Question

The natural frequency of an LC circuit is 120 kHz. When the capacitor in the circuit is totally filled with a dielectric material, the natural frequency of the circuit decreases by 20 kHz. The dielectric constant of the material is:

• A. \( 3.33 \)
• B. \( 1.44 \)
• C. \( 2.12 \)
• D. \( 1.91 \)

Hint:

The frequency of an LC circuit changes when a dielectric is introduced, following \( f' = \frac{f}{\sqrt{\kappa}} \).
- The dielectric constant is found using the ratio \( \frac{f'}{f} \).

Answer 1

The Correct Option is B

The natural frequency of an LC circuit is given by: \[ f = \frac{1}{2\pi} \frac{1}{\sqrt{LC}} \] When a dielectric of constant \( \kappa \) is introduced, the new capacitance becomes: \[ C' = \kappa C \] and the new frequency is: \[ f' = \frac{1}{2\pi} \frac{1}{\sqrt{L \kappa C}} \] 1. Given data: - Initial frequency: \( f = 120 \) kHz - Final frequency after dielectric: \( f' = 100 \) kHz (since frequency decreases by 20 kHz) 2. Finding \( \kappa \): \[ \frac{f'}{f} = \frac{1}{\sqrt{\kappa}} \] \[ \frac{100}{120} = \frac{1}{\sqrt{\kappa}} \] \[ \sqrt{\kappa} = \frac{120}{100} = 1.2 \] \[ \kappa = (1.2)^2 = 1.44 \] Thus, the correct answer is \(\boxed{1.44}\).