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 :

Options:

$e^{-1}$

$1 - e^{-1}$

1 - e

Correct Answer:

$1 - e^{-1}$

Explanation:

$I = I_0 (1 - e^\frac{-t}{\tau})$

$I_0 = \frac{\xi}{R} = 1A$

$\tau = \frac{L}{R} = 2 Sec$

$\Rightarrow I = (1 - e^\frac{-2}{2}) = 1- e^{-1}$

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 :