JEE Main 2024 — Alternating Current Question with Solution

The resonance frequency $$f_0$$ of an LCR circuit is given by:

where:

• $$L$$ is the inductance.
• $$C$$ is the capacitance.

To keep the resonance frequency unchanged when the capacitance is changed from $$C$$ to $$4C$$, we must adjust the inductance $$L$$ to a new value $$L^{\prime }$$ such that:

Now solving for $$L^{\prime }$$:

Squaring both sides, we have:

Dividing both sides by $$4C$$, we get:

Thus, the new inductance $$L^{\prime }$$ is one fourth of the original inductance $$L$$. Therefore, to achieve the same resonance frequency with the capacitance increased to $$4C$$, the inductance should be reduced to a quarter of its initial value:

This means we have reduced the inductance by $$\frac{3}{4}L$$, so the correct answer is:

Option C: reduced by $$\frac{3}{4}L$$