Question

In an LC oscillator, if the values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of the oscillator becomes \( x \) times its initial resonant frequency \( \omega_0 \). The value of \( x \) is:

• A. \( \frac{1}{4} \)
• B. \( 16 \)
• C. \( \frac{1}{16} \)
• D. \( 4 \)

Hint:

For LC oscillators: \[ \omega = \frac{1}{\sqrt{LC}} \] If \( L \) is multiplied by \( a \) and \( C \) by \( b \), the new frequency is: \[ \omega' = \frac{\omega_0}{\sqrt{a b}} \]

Answer 1

The Correct Option is A

Step 1: Understanding the formula for resonant frequency. The resonant frequency \( \omega \) of an LC oscillator is given by: \[ \omega = \frac{1}{\sqrt{LC}} \] where: - \( L \) is the inductance, - \( C \) is the capacitance. 
Step 2: Initial resonant frequency. \[ \omega_0 = \frac{1}{\sqrt{L_0 C_0}} \] 
Step 3: New resonant frequency after changes. Given: - \( L' = 2L_0 \), - \( C' = 8C_0 \), the new resonant frequency is: \[ \omega' = \frac{1}{\sqrt{L' C'}} \] \[ \omega' = \frac{1}{\sqrt{(2L_0)(8C_0)}} \] \[ \omega' = \frac{1}{\sqrt{16 L_0 C_0}} \] \[ \omega' = \frac{1}{4} \times \frac{1}{\sqrt{L_0 C_0}} \] \[ \omega' = \frac{\omega_0}{4} \] 
Step 4: Finding \( x \). Since \( \omega' = \frac{\omega_0}{4} \), we get: \[ x = \frac{1}{4} \] 
Final Answer: \[ \boxed{\frac{1}{4}} \]