Capacitors

JEE Main level, numerical answer, medium

150 of 180

Question 150 of 180

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 )

Show answer and solution

Answer: 4

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}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Capacitors, dielectrics and RC circuits simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Capacitors (Charging and Discarging of capacitors), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.