The current flowing through an AC circuit is given by I = 10 (100 t) A . What is the shortest time interval required for the current to increase from 5 A to 5 2 A ?

Step-by-step solution

The given equation for the current is I = 10 (100 t) . Let t₁ be the time when the current is 5 A . 10 (100 t₁) = 5 (100 t₁) = 1 2 100 t₁ = 6 t₁ = 1 600 s Let t₂ be the time when the current is 5 2 A . 10 (100 t₂) = 5 2 (100 t₂) = 1 2 100 t₂ = 4 t₂ = 1 400 s The shortest time interval required is t = t₂ - t₁ . t = 1 400 - 1 600 t = 3 - 2 1200 = 1 1200 s Answer: 1 1200 s