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, $\log 3=1.1$ )

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{E}=\frac{\mathrm{E}_0}{3} \Rightarrow \mathrm{V}=\frac{\mathrm{V}_0}{3} \ & \frac{\mathrm{V}_0}{3}=\mathrm{V}_0 \mathrm{e}^{-\frac{\mathrm{t}}{	au}} \ & \mathrm{t}=	au \ell \mathrm{n} 3 \ & 6.6 	imes 10^{-6}=\mathrm{R}\left(1.5 	imes 10^{-6}ight)(1.1) \ & \mathrm{R}=\frac{6}{1.5}=4 \Omega\end{aligned}$

The electric field between the two parallel plates of a capacitor of $1.5 \mu \mathrm{F}$ capacitance drops…

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