Question

The natural frequency of an $L-C$ circuit is $125000 \, cycles/s$ . When a dielectric medium of dielectric constant $K$ is introduced between the plates of the capacitor, the frequency decreases by $25 \, kHz$ . The value of $K$ is

• A. $3.05$
• B. $2.18$
• C. $1.56$
• D. $1.74$

Answer 1

Answer: (C)

$f=\frac{1}{2 \pi \sqrt{L C}}$
When a dielectric is introduced the capacitance becomes
$C ^{'} = K ⁡ C ⁡$
$\Rightarrow \frac{f ^{'}}{f}=\sqrt{\frac{C ⁡}{C ⁡^{'}}}$
$\frac{125000 - 25000}{125000}=\sqrt{\frac{C }{\textit{KC}}}$
$\frac{100}{125}=\frac{1}{\sqrt{K}}$
$K=\left(\frac{125}{100}\right)^{2}=1.56$