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Q. The LCR circuit shown below is driven by an ideal AC voltage source.
image
At frequency $f=\frac{1}{2 \pi \sqrt{ LC }}$, the correct statement is

TS EAMCET 2020

Solution:

The $A C$ circuit is shown in the following figure,
image
Frequency of $A C$ source is given as
$f=\frac{1}{2 \pi \sqrt{L C}}$
Clearly, it is resonant frequency.
Hence, $X_{L}=X_{C}$
In this case, voltages developed across $L$ and $C$ are equal in magnitude but opposite in pollaries.
$\therefore Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$
$=\sqrt{R^{2}+0^{2}} $
$Z=R $
$\therefore I=\frac{V}{Z}=\frac{V}{R}$
Hence, the current through $R$ depends only on the value of $R$ and not $L$ and $C$.