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In an $L R$-circuit, time constant is that time in which current grows from zero to the value (where $I_0$ is the steady state current)
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The correct answer is:
$0.63$ $I_0$
Current at any instant of time $t$ after closing an $L-R$ circuit is given by
$I=I_0\left[1-e^{\frac{-R}{L} t}\right]$
Time constant $t=\frac{L}{R}$
$\therefore I=I_0\left[1-e^{\frac{-R}{L} \times \frac{L}{R}}\right]=I_0\left(1-e^{-1}\right)=I_0\left(1-\frac{1}{e}\right)$
$=I_0\left(1-\frac{1}{2.718}\right)=0.63 I_0=63 \%$ of $I_0$
$I=I_0\left[1-e^{\frac{-R}{L} t}\right]$
Time constant $t=\frac{L}{R}$
$\therefore I=I_0\left[1-e^{\frac{-R}{L} \times \frac{L}{R}}\right]=I_0\left(1-e^{-1}\right)=I_0\left(1-\frac{1}{e}\right)$
$=I_0\left(1-\frac{1}{2.718}\right)=0.63 I_0=63 \%$ of $I_0$
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