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JEE Main 20245 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Morning Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2024 · 5 April · Morning Shift · Physics · Capacitance

The electric field between the two parallel plates of a capacitor...

The electric field between the two parallel plates of a capacitor of 1.5μF1.5 \mu \mathrm{F} capacitance drops to one third of its initial value in 6.6μs6.6 \mu \mathrm{s} when the plates are connected by a thin wire. The resistance of this wire is _____Ω\Omega. (Given, log⁡3=1.1\log 3=1.1 )

Correct answer

0

Step-by-step solution

E=E03⇒V=V03V03=V0e−tτt=τℓn36.6×10−6=R(1.5×10−6)(1.1)R=61.5=4Ω\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}
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