Transient solution of current in circuit with two coupled inductors, with resistances
The primary side inductor at \$t=0s\$ will behave just like an open circuit but &quot;transformer action&quot; will occur at \$t=0s\$ as well. So we cannot ignore it.
Thus for \$L_1 = 50mH\$ and \$L_2 = 100mH\$ the turn ratio is:
$$N= \sqrt{\frac{L_1}{L_2}}= \sqrt{\frac{50mH}{100mH}} = \frac{1}{\sqrt{2}} \approx 0.707$$
thus the resistance \$R_2\$ is seen at the primary side ( reflected resistance) as a resistor equal to:
$$R_r = N^2*R_2 = 0.5k\Omega$$
Therefore the voltage across the primary inductance at \$t=0s\$ is
$$V_{L1(0s)} = 12V \frac{0.5k\Omega}{1k\Omega +0.5k\Omega} = 4V$$
And the current is
$$I_{L1(0s)} = \frac{12V - 4V}{1k\Omega} = 8mA$$
And the current will start to rise exponential from \$8mA\$ to \$12mA\$ with a time constant equal to:
$$\tau = \frac{L_1}{R_1||(N^2R_2)}$$
And all of this is true for \$K = 1\$ at least I hope so.