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If the conductance and capacitance are both doubled in $L-C$ - $R$ circuit, the resonant frequency of the circuit will
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Verified Answer
The correct answer is:
decrease to one-half the original value
The resonant frequency
$$
v_{0}=\frac{1}{2 \pi \sqrt{L C}} \Rightarrow v_{0} \propto \frac{1}{\sqrt{L C}}
$$
If conductance and capacitance both are doubled, then
$$
v_{0}=\frac{1}{2}\left(\frac{1}{2 \pi} \sqrt{L C}\right)
$$
So, the resonant frequency will decrease to one-half of the original value.
$$
v_{0}=\frac{1}{2 \pi \sqrt{L C}} \Rightarrow v_{0} \propto \frac{1}{\sqrt{L C}}
$$
If conductance and capacitance both are doubled, then
$$
v_{0}=\frac{1}{2}\left(\frac{1}{2 \pi} \sqrt{L C}\right)
$$
So, the resonant frequency will decrease to one-half of the original value.
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