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Question

(C) In the given circuit,two inductors of inductance $L$ are connected in parallel. The equivalent inductance $L_{eq}$ is given by:$\frac{1}{L_{eq}} =

Vedclass · Accepted Answer

$\frac{1}{\pi \sqrt{LC}}$

Vedclass · Answer

(C) In the given circuit,two inductors of inductance $L$ are connected in parallel. The equivalent inductance $L_{eq}$ is given by:$\frac{1}{L_{eq}} = \frac{1}{L} + \frac{1}{L} = \frac{2}{L} \implies L_{eq} = \frac{L}{2}$Two capacitors of capacitance $C$ are connected in series. The equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C} \implies C_{eq} = \frac{C}{2}$The resonant frequency $f$ of an $LC$ circuit is given by the formula:$f = \frac{1}{2 \pi \sqrt{L_{eq} C_{eq}}}$Substituting the values of $L_{eq}$ and $C_{eq}$:$f = \frac{1}{2 \pi \sqrt{(\frac{L}{2}) (\frac{C}{2})}} = \frac{1}{2 \pi \sqrt{\frac{LC}{4}}} = \frac{1}{2 \pi \frac{\sqrt{LC}}{2}} = \frac{1}{\pi \sqrt{LC}}$Thus,the correct option is $C$.

The frequency at resonance for the circuit shown in the figure is:

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