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(C) The resonance frequency of a series $LCR$ circuit is given by $\omega = \frac{1}{\sqrt{LC}}$.To keep the resonance frequency unchanged,we must hav

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reduced by $\frac{3}{4} L$

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(C) The resonance frequency of a series $LCR$ circuit is given by $\omega = \frac{1}{\sqrt{LC}}$.To keep the resonance frequency unchanged,we must have $\omega' = \omega$,which implies $\frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{LC}}$.Squaring both sides,we get $L'C' = LC$.Given that the new capacitance $C' = 4C$,we substitute this into the equation: $L'(4C) = LC$.Solving for $L'$,we get $L' = \frac{L}{4}$.The change in inductance is $\Delta L = L - L' = L - \frac{L}{4} = \frac{3L}{4}$.Therefore,the inductance must be reduced by $\frac{3}{4} L$.

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

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