Resonance frequency of $L-C-R$ series AC circuit is $f_0$. Now inductance is reduced to $\frac{1}{4}$ times…
Resonance frequency of $L-C-R$ series AC circuit is $f_0$. Now inductance is reduced to $\frac{1}{4}$ times and capacitance is increased to 16 times, then the resonance frequency becomes.
$\frac{f_0}{4}$
$\frac{t_0}{2}$
$2 f_0$
$4 f_0$
Solution
Resonance frequency of series $L-C-R$ circuit,
$
f_0=\frac{1}{2 \pi \sqrt{L C}}...(i)
$
$\Rightarrow$ When inductance is reduced to $\frac{L}{4}$ and capacitance is increased to $16 \mathrm{C}$, then new value of resonance frequency,
$
\begin{aligned}
f_0^{\prime} & =\frac{1}{2 \pi \sqrt{\frac{L}{4} \times 16 C}} \\
& =\frac{1}{2 \pi \sqrt{4 L C}}=\frac{1}{2} \cdot \frac{1}{2 \pi \sqrt{L C}} \\
& =\frac{f_0}{2}
[from Eq (i)]\end{aligned}
$