30. Charge and Heat in Parallel Circuit
Step 1: Final Currents (Trapped Flux)
The inductors $L$ and $2L$ form a superconducting loop with each other. Since they are in parallel, $V_L = V_{2L}$, which implies $\frac{d\Phi_1}{dt} = \frac{d\Phi_2}{dt}$.
$$ L i_1 – 2L i_2 = \text{constant} $$
$$ L(I_0) – 2L(I_0) = -L I_0 $$
Eventually, current stops flowing through $R$, so the net current leaving the inductor block is zero: $i_1 + i_2 = 0 \implies i_2 = -i_1$.
$$ L i_1 – 2L (-i_1) = -L I_0 \implies 3L i_1 = -L I_0 $$
$$ i_{1,final} = -\frac{I_0}{3}, \quad i_{2,final} = \frac{I_0}{3} $$
Step 2: Total Charge Flow
The voltage across the resistor is $V_R = -L \frac{di_1}{dt}$. Charge $q = \int i_R dt = \int \frac{V_R}{R} dt$.
$$ q = -\frac{L}{R} \int_{I_0}^{-I_0/3} di_1 = -\frac{L}{R} \left( -\frac{I_0}{3} – I_0 \right) $$
$$ q = \frac{4 L I_0}{3 R} $$
Step 3: Total Heat Dissipated
Heat is the loss in stored magnetic energy.
$$ U_{initial} = \frac{1}{2}L I_0^2 + \frac{1}{2}(2L) I_0^2 = \frac{3}{2} L I_0^2 $$
$$ U_{final} = \frac{1}{2}L \left(-\frac{I_0}{3}\right)^2 + \frac{1}{2}(2L) \left(\frac{I_0}{3}\right)^2 = \frac{1}{6} L I_0^2 $$
$$ \text{Heat} = U_i – U_f = \frac{3}{2} L I_0^2 – \frac{1}{6} L I_0^2 = \frac{4}{3} L I_0^2 $$
Charge = $\frac{4 L I_0}{3 R}$, Heat = $\frac{4}{3} L I_0^2$