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In a $L-C-R$ circuit, the capacitance is made one-fourth, then what should be change in inductance, so that the circuit remains in resonance?
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The correct answer is:
4 times
Resonance frequency $f=\frac{1}{2 \pi \sqrt{L C}}$
For the circuit to remain in resonance, $\sqrt{L C}=$ constant
$\begin{aligned} or \quad L_1 C_1 & =L_2 C_2 \\ or \quad L C & =L_2 \frac{C}{4} \\ \therefore L_2 & =4 L\end{aligned}$
$\therefore$ Inductance should be change to four times.
For the circuit to remain in resonance, $\sqrt{L C}=$ constant
$\begin{aligned} or \quad L_1 C_1 & =L_2 C_2 \\ or \quad L C & =L_2 \frac{C}{4} \\ \therefore L_2 & =4 L\end{aligned}$
$\therefore$ Inductance should be change to four times.
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