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

$$ f_0 = \frac{1}{2\pi\sqrt{LC}} $$

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 $$ L' $$ such that:

$$ \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{L' \cdot 4C}} $$

Now solving for $$ L' $$:

$$ \sqrt{LC} = \sqrt{4CL'} $$

Squaring both sides, we have:

$$ LC = 4CL' $$

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

$$ \frac{L}{4} = L' $$

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:

$$ L' = \frac{L}{4} $$

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

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