An ideal coil of 10 Henry is joined in series with a resistance of 5 ohm and a battery of 5 volt. 2 second after joining, the current flowing in ampere in the circuit will be :
Answer & explanation
Correct answer: option 2
$I = I_0 (1 - e^\frac{-t}{\tau})$
$I_0 = \frac{\xi}{R} = 1A$
$\tau = \frac{L}{R} = 2 Sec$
$\Rightarrow I = (1 - e^\frac{-2}{2}) = 1- e^{-1}$