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A coil of inductance 300 mH and resistance $2 \Omega$ is connected to a source of voltage 2 V . The current reaches half of its steady state value in
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Verified Answer
The correct answer is:
0.1 s
The current at any instant is given by
$I=I_0\left(1-e^{-R t / L}\right)$
$\begin{aligned} \frac{I_0}{2} & =I_0\left(1-e^{-R t / L}\right) \\ \frac{1}{2} & =\left(1-e^{-R t / L}\right) \\ e^{-R t / L} & =1 / 2 \\ \frac{R t}{L} & =\ln 2\end{aligned}$
$\begin{aligned}
\therefore \quad t=\frac{L}{R} \ln 2 & =\frac{300 \times 10^{-3}}{2} \times 0.693 \\
& =150 \times 0.693 \times 10^{-3} \\
& =0.10395 \mathrm{~s}=0.1 \mathrm{~s}
\end{aligned}$
$I=I_0\left(1-e^{-R t / L}\right)$
$\begin{aligned} \frac{I_0}{2} & =I_0\left(1-e^{-R t / L}\right) \\ \frac{1}{2} & =\left(1-e^{-R t / L}\right) \\ e^{-R t / L} & =1 / 2 \\ \frac{R t}{L} & =\ln 2\end{aligned}$
$\begin{aligned}
\therefore \quad t=\frac{L}{R} \ln 2 & =\frac{300 \times 10^{-3}}{2} \times 0.693 \\
& =150 \times 0.693 \times 10^{-3} \\
& =0.10395 \mathrm{~s}=0.1 \mathrm{~s}
\end{aligned}$
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