An ideal coil of 10, H is connected in series | vedclass

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(D) The growth of current in an $L-R$ circuit is given by the formula:$I = I_{0}(1 - e^{-t/	au})$where $I_{0}$ is the steady-state current and $	au$

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$(1-e^{-1})$

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(D) The growth of current in an $L-R$ circuit is given by the formula:$I = I_{0}(1 - e^{-t/	au})$where $I_{0}$ is the steady-state current and $	au$ is the time constant.First,calculate the steady-state current $I_{0}$:$I_{0} = \frac{E}{R} = \frac{5\, V}{5\, \Omega} = 1\, A$Next,calculate the time constant $	au$:$	au = \frac{L}{R} = \frac{10\, H}{5\, \Omega} = 2\, s$Given the time $t = 2\, s$,substitute the values into the growth equation:$I = 1 \cdot (1 - e^{-2/2})$$I = (1 - e^{-1})\, A$

An ideal coil of $10\, H$ is connected in series with a resistance of $5\, \Omega$ and a battery of $5\, V$. After $2\, s$,after the connection is made,the current flowing (in ampere) in the circuit is.

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