In the circuit shown in the figure,R = sqrt{f | vedclass

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(B) In the given circuit,the current through the $LR$ branch is $I_L = \frac{V}{R}(1 - e^{-Rt/L})$.The current through the $CR$ branch is $I_C = \frac

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$CR \ln(2)$

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(B) In the given circuit,the current through the $LR$ branch is $I_L = \frac{V}{R}(1 - e^{-Rt/L})$.The current through the $CR$ branch is $I_C = \frac{V}{R} e^{-t/CR}$.Given that $R = \sqrt{\frac{L}{C}}$,we have $R^2 = \frac{L}{C}$,which implies $\frac{L}{R} = CR$.Equating the two currents: $\frac{V}{R}(1 - e^{-Rt/L}) = \frac{V}{R} e^{-t/CR}$.Substituting $\frac{L}{R} = CR$,we get $1 - e^{-t/CR} = e^{-t/CR}$.$1 = 2 e^{-t/CR}$.$e^{t/CR} = 2$.Taking the natural logarithm on both sides: $\frac{t}{CR} = \ln(2)$.Therefore,$t = CR \ln(2)$.

Column-$I$	Column-$II$
$A$. Halophiles	$i$. Gram-negative bacteria
$B$. Thermoacidophiles	$ii$. Inhabit extremely salty areas
$C$. Spirochetes	$iii$. Responsible for methane production
$D$. Firmicutes	$iv$. Survive in acidic and high-temperature conditions
	$v$. Gram-positive bacteria

In the circuit shown in the figure,$R = \sqrt{\frac{L}{C}}$. Switch $S$ is closed at time $t = 0$. The current through the inductor $L$ and the capacitor $C$ would be equal after a time $t$ equal to:

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