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, \(\log 3=1.1)\)
- A \(2\)
- B \(3\)
- C \(4\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\frac{\mathrm{E}_0}{3} \Rightarrow \mathrm{V}=\frac{\mathrm{V}_0}{3}\) \(\frac{\mathrm{V}_0}{3}=\mathrm{V}_0 \mathrm{e}^{-\frac{\mathrm{t}}{\mathrm{t}}}\) \(\mathrm{t}=\tau \ell \mathrm{n} 3\) \(6.6 \times 10^{-6}=\mathrm{R}\left(1.5 \times 10^{-6}\right)(1.1)\)…
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