EMI BYU 37

Solution 37: Parallel Inductors (Final)

37. Max Charge and Current

Part 1: Maximum Charge on Capacitor

We use the Conservation of Energy. The energy initially stored in the magnetic field of the parallel inductors transfers entirely to the electric field of the capacitor at the moment of maximum charge.

Initial State: Total current $I_{tot} = \mathcal{E}/r$. Equivalent inductance $L_{eq} = \frac{L_1 L_2}{L_1 + L_2}$.

$$ U_{mag} = \frac{1}{2} L_{eq} I_{tot}^2 = \frac{1}{2} \left( \frac{L_1 L_2}{L_1 + L_2} \right) \left( \frac{\mathcal{E}}{r} \right)^2 $$

Max Charge State: Current is momentarily zero (all energy in capacitor).

$$ U_{elec} = \frac{Q^2}{2C} $$

Equating energies ($U_{mag} = U_{elec}$):

$$ \frac{Q^2}{2C} = \frac{1}{2} \frac{L_1 L_2}{L_1 + L_2} \frac{\mathcal{E}^2}{r^2} $$ $$ Q = \frac{\mathcal{E}}{r} \sqrt{\frac{C L_1 L_2}{L_1 + L_2}} $$

Part 2: Maximum Current in L2

The system undergoes LC oscillations. The total current oscillates between $+I_{start}$ and $-I_{start}$.

Initial Current in $L_2$: In the parallel configuration, current splits inversely to inductance.

$$ I_{2, initial} = I_{tot} \left( \frac{L_1}{L_1 + L_2} \right) = \frac{\mathcal{E}}{r} \left( \frac{L_1}{L_1 + L_2} \right) $$

Total Swing (Maximum Current): Taking the oscillation into account, the current swings from positive peak to negative peak. The maximum current range (peak-to-peak amplitude) is twice the initial value.

$$ I_{2, max} = 2 \frac{\mathcal{E}}{r} \left( \frac{L_1}{L_1 + L_2} \right) $$