CAPACITANCE_JEE-Main_050424_(S1) - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2024

05-04-2024 S1

Question

The electric field between the two parallel plates of a capacitor of 1.5μF capacitance drops to one third of its initial value in 6.6μs when the plates are connected by a thin wire. The resistance of this wire is.........Ω. (Given, log⁡3=1.1 )

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Solution

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

Question Tags

JEE Main

Physics

Medium

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