Energy Stored in a Magnetic Field — Question 8

Question 8

An unequal current split stores extra energy in uncoupled parallel inductors

Two uncoupled ideal inductors L1 and L2 carry branch currents i1 and i2 with prescribed total $i_1+i_2=I$. Do not assume a particular preparation history.

1. Find the current split that minimizes total magnetic energy.

2. Write the energy for a deviation delta from that split and identify the excess.

Question 8 - Solution

An unequal current split stores extra energy in uncoupled parallel inductors

(a) Method and calculation

$$U=\tfrac12L_1i_1^2+\tfrac12L_2(I-i_1)^2,\quad\frac{dU}{di_1}=(L_1+L_2)i_1-L_2I.$$$$\boxed{i_{1*}=\frac{L_2I}{L_1+L_2},\quad i_{2*}=\frac{L_1I}{L_1+L_2}}.$$

(b) Interpretation and further calculation

$$i_1=i_{1*}+\delta,\quad i_2=i_{2*}-\delta.$$$$\boxed{U=\frac12\frac{L_1L_2}{L_1+L_2}I^2+\frac12(L_1+L_2)\delta^2}.$$

Physical check

The calculation identifies an energy minimum; it does not claim that ideal lossless branches automatically relax to that split.