ExamBro
ExamBro
JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

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, \(\log 3=1.1)\)

  1. A \(2\)
  2. B \(3\)
  3. C \(4\)
  4. D \(6\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4\)

Step-by-step Solution

Detailed explanation

\(\mathrm{E}=\frac{\mathrm{E}_0}{3} \Rightarrow \mathrm{V}=\frac{\mathrm{V}_0}{3}\) \(\frac{\mathrm{V}_0}{3}=\mathrm{V}_0 \mathrm{e}^{-\frac{\mathrm{t}}{\mathrm{t}}}\) \(\mathrm{t}=\tau \ell \mathrm{n} 3\) \(6.6 \times 10^{-6}=\mathrm{R}\left(1.5 \times 10^{-6}\right)(1.1)\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app