Question

In an LCR series circuit, the resonance frequency is \( f \). If the capacitance is made 4 times, what will be the new resonance frequency?

• A. \(4f\)
• B. \(2f\)
• C. \(f/2\)
• D. \(f/4\)

Hint:

In an LCR circuit, \[ f \propto \frac{1}{\sqrt{C}} \] So if capacitance increases \(k\) times, the frequency becomes \( \frac{1}{\sqrt{k}} \) times the original value.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to determine how the resonance frequency of an AC circuit changes when the physical parameters of the components (specifically capacitance) are altered.
Step 2: Key Formula or Approach:
The resonance frequency \( f \) of a series LCR circuit is:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]
This shows that \( f \propto \frac{1}{\sqrt{C}} \) when inductance \( L \) remains constant.
Step 3: Detailed Explanation:
Let the initial frequency be \( f_1 = \frac{1}{2\pi \sqrt{LC}} \).
The new capacitance is \( C_2 = 4C \).
The new frequency \( f_2 \) will be:
\[ f_2 = \frac{1}{2\pi \sqrt{L(4C)}} \]
\[ f_2 = \frac{1}{2\pi \cdot 2\sqrt{LC}} \]
\[ f_2 = \frac{1}{2} \left( \frac{1}{2\pi \sqrt{LC}} \right) \]
\[ f_2 = \frac{f_1}{2} \]
Step 4: Final Answer:
The new resonance frequency will be \( f/2 \).