A

Given the circuit with an inductor L and a resistor R, when the switch S shifts from position 1 to position 2, the circuit configuration changes. We can assume the voltage across the inductor is initially zero because the switch has been in position 1 for a long time.

Now, applying Kirchhoff's loop law, when the switch is moved to position 2, we have:
$$ V = L \frac{di}{dt} + Ri $$
where:
- V is the voltage of the source
- R is the resistance
- L is the inductance
- i is the current

Rearranging gives:
$$ L \frac{di}{dt} + Ri = V $$

To solve this first-order linear differential equation, we can use the standard method:
1. Rewrite the equation:
$$ \frac{di}{dt} + \frac{R}{L} i = \frac{V}{L} $$
2. The integrating factor \( \mu(t) \) is given by:
$$ \mu(t) = e^{\frac{R}{L}t} $$
3. Multiply through by the integrating factor:
$$ e^{\frac{R}{L}t} \frac{di}{dt} + \frac{R}{L} e^{\frac{R}{L}t} i = \frac{V}{L} e^{\frac{R}{L}t} $$
4. Integrate both sides to find i(t):
$$ i(t) = \left(\frac{V}{R}\right) \left(1 - e^{-\frac{R}{L}t}\right) + C e^{-\frac{R}{L}t} $$
5. At t = 0, assuming the current i(0) = 0, we find C = -\( \frac{V}{R} \) and thus:
$$ i(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right) $$
This expression describes the growth of current in the circuit over time after the switch is moved to position 2. Therefore, the final solution is:
$$ i(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right) $$.