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

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(A) The current in an $LR$ circuit during growth is given by the formula: $I(t) = I_0(1 - e^{-\frac{R}{L}t})$Here,$I_0 = \frac{E}{R}$ is the maximum s

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

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(A) The current in an $LR$ circuit during growth is given by the formula: $I(t) = I_0(1 - e^{-\frac{R}{L}t})$Here,$I_0 = \frac{E}{R}$ is the maximum steady-state current.Given values are: $L = 10\,H$,$R = 5\,\Omega$,$E = 5\,V$,and $t = 2\,s$.First,calculate $I_0 = \frac{5\,V}{5\,\Omega} = 1\,A$.Now,substitute the values into the growth equation:$I(2) = 1 	imes (1 - e^{-\frac{5}{10} 	imes 2})$$I(2) = 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 $2\,s$ the connection is made,the current flowing in ampere in the circuit is

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