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 .............. Ω. (Given, log 3 = 1.1)

Answer 1

$E = \frac{E_0}{3} \implies V = \frac{V_0}{3}$

$\frac{V_0}{3} = V_0 e^{-\frac{t}{\tau}}$

$t = \tau \ln 3$

6.6 \times 10^{-6} = R (1.5 \times 10^{-6})(1.1) $$$$ R = \frac{6.6}{1.5} = 4 \, \Omega