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**Hint:**when switch ${k_1}$ is closed an emf is induced in the circuit. Applying the formula of emf for the inductor inducing it, we will establish a relation between inductor current and time.

In the above relation, when the boundary conditions of the second circuit alignment is inputted, we will obtain another relation between current and time.

**Formulae used:**emf is induced in the circuit: $E = L\dfrac{{di}}{{dt}}$.

Where $E$ is the induced emf and is expressed in Volts $(V)$, $L$ is the inductance and is expressed in Henry $(H)$, $di$ is the change in current and is expressed in Ampere $(A)$ and $dt$ is the time taken for the current to change and is expressed in seconds $(s)$.

**Step by step solution:**When ${k_1}$ is closed, current flows in the circuit due to the battery connected to it. The presence of the inductance coil creates a magnetic flux, which on changing, results in the induction of an electromagnetic force. This is represented by $E$ and is equal to $L\dfrac{{di}}{{dt}}$.

Applying the boundary condition at $t = 0$ we get,

$

E = L\dfrac{{di}}{{dt}} \\

\Rightarrow di = \dfrac{E}{L} \times dt \\

$

Upon integration within the limits of $0 \to {t_o}$ we get,

$

\Rightarrow {\smallint _0}^{{I_{{t_o}}}}di = {\smallint _0}^{{t_o}}\dfrac{E}{L} \times dt \\

\Rightarrow {I_{{t_o}}} = \dfrac{E}{L}{t_o} \\

$

This linear variation is from $t$ to ${t_o}$.

Now, applying the boundary conditions at $t > {t_o}$ having limits of ${t_o} \to \infty $ we get,

$

E = L\dfrac{{di}}{{dt}} \\

\Rightarrow di = \dfrac{E}{L} \times dt \\

\Rightarrow {\smallint _{{I_{{t_o}}}}}^\infty di = {\smallint ^\infty }_{{t_o}}\dfrac{E}{L} \times dt \\

\Rightarrow {I_\infty } - {I_{{t_o}}} = \dfrac{E}{L}({t_\infty } - {t_o}) \\

$

Which is impossible.

Therefore $L\dfrac{{di}}{{dt}} = 0$ is considered.

This is a constant and therefore there is no variation in inductor current with time.

In conclusion, the correct graph is option A.

**Note:**The first boundary condition is not $t = 0$. It ranges from $t = 0$ to $t = {t_o}$. Similarly, the second one is also a range, that is, $t = {t_o}$ to $t = {t_\infty }$. Calculations are made to be considered in a given time period and not just at a particular instant of time.

**Additional information:**A LR circuit is a circuit having a combination of inductor(s) and resistor(s). In AC circuits, they reduce voltage and in DC circuits, the inductor acts as a static resistance. Therefore, the circuit given in the above problem has DC connection because no resistor is present.

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