An ideal coil of 10 H is connected in series with a resistance of 5 Ω and a battery of 5 ~V . 2 second after the connection is made the current flowing in amperes in the circuit is

Correct answer: (D) (1-e^-1 ). i=i_0 (1-e^Rt/L ) = (1-e^-1 )

Q83 / 120 · JEE Main 2007 · Physics

An ideal coil of $10 \mathrm{H}$ is connected in series with a resistance of $5 \Omega$ and a battery of $5 \mathrm{~V}$. $2$ second after the connection is made the current flowing in amperes in the circuit is

$\mathrm{i}=\mathrm{i}_0\left(1-\mathrm{e}^{\frac{\mathrm{Rt}}{\mathrm{L}}}\right)$ $=\left(1-\mathrm{e}^{-1}\right)$

Hence, the correct answer is option: 'D'.