Question

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

• A. 3.33
• B. 2.12
• C. 1.56
• D. 1.91

Answer 1

Answer: (C)

Given, natural frequency of an $L-C$ circuit
$=125kHz=125\times 10^{3}Hz$
Let dielectric constant $=K$
The natural frequency, $f_N=\frac{1}{2\pi \sqrt{LC}}$
After filling the dielectric material of dielectric constant $K$
$f_M=\frac{1}{2\pi \sqrt{KLC}}\left(\because C^{{}^{\prime }}=KC\right)$
$\therefore \frac{f_N}{f_M}=\frac{1/2\pi \sqrt{LC}}{1/2\pi \sqrt{KLC}}=\sqrt{K}$
$\Rightarrow \frac{125\times 10^3}{100\times 10^3}=\sqrt{K}$
$\left[f_M=125kHz-25kHz=100kHz=100\times 10^{3}Hz\right]$
$\Rightarrow \frac{5}{4}=\sqrt{K}$
$\Rightarrow K={\left(\frac{5}{4}\right)}^2=\frac{25}{16}=1.562$