The electric field between the two parallel plates of a capacitor of $1.5 \mu \mathrm{F}$ capacitance drops to one third of its initial value in $6.6 \mu \mathrm{s}$ when the plates are connected by a thin wire. The resistance of this wire is. . . . . . . .$\Omega$. (Given, $\log 3=1.1)$
JEE MAIN 2024, Diffcult
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$\mathrm{E}=\frac{\mathrm{E}_0}{3} \Rightarrow \mathrm{V}=\frac{\mathrm{V}_0}{3}$
$\frac{\mathrm{V}_0}{3}=\mathrm{V}_0 \mathrm{e}^{-\frac{\mathrm{t}}{\mathrm{t}}}$
$\mathrm{t}=\tau \ell \mathrm{n} 3$
$6.6 \times 10^{-6}=\mathrm{R}\left(1.5 \times 10^{-6}\right)(1.1)$
$\mathrm{R}=\frac{6}{1.5}=4 \Omega$
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