Answer & explanation
Correct answer: option 4
$\text{Rate of increase of current is } \frac{dI}{dt} = \frac{I_0}{\tau} e^\frac{-t}{\tau}$
$\tau = \frac{L}{R}= \frac{5\times 10^{-3}}{180}$
$I_0 = \frac{50}{180}$
Correct answer: option 4
$\text{Rate of increase of current is } \frac{dI}{dt} = \frac{I_0}{\tau} e^\frac{-t}{\tau}$
$\tau = \frac{L}{R}= \frac{5\times 10^{-3}}{180}$
$I_0 = \frac{50}{180}$