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.
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