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, $\log 3=1.1$)

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Current Electricity

The electric field between the two parallel plates of a capacitor of

1.5 μ F

capacitance drops to one third of its initial value in

6.6 μ s

when the plates are connected by a thin wire. The resistance of this wire is ______

Ω

. (Given,

lo g 3 = 1.1

)

Answer: 4

During discharge the field (and voltage) decays as

E = E_{0} e^{- t / R C}

. It falls to one third when

\frac{E _{0}}{3} = E_{0} e^{- t / R C} \Rightarrow t = R C ln 3.

With

t = 6.6 μ s

C = 1.5 μ F

ln 3 \approx lo g 3 (as given) = 1.1

6.6 \times 1 0^{- 6} = R (1.5 \times 1 0^{- 6}) (1.1) \Rightarrow R = \frac{6.6}{1.65} = 4 Ω.

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Integer answer