there is nothing surprising here  the inductance L2 is less than L1, therefore the current amplitude is higher. How to find the energy stored in an inductor? According to the following formula: ML = I^2 * L / 2 If losses are not taken into account, then ML=I(L1)^2 * L1 / 2 = I(L2)^2 * L2 / 2 From here I(L1)^2 * L1 = I(L2)^2 * L2 Further I(L2) = I(L1) * sqrt(L1 / L2) Therefore, if L2 < L1, then L1 / L2 > 1, and I(L2) > I(L1). Let's reduce L2 to, say, 10mH and the difference in amplitudes will be noticeable to the naked eye. However, the energy from this no larger becomes.
