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

Physics · Capacitors and R-C Circuits

JEE Main 2024 — 5 April, Shift 1 — Question 51

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 $\qquad$ $\Omega$ . (Given, $\log 3=1.1)$

Numerical answer — enter this value.

$\mathrm{E}=\frac{\mathrm{E}_{0}}{3} \Rightarrow \mathrm{~V}=\frac{\mathrm{V}_{0}}{3}$

$\frac{\mathrm{V}_{0}}{3}=\mathrm{V}_{0} \mathrm{e}^{-\frac{\mathrm{t}}{\tau}}$

$\mathrm{t}=\tau \ell \mathrm{n} 3$

$6.6 \times 10^{-6}=\mathrm{R}\left(1.5 \times 10^{-6}\right)(1.1)$

$\mathrm{R}=\frac{6}{1.5}=4 \Omega$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.