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Step-by-Step Solution
Step 1: Identify the given circuit and parameters
We have a series circuit with a resistor of resistance $R = 5\,\Omega$ and an inductor of inductance $L = 10\,\text{mH} = 0.01\,\text{H}$. A constant emf of $20\,\text{V}$ is applied at $t = 0$.
Step 2: Write the expression for the current in an LR circuit
For a DC source applied to an LR circuit at $t=0$, the current $i(t)$ grows according to the formula:
$$
i(t) = i_0\Bigl(1 - e^{-\tfrac{Rt}{L}}\Bigr),
$$
where $i_0 = \dfrac{V}{R}$ is the steady-state (or final) current.
Step 3: Calculate the steady-state current $i(\infty)$
As $t \to \infty$, $e^{-\tfrac{Rt}{L}} \to 0$, so the current becomes
$$
i(\infty) = i_0 = \frac{V}{R}.
$$
Step 4: Calculate the current at $t = 40\,\text{s}$
At $t = 40\,\text{s}$, the current is
$$
i(40) = i_0 \Bigl(1 - e^{-\tfrac{R \times 40}{L}}\Bigr).
$$
Step 5: Form the ratio $\,\dfrac{i(\infty)}{i(40)}$
The ratio of the currents is
$$
\frac{i(\infty)}{i(40)}
\;=\;
\frac{i_0}{\,i_0\bigl(1 - e^{-\tfrac{R \times 40}{L}}\bigr)}
\;=\;
\frac{1}{\,1 - e^{-\tfrac{5 \times 40}{0.01}}}.
$$
Step 6: Evaluate the exponential term
We have
$$
e^{-\tfrac{5 \times 40}{0.01}} \;=\; e^{-2000}.
$$
This value is extremely small, effectively making $1 - e^{-2000} \approx 1.$ As a result,
$$
\frac{1}{1 - e^{-2000}} \approx 1.
$$
Step 7: Match with the closest given option
The value is very close to 1. Among the provided choices, the nearest answer is 1.06.
