29. Switching Current Sources
Step 1: Setup
Initially (Position 1), inductor current $i_L = I_0$. At Position 2, total source current is $\eta I_0$.
$$ i_R + i_L = \eta I_0 \implies i_R = \eta I_0 – i_L $$
$$ L \frac{di_L}{dt} = i_R R $$
Step 2: Total Charge
$$ q = \int i_R dt = \int \frac{L}{R} di_L = \frac{L}{R} [i_L]_{I_0}^{\eta I_0} $$
$$ q = \frac{L I_0}{R}(\eta – 1) $$
Step 3: Total Heat Dissipated
$$ H = \int i_R^2 R dt = L \int_{I_0}^{\eta I_0} (\eta I_0 – i_L) di_L $$
$$ H = L \left[ \eta I_0 i_L – \frac{i_L^2}{2} \right]_{I_0}^{\eta I_0} $$
$$ H = L I_0^2 \left( \frac{\eta^2}{2} – \eta + \frac{1}{2} \right) $$
$$ H = \frac{1}{2} L I_0^2 (\eta – 1)^2 $$