Question

The electric field between the two parallel plates of a capacitor of 1.5 µF capacitance drops to one third of its initial value in 6.6 µs when the plates are connected by a thin wire. The resistance of this wire is .............. \(\Omega\). (Given, log 3=1.1)

Answer 1

Correct answer is : 4 

\(\mathrm{E}=\frac{\mathrm{E}_{0}}{3} \Rightarrow \mathrm{V}=\frac{\mathrm{V}_{0}}{3}\)

\(\frac{\mathrm{V}_{0}}{3}=\mathrm{V}_{0} \mathrm{e}^{-\frac{\mathrm{t}}{\tau}}\)

\(\mathrm{t}=\tau \mathrm{\ln}\ 3\)

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

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