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Step-by-Step Solution
Step 1: Identify the Steady-State Current
When the inductor reaches steady state (long time after the circuit is switched on), the inductor behaves like a simple wire and the current is given by Ohm’s law. The steady-state current, denoted by $i_{0}$, is:
$i_{0} = \frac{V}{R} \;=\; \frac{2\,\text{V}}{2\,\Omega} \;=\; 1\,\text{A}$
Step 2: Write Down the Growth of Current in an LR Circuit
For a series LR circuit connected to a DC source, the instantaneous current $i(t)$ grows according to the formula:
$i(t) = i_{0}\Bigl(1 - e^{-\frac{R\,t}{L}}\Bigr)$
Step 3: Apply the Condition for Half of the Steady-State Current
We want the time $t$ when $i(t)$ is half of its maximum value $i_{0}$. So, set $i(t) = \frac{i_{0}}{2}$:
$\frac{i_{0}}{2} = i_{0}\Bigl(1 - e^{-\frac{R\,t}{L}}\Bigr)$
Divide both sides by $i_{0}$:
$\frac{1}{2} = 1 - e^{-\frac{R\,t}{L}} \quad\Rightarrow\quad e^{-\frac{R\,t}{L}} = \frac{1}{2}$
Step 4: Solve for the Time $t$
Take the natural logarithm on both sides:
$-\frac{R\,t}{L} = \ln\Bigl(\frac{1}{2}\Bigr) = -\ln(2)
Hence,
$\frac{R\,t}{L} = \ln(2) \quad \Rightarrow \quad t = \frac{L}{R}\,\ln(2)
Step 5: Substitute the Values
Given: $L = 300\,\text{mH} = 0.3\,\text{H}$ and $R = 2\,\Omega$. So,
$t = \frac{0.3\,\text{H}}{2\,\Omega}\,\ln(2) \approx 0.15 \times 0.693 \,\approx\, 0.10\,\text{s}
Final Answer
The current in the circuit reaches half of its steady-state value in $0.1\,\text{s}$.
