JEE Main 2024 Physics 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: 4
Solution
E ∝ : t = RC ln 3 ⇒ R = 6.6 × 10⁻⁶/(1.5 × 10⁻⁶ × 1.1) = 4 Ω (ln 3 taken as 1.1).