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$L C R$ circuit, the resonance frequency of circuit increases two times of the initial circuit by changing $C$ and $C^{\prime}$ and $R$ from $100 \Omega$ to $400 \Omega$, while the inductance was kept the same. The ratio $C / C^{\prime}$, is
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The correct answer is:
4
Given,
in series $L C R$ circuit,
initial capacitance, $C_1=C$
final value of capacitance, $C_2=C^{\prime}$
inductance, $L_1=L_2=L$
and resistance, $R_1=100 \Omega$
$R_2=400 \Omega$
$\therefore$ Resonance frequency,
and
Given,
$$
f_1=\frac{1}{2 \pi \sqrt{L_1 C_1}} \Rightarrow f_1=\frac{1}{2 \pi \sqrt{L C}}
$$
and $\quad f_2=\frac{1}{2 \pi \sqrt{L_2 C_2}} \Rightarrow f_2=\frac{1}{2 \pi \sqrt{L \cdot C^{\prime}}}$
Given, $\quad f_2=2 f_1 \Rightarrow \frac{1}{2 \pi \sqrt{L C^{\prime}}}=\frac{2}{2 \pi \sqrt{L C}}$
$$
\frac{1}{\sqrt{C^{\prime}}}=\frac{2}{\sqrt{C}} \Rightarrow \frac{1}{C^{\prime}}=\frac{4}{C} \Rightarrow \frac{C}{C^{\prime}}=4
$$
in series $L C R$ circuit,
initial capacitance, $C_1=C$
final value of capacitance, $C_2=C^{\prime}$
inductance, $L_1=L_2=L$
and resistance, $R_1=100 \Omega$
$R_2=400 \Omega$
$\therefore$ Resonance frequency,
and
Given,
$$
f_1=\frac{1}{2 \pi \sqrt{L_1 C_1}} \Rightarrow f_1=\frac{1}{2 \pi \sqrt{L C}}
$$
and $\quad f_2=\frac{1}{2 \pi \sqrt{L_2 C_2}} \Rightarrow f_2=\frac{1}{2 \pi \sqrt{L \cdot C^{\prime}}}$
Given, $\quad f_2=2 f_1 \Rightarrow \frac{1}{2 \pi \sqrt{L C^{\prime}}}=\frac{2}{2 \pi \sqrt{L C}}$
$$
\frac{1}{\sqrt{C^{\prime}}}=\frac{2}{\sqrt{C}} \Rightarrow \frac{1}{C^{\prime}}=\frac{4}{C} \Rightarrow \frac{C}{C^{\prime}}=4
$$
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