The problem involves two distinct phases. First, Section A (an $LC$ oscillator) oscillates until its capacitor voltage drops to half its initial value. Second, a switch connects Section A to an identical Section B. We must analyze the energy distribution and current dynamics to find the maximum current in the inductor of Section B.

Initially, the capacitor in Section A has voltage $V_0$. The total energy in Section A is: $$ E_{total} = \frac{1}{2} C V_0^2 $$ The circuit oscillates. At the instant the switch is closed (let’s call this $t=0$), the voltage across the capacitor has dropped to $V_A = \frac{V_0}{2}$. We can find the current $I_0$ flowing through the inductor $L$ of Section A at this instant using conservation of energy:

$$ U_{initial} = U_{C,final} + U_{L,final} $$ $$ \frac{1}{2} C V_0^2 = \frac{1}{2} C \left( \frac{V_0}{2} \right)^2 + \frac{1}{2} L I_0^2 $$ $$ \frac{1}{2} C V_0^2 = \frac{1}{8} C V_0^2 + \frac{1}{2} L I_0^2 $$ $$ \frac{3}{8} C V_0^2 = \frac{1}{2} L I_0^2 \implies I_0^2 = \frac{3C V_0^2}{4L} $$ $$ I_0 = \sqrt{\frac{3C}{4L}} V_0 $$

When the switch closes, the capacitor of Section A ($C_A$) and the capacitor of Section B ($C_B$) are connected in parallel.
Note: Charge sharing between capacitors happens almost instantaneously. However, due to the inertia of inductance, the currents through the inductors cannot change instantaneously.

• Currents just after switch ($t=0^+$): $i_1 = I_0$ and $i_2 = 0$.
• Voltages just after switch: Using conservation of charge for the capacitors: $$ Q_{total} = C \left( \frac{V_0}{2} \right) + C(0) = \frac{CV_0}{2} $$ Since the capacitors are identical ($C$) and in parallel, the total capacitance is $2C$. The common voltage $V’$ becomes: $$ V’ = \frac{Q_{total}}{2C} = \frac{CV_0/2}{2C} = \frac{V_0}{4} $$

We calculate the total energy of the system immediately after the redistribution (accounting for heat loss in the wire during charge sharing):

After switching, the two inductors $L_A$ and $L_B$ are connected in parallel across the capacitor bank. Therefore, the voltage across them is always equal: $$ V_L = L \frac{di_1}{dt} = L \frac{di_2}{dt} $$ Integrating this relationship: $$ \frac{d}{dt}(i_1 – i_2) = 0 \implies i_1 – i_2 = \text{constant} $$ Using the initial values at $t=0^+$ ($i_1 = I_0, i_2 = 0$): $$ i_1 – i_2 = I_0 \implies i_1 = i_2 + I_0 $$

We need to find the maximum current in the inductor of Section B, denoted as $i_{2,max}$. In an LC oscillation, current is maximum when the potential energy (voltage across capacitors) is zero ($V_C = 0$). At this moment, all the energy of the system is stored in the magnetic field of the inductors.

Applying Conservation of Energy between $t=0^+$ and the moment of maximum current: $$ E_{sys} = \frac{1}{2} L i_1^2 + \frac{1}{2} L i_2^2 $$ Substitute $i_1 = i_2 + I_0$: $$ \frac{7}{16} C V_0^2 = \frac{1}{2} L (i_2 + I_0)^2 + \frac{1}{2} L i_2^2 $$ Multiply by $\frac{2}{L}$: $$ \frac{7}{8} \frac{C V_0^2}{L} = (i_2 + I_0)^2 + i_2^2 $$ $$ \frac{7}{8} \frac{C V_0^2}{L} = i_2^2 + I_0^2 + 2 i_2 I_0 + i_2^2 $$ $$ \frac{7}{8} \frac{C V_0^2}{L} = 2i_2^2 + 2 i_2 I_0 + I_0^2 $$ Substitute $I_0^2 = \frac{3}{4} \frac{C V_0^2}{L}$ and let $K = \frac{C V_0^2}{L}$: $$ \frac{7}{8} K = 2i_2^2 + 2 i_2 \sqrt{\frac{3}{4}K} + \frac{3}{4} K $$ $$ \frac{7}{8} K – \frac{6}{8} K = 2i_2^2 + \sqrt{3K} i_2 $$ $$ \frac{1}{8} K = 2i_2^2 + \sqrt{3}\sqrt{K} i_2 $$ Rearranging into a standard quadratic form $Ax^2 + Bx + C = 0$: $$ 16i_2^2 + 8\sqrt{3K} i_2 – K = 0 $$ Solving for $i_2$: $$ i_2 = \frac{-8\sqrt{3K} \pm \sqrt{ (8\sqrt{3K})^2 – 4(16)(-K) }}{2(16)} $$ $$ i_2 = \frac{-8\sqrt{3}\sqrt{K} \pm \sqrt{ 192K + 64K }}{32} $$ $$ i_2 = \frac{-8\sqrt{3}\sqrt{K} \pm \sqrt{ 256K }}{32} $$ $$ i_2 = \frac{-8\sqrt{3}\sqrt{K} \pm 16\sqrt{K}}{32} $$ Considering the magnitude of the current (and noting that the oscillation allows for negative values representing direction): $$ i_2 = \left( \frac{\pm 16 – 8\sqrt{3}}{32} \right) \sqrt{K} $$ The two extreme values are $\frac{2 – \sqrt{3}}{4}\sqrt{K}$ and $\frac{-2 – \sqrt{3}}{4}\sqrt{K}$. The maximum current corresponds to the largest magnitude: $$ |i_{2,max}| = \frac{2 + \sqrt{3}}{4} \sqrt{K} $$