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In an $L-C-R$ circuit, the capacitance is changed from $C$ to $4 C$. For the same resonant frequency, the inductance should be changed from $L$ to
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Verified Answer
The correct answer is:
$\frac{L}{4}$
In $L-C-R$ circuit,
Resonant frequency, $f_0=\frac{1}{2 \pi \sqrt{L C}}$
When, $C^{\prime}=4 C$, then resonant frequency is given as
$$
\begin{aligned}
{f^{\prime}}_0^{\prime} & =\frac{1}{2 \pi \sqrt{L^{\prime} C^{\prime}}} \\
& =\frac{1}{2 \pi \sqrt{L^{\prime} \cdot 4 C}}=\frac{1}{4 \pi \sqrt{L^{\prime} C}}
\end{aligned}
$$
According to question,
$$
\begin{aligned}
& f_0=f^{\prime}{ }_0 \\
& \frac{1}{2 \pi \sqrt{L C}}=\frac{1}{4 \pi \sqrt{L^{\prime} C}} \\
& \Rightarrow \quad \frac{1}{\sqrt{L C}}=\frac{1}{2 \sqrt{L^{\prime} C}} \Rightarrow \frac{1}{L C}=\frac{1}{4 L^{\prime} C} \\
& \Rightarrow \quad 4 L^{\prime}=L \quad \Rightarrow \quad L^{\prime}=\frac{L}{4} \\
&
\end{aligned}
$$
Resonant frequency, $f_0=\frac{1}{2 \pi \sqrt{L C}}$
When, $C^{\prime}=4 C$, then resonant frequency is given as
$$
\begin{aligned}
{f^{\prime}}_0^{\prime} & =\frac{1}{2 \pi \sqrt{L^{\prime} C^{\prime}}} \\
& =\frac{1}{2 \pi \sqrt{L^{\prime} \cdot 4 C}}=\frac{1}{4 \pi \sqrt{L^{\prime} C}}
\end{aligned}
$$
According to question,
$$
\begin{aligned}
& f_0=f^{\prime}{ }_0 \\
& \frac{1}{2 \pi \sqrt{L C}}=\frac{1}{4 \pi \sqrt{L^{\prime} C}} \\
& \Rightarrow \quad \frac{1}{\sqrt{L C}}=\frac{1}{2 \sqrt{L^{\prime} C}} \Rightarrow \frac{1}{L C}=\frac{1}{4 L^{\prime} C} \\
& \Rightarrow \quad 4 L^{\prime}=L \quad \Rightarrow \quad L^{\prime}=\frac{L}{4} \\
&
\end{aligned}
$$
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