric constant of the material is nearly

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