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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2024
  • Hard
  • Numerical

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, \(\log 3=1.1)\)

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Answer & Solution

Correct Answer

(C) \(4\)

Step-by-step Solution

Detailed explanation

\(\mathrm{E}=\frac{\mathrm{E}_0}{3} \Rightarrow \mathrm{V}=\frac{\mathrm{V}_0}{3}\)

\(\frac{\mathrm{V}_0}{3}=\mathrm{V}_0 \mathrm{e}^{-\frac{\mathrm{t}}{\mathrm{t}}}\)

\(\mathrm{t}=\tau \ell \mathrm{n} 3\)

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

\(\mathrm{R}=\frac{6}{1.5}=4 \Omega\)

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