In series LCR circuit, the capacitance is changed from $C$ to $4 C$. To keep the resonance frequency unchanged, the new inductance should be:

<p>The resonance frequency $f_0$ of an LCR circuit is given by:</p> $f_0 = \frac{1}{2\pi\sqrt{LC}}$ <p>where:</p> <ul> <li>$L$ is the inductance.</li> <li>$C$ is the capacitance.</li> </ul> <p>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 $L'$ such that:</p> $\frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{L' \cdot 4C}}$ <p>Now solving for $L'$:</p> $\sqrt{LC} = \sqrt{4CL'}$ <p>Squaring both sides, we have:</p> $LC = 4CL'$ <p>Dividing both sides by $4C$, we get:</p> $\frac{L}{4} = L'$ <p>Thus, the new inductance $L'$ 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:</p> $L' = \frac{L}{4}$ <p>This means we have reduced the inductance by $\frac{3}{4}L$, so the correct answer is:</p> <p>Option C: reduced by $\frac{3}{4}L$</p>