electromagnetic induction - Secondary current in a loosely coupled transformer

How does the secondary current depend on the resistive load placed on the secondary winding of a loosely coupled transformer, in which the primary winding is supplied with a linear current ramp ?

Answer

The easiest way to analyze such problem is to write Kirchhoff's equations for a pair of coupled coils, the way EEs do it: $$V_1 = L_1 \frac{dI_1}{dt} + M \frac{dI_2}{dt}\\ V_2 = M \frac{dI_1}{dt} + L_2 \frac{dI_2}{dt}$$ with the termination condition : $$V_2=-I_2R_2$$

You have also specified that the input current is a ramp $I_1=k_1t$ so that the coupled equations become $$V_1 = L_1 k_1 + M \frac{dI_2}{dt}\\ -I_2R_2 = M k_1 + L_2 \frac{dI_2}{dt}$$ and after rearrangement $L_2 \frac{dI_2}{dt}+I_2R_2 + M k_1=0$ for $t>0$ the solution of which is an increasing exponential $I_2(t)=-\frac{Mk_1}{R_2} (1-e^{-t\frac{R_2}{L_2}})$. Notice that this approaches a constant current $I_{max}=-\frac{Mk_1}{R_2}$ as $t \to \infty$