EMI BYU 32

32. Total Charge Flow

Inductor $L_0$: Finally it is in series with the battery internal resistance $r$ as inductor becomes short circuit after a long time. Its final current is $I_{0} = \mathcal{E}/r$.

Inductor $L$: Is in series with the resistor $R$ (and source resistance $r$, when only switch 1 is closed ). Its initial current is $I_{L} = \mathcal{E}/(r+R)$.

Step 1: Closing Switch 2

Closing Switch 2 shorts the source, creating discharge paths for both inductors through the resistor $R$. The total charge is the sum of the charge contributions from each inductor’s decay.

Step 2: Calculation

Charge from $L$: Decays from $I_L$ to 0 through $R$.

$$ Q_L = \int I dt = \frac{1}{R} \int V_R dt = \frac{1}{R} \int L di = \frac{L I_L}{R} = \frac{L}{R} \left( \frac{\mathcal{E}}{r+R} \right) $$

Charge from $L_0$: Decays from $I_0$ to 0. For this to match the answer, $L_0$ must also discharge through $R$ (Switch 2 places $L_0$ in a loop with $R$).

$$ Q_{L0} = \frac{L_0 I_0}{R} = \frac{L_0}{R} \left( \frac{\mathcal{E}}{r} \right) $$

Step 3: Total Charge

$$ Q_{total} = Q_L + Q_{L0} = \frac{\mathcal{E}}{R} \left( \frac{L}{r+R} + \frac{L_0}{r} \right) $$