Zes,
Kirchoff says that the sum of emf's around a circuit loop will be zero.
For your circuit, the terminal voltage of your coil will be equal to the supply voltage when the key is closed.
From Lenz law e =-l di/dt.
If the supply voltage is a constant amplitude, the the coil voltage will be constant and the rate of rise of current will be constant.
Note that Lenz law has a minus sign. This sign says that the voltage produce in the coil is of the opposite polarity to that of the voltage which caused it. Doing a Kirchoff shows that the sum of the voltage around the loop is zero.
So, while the switch is closed, there is a constant rate of change of current through the coil. That is, the current increases linearly with time, and continues to increase, and in your case, because there is no resistance, the current increases to infinity.
If at some point, the switch is opened, then the rate of change of current becomes infinite (dI/0 = infinity). The voltage across the coil becomes infinitely large.
A change to your circuit to short circuit the coil when the switch opens, will result in the terminal voltage of the coil becoming zero. In the is case, the rate of change of current will be zero; that is, the current will be maintained. So, with a perfect coil, once a current is flowing through a coil, by short circuiting its terminals, the current will be maintained forever. This principle is used in magnetic memory devices which operate under superconducting conditions.
In the above, we used a step voltage to drive the coil. If we used a source of alternating voltage, the with the switch closed, the coil terminal voltage would again be equal to the source voltage. So, for e = E sin wt, the coil voltage would be E sin wt = - L * di/dt . So dI/dt = (E sin wt)/L. Integrating gives; I = - (E/(L*w)) * cos wt . Since cos wt = sin(90-wt), the we can say that; I=(E/(L*w)) * sin(wt-90).
Thus, with a sine wave source, the current lags the voltage by 90 degree.
You can substitute your constants for L and wt and calculate current.
Part of your question is about transformers. In a perfect transformer, the transformer simply transforms the load side impedance into a different impedance at the primary side. The transformer is only an impedance changing device. to deal with a transformer, change the load impedance by the ratio (N1/N2)squared.
In practical circuits, the transformer has leakage inductance between the primary and secondary and this leakage inductance is used for timing effects and other application requirements. For example, the Kettering spark ignition coil has a loosely coupled secondary winding. When the coil current is interrupted, the primary voltage rises to say 250 volt. the turns ratio of the coil then multiplies this voltage to say 20,00 volts and generates the spark.if the primary and secondary were closely coupled with no leakage inductance, the primary voltage would rise to only 24 volt or so and the turns ratio of the coil would need to be greater to get a spark. Note on the newer cars that the coil structure looks different. These modern coils are designed to have closer coupling than the old Kettering types.
Hope this helps.