Question

In an \(LC\) circuit, angular frequency at resonance is \(\omega\). The new angular frequency when inductance is made four times and capacitance is made eight times is

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

Hint:

Resonant frequency is inversely proportional to the square root of the product of L and C.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the change in resonant frequency of an \(LC\) circuit when inductance and capacitance are scaled by certain factors.
Step 2: Key Formula or Approach:
The resonant angular frequency of an \(LC\) circuit is given by:
\[ \omega = \frac{1}{\sqrt{LC}} \]
Step 3: Detailed Explanation:
Let the initial inductance be \(L\) and initial capacitance be \(C\).
The initial resonant frequency is \(\omega = \frac{1}{\sqrt{LC}}\).
According to the problem:
New inductance \(L' = 4L\)
New capacitance \(C' = 8C\)
The new angular frequency \(\omega'\) is:
\[ \omega' = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(4L)(8C)}} \]
\[ \omega' = \frac{1}{\sqrt{32LC}} = \frac{1}{\sqrt{16 \times 2 \times LC}} \]
\[ \omega' = \frac{1}{4\sqrt{2} \cdot \sqrt{LC}} \]
Substituting \(\omega = \frac{1}{\sqrt{LC}}\):
\[ \omega' = \frac{\omega}{4\sqrt{2}} \]
Step 4: Final Answer:
The new frequency is \(\frac{\omega}{4\sqrt{2}}\).