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+### Statement
+
+$11.4.22.$ [Insert the problem statement]
+
+### Solution
+
+a) Heat dissipated in R when closing K
+
+Initially: current I in $L_1 $ zero in $L_2$
+Magnetic fluxes cannot change abruptly. Upon closing K, the following is conserved:
+
+$L_1 I_1 + L_2 I_2 = L_1 I$
+
+In the final state,$ I_1 = I_2 = I_0, so I_0 = \frac{L_1}{L_1+L_2}I$
+The dissipated energy is the difference between the initial and final magnetic energy:
+
+$W = \frac{1}{2}L_1 I^2 - \frac{1}{2}(L_1+L_2)I_0^2
+ = \frac{L_1 L_2}{2(L_1+L_2)} I^2$
+
+b) Current limits after opening K
+
+With K closed, currents are$ I_1 $in $L_1$ and$ I_2 $in $L_2$
+Upon opening K, the total flux remains constant.
+Since they are in series, $I_1' = I_2' = I'$
+Therefore:
+
+$(L_1 + L_2) I' = L_1 I_1 + L_2 I_2 \quad\Rightarrow\quad I' = \frac{L_1 I_1 + L_2 I_2}{L_1 + L_2}$
+
+The current in L_1 varies between:
+
+$I_1 \quad\text{and}\quad I_1 - \frac{2(I_1 - I_2)}{1 + L_1/L_2}$
+
+The current in L_2 varies between:
+
+$I_2 \quad\text{and}\quad I_2 + \frac{2(I_1 - I_2)}{1 + L_2/L_1}$
+
+#### Answer
+
+[Insert a concise answer or boxed result]