28. Parallel L-R with Internal Resistance
Step 1: Circuit Analysis
Let $i_L$ be the inductor current and $i_R$ the resistor current. The internal resistance $r$ is in series with the parallel combination.
$$ \text{Loop Rule: } \mathscr{E} – (i_L + i_R)r – i_R R = 0 $$
$$ i_R = \frac{\mathscr{E} – i_L r}{R+r} $$
$$ \text{Also: } V_L = V_R \implies L \frac{di_L}{dt} = i_R R $$
Step 2: Heat Integration
We calculate heat using $H = \int i_R^2 R dt$. Substitute $dt = \frac{L di_L}{i_R R}$:
$$ H = \int_{0}^{I_{\text{max}}} i_R^2 R \left( \frac{L di_L}{i_R R} \right) = L \int_{0}^{\mathscr{E}/r} i_R di_L $$
$$ H = L \int_{0}^{\mathscr{E}/r} \frac{\mathscr{E} – i_L r}{R+r} di_L $$
$$ H = \frac{L}{R+r} \left[ \mathscr{E} i_L – \frac{r i_L^2}{2} \right]_0^{\mathscr{E}/r} $$
$$ H = \frac{L \mathscr{E}^2}{2r(R+r)} $$