Question

The resonant frequency of an LC circuit is \( f_0 \). If a dielectric slab of constant 16 is inserted completely between the plates of the capacitor, then the resonant frequency is

• A. \(\frac{f_0}{2}\)
• B. \(2 f_0\)
• C. \(\frac{f_0}{4}\)
• D. \(4 f_0\)

Hint:

Dielectric increases capacitance, thus decreasing resonant frequency by \(\sqrt{K}\).

Answer 1

The Correct Option is C

Step 1: Formula for resonant frequency
\[ f_0 = \frac{1}{2\pi \sqrt{LC}} \] Step 2: Effect of dielectric
Inserting a dielectric increases capacitance by factor \(K\): \[ C' = K C, \quad K = 16 \] Step 3: New resonant frequency
\[ f = \frac{1}{2 \pi \sqrt{L C'}} = \frac{1}{2 \pi \sqrt{L \times 16 C}} = \frac{1}{4} \times \frac{1}{2 \pi \sqrt{L C}} = \frac{f_0}{4} \] Step 4: Conclusion
New resonant frequency is \(\frac{f_0}{4}\).